Linear Algebra Examples

$A = [\begin{array}{c} 2 & 0 & 0 & 3 \\ 3 & - 1 & 0 & 0 \\ 8 & 7 & \frac{1}{2} & 0 \\ - 1 & 9 & 6 & 0 \end{array}]$

Step 1

Set up the formula to find the characteristic equation $p (λ)$ .

$p (λ) = determinant (A - λ I_{4})$

Step 2

The identity matrix or unit matrix of size $4$ is the $4 \times 4$ square matrix with ones on the main diagonal and zeros elsewhere.

$[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}]$

Step 3

Substitute the known values into $p (λ) = determinant (A - λ I_{4})$ .

Tap for more steps...

Step 3.1

Substitute $[\begin{array}{c} 2 & 0 & 0 & 3 \\ 3 & - 1 & 0 & 0 \\ 8 & 7 & \frac{1}{2} & 0 \\ - 1 & 9 & 6 & 0 \end{array}]$ for $A$ .

$p (λ) = determinant ([\begin{array}{c} 2 & 0 & 0 & 3 \\ 3 & - 1 & 0 & 0 \\ 8 & 7 & \frac{1}{2} & 0 \\ - 1 & 9 & 6 & 0 \end{array}] - λ I_{4})$

Step 3.2

Substitute $[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}]$ for $I_{4}$ .

$p (λ) = determinant ([\begin{array}{c} 2 & 0 & 0 & 3 \\ 3 & - 1 & 0 & 0 \\ 8 & 7 & \frac{1}{2} & 0 \\ - 1 & 9 & 6 & 0 \end{array}] - λ [\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}])$

Step 4

Simplify.

Tap for more steps...

Step 4.1

Simplify each term.

Tap for more steps...

Step 4.1.1

Multiply $- λ$ by each element of the matrix.

$p (λ) = determinant ([\begin{array}{c} 2 & 0 & 0 & 3 \\ 3 & - 1 & 0 & 0 \\ 8 & 7 & \frac{1}{2} & 0 \\ - 1 & 9 & 6 & 0 \end{array}] + [\begin{array}{c} - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2

Simplify each element in the matrix.

Tap for more steps...

Step 4.1.2.1

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} 2 & 0 & 0 & 3 \\ 3 & - 1 & 0 & 0 \\ 8 & 7 & \frac{1}{2} & 0 \\ - 1 & 9 & 6 & 0 \end{array}] + [\begin{array}{c} - λ & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.2

Multiply $- λ \cdot 0$ .

Tap for more steps...

Step 4.1.2.2.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 2 & 0 & 0 & 3 \\ 3 & - 1 & 0 & 0 \\ 8 & 7 & \frac{1}{2} & 0 \\ - 1 & 9 & 6 & 0 \end{array}] + [\begin{array}{c} - λ & 0 λ & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.2.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 2 & 0 & 0 & 3 \\ 3 & - 1 & 0 & 0 \\ 8 & 7 & \frac{1}{2} & 0 \\ - 1 & 9 & 6 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.3

Multiply $- λ \cdot 0$ .

Tap for more steps...

Step 4.1.2.3.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 2 & 0 & 0 & 3 \\ 3 & - 1 & 0 & 0 \\ 8 & 7 & \frac{1}{2} & 0 \\ - 1 & 9 & 6 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 λ & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.3.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 2 & 0 & 0 & 3 \\ 3 & - 1 & 0 & 0 \\ 8 & 7 & \frac{1}{2} & 0 \\ - 1 & 9 & 6 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.4

Multiply $- λ \cdot 0$ .

Tap for more steps...

Step 4.1.2.4.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 2 & 0 & 0 & 3 \\ 3 & - 1 & 0 & 0 \\ 8 & 7 & \frac{1}{2} & 0 \\ - 1 & 9 & 6 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 λ \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.4.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 2 & 0 & 0 & 3 \\ 3 & - 1 & 0 & 0 \\ 8 & 7 & \frac{1}{2} & 0 \\ - 1 & 9 & 6 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.5

Multiply $- λ \cdot 0$ .

Tap for more steps...

Step 4.1.2.5.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 2 & 0 & 0 & 3 \\ 3 & - 1 & 0 & 0 \\ 8 & 7 & \frac{1}{2} & 0 \\ - 1 & 9 & 6 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 λ & - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.5.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 2 & 0 & 0 & 3 \\ 3 & - 1 & 0 & 0 \\ 8 & 7 & \frac{1}{2} & 0 \\ - 1 & 9 & 6 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.6

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} 2 & 0 & 0 & 3 \\ 3 & - 1 & 0 & 0 \\ 8 & 7 & \frac{1}{2} & 0 \\ - 1 & 9 & 6 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.7

Multiply $- λ \cdot 0$ .

Tap for more steps...

Step 4.1.2.7.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 2 & 0 & 0 & 3 \\ 3 & - 1 & 0 & 0 \\ 8 & 7 & \frac{1}{2} & 0 \\ - 1 & 9 & 6 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 λ & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.7.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 2 & 0 & 0 & 3 \\ 3 & - 1 & 0 & 0 \\ 8 & 7 & \frac{1}{2} & 0 \\ - 1 & 9 & 6 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.8

Multiply $- λ \cdot 0$ .

Tap for more steps...

Step 4.1.2.8.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 2 & 0 & 0 & 3 \\ 3 & - 1 & 0 & 0 \\ 8 & 7 & \frac{1}{2} & 0 \\ - 1 & 9 & 6 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 λ \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.8.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 2 & 0 & 0 & 3 \\ 3 & - 1 & 0 & 0 \\ 8 & 7 & \frac{1}{2} & 0 \\ - 1 & 9 & 6 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.9

Multiply $- λ \cdot 0$ .

Tap for more steps...

Step 4.1.2.9.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 2 & 0 & 0 & 3 \\ 3 & - 1 & 0 & 0 \\ 8 & 7 & \frac{1}{2} & 0 \\ - 1 & 9 & 6 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 λ & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.9.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 2 & 0 & 0 & 3 \\ 3 & - 1 & 0 & 0 \\ 8 & 7 & \frac{1}{2} & 0 \\ - 1 & 9 & 6 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.10

Multiply $- λ \cdot 0$ .

Tap for more steps...

Step 4.1.2.10.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 2 & 0 & 0 & 3 \\ 3 & - 1 & 0 & 0 \\ 8 & 7 & \frac{1}{2} & 0 \\ - 1 & 9 & 6 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 & 0 λ & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.10.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 2 & 0 & 0 & 3 \\ 3 & - 1 & 0 & 0 \\ 8 & 7 & \frac{1}{2} & 0 \\ - 1 & 9 & 6 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 & 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.11

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} 2 & 0 & 0 & 3 \\ 3 & - 1 & 0 & 0 \\ 8 & 7 & \frac{1}{2} & 0 \\ - 1 & 9 & 6 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 & 0 & - λ & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.12

Multiply $- λ \cdot 0$ .

Tap for more steps...

Step 4.1.2.12.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 2 & 0 & 0 & 3 \\ 3 & - 1 & 0 & 0 \\ 8 & 7 & \frac{1}{2} & 0 \\ - 1 & 9 & 6 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 & 0 & - λ & 0 λ \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.12.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 2 & 0 & 0 & 3 \\ 3 & - 1 & 0 & 0 \\ 8 & 7 & \frac{1}{2} & 0 \\ - 1 & 9 & 6 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 & 0 & - λ & 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.13

Multiply $- λ \cdot 0$ .

Tap for more steps...

Step 4.1.2.13.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 2 & 0 & 0 & 3 \\ 3 & - 1 & 0 & 0 \\ 8 & 7 & \frac{1}{2} & 0 \\ - 1 & 9 & 6 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 & 0 & - λ & 0 \\ 0 λ & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.13.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 2 & 0 & 0 & 3 \\ 3 & - 1 & 0 & 0 \\ 8 & 7 & \frac{1}{2} & 0 \\ - 1 & 9 & 6 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 & 0 & - λ & 0 \\ 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.14

Multiply $- λ \cdot 0$ .

Tap for more steps...

Step 4.1.2.14.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 2 & 0 & 0 & 3 \\ 3 & - 1 & 0 & 0 \\ 8 & 7 & \frac{1}{2} & 0 \\ - 1 & 9 & 6 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 & 0 & - λ & 0 \\ 0 & 0 λ & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.14.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 2 & 0 & 0 & 3 \\ 3 & - 1 & 0 & 0 \\ 8 & 7 & \frac{1}{2} & 0 \\ - 1 & 9 & 6 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 & 0 & - λ & 0 \\ 0 & 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.15

Multiply $- λ \cdot 0$ .

Tap for more steps...

Step 4.1.2.15.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 2 & 0 & 0 & 3 \\ 3 & - 1 & 0 & 0 \\ 8 & 7 & \frac{1}{2} & 0 \\ - 1 & 9 & 6 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 & 0 & - λ & 0 \\ 0 & 0 & 0 λ & - λ \cdot 1 \end{array}])$

Step 4.1.2.15.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 2 & 0 & 0 & 3 \\ 3 & - 1 & 0 & 0 \\ 8 & 7 & \frac{1}{2} & 0 \\ - 1 & 9 & 6 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 & 0 & - λ & 0 \\ 0 & 0 & 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.16

Multiply $- 1$ by $1$ .

Step 4.2

Add the corresponding elements.

$p (λ) = determinant [\begin{array}{c} 2 - λ & 0 + 0 & 0 + 0 & 3 + 0 \\ 3 + 0 & - 1 - λ & 0 + 0 & 0 + 0 \\ 8 + 0 & 7 + 0 & \frac{1}{2} - λ & 0 + 0 \\ - 1 + 0 & 9 + 0 & 6 + 0 & 0 - λ \end{array}]$

Step 4.3

Simplify each element.

Tap for more steps...

Step 4.3.1

Add $0$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 2 - λ & 0 & 0 + 0 & 3 + 0 \\ 3 + 0 & - 1 - λ & 0 + 0 & 0 + 0 \\ 8 + 0 & 7 + 0 & \frac{1}{2} - λ & 0 + 0 \\ - 1 + 0 & 9 + 0 & 6 + 0 & 0 - λ \end{array}]$

Step 4.3.2

Add $0$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 2 - λ & 0 & 0 & 3 + 0 \\ 3 + 0 & - 1 - λ & 0 + 0 & 0 + 0 \\ 8 + 0 & 7 + 0 & \frac{1}{2} - λ & 0 + 0 \\ - 1 + 0 & 9 + 0 & 6 + 0 & 0 - λ \end{array}]$

Step 4.3.3

Add $3$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 2 - λ & 0 & 0 & 3 \\ 3 + 0 & - 1 - λ & 0 + 0 & 0 + 0 \\ 8 + 0 & 7 + 0 & \frac{1}{2} - λ & 0 + 0 \\ - 1 + 0 & 9 + 0 & 6 + 0 & 0 - λ \end{array}]$

Step 4.3.4

Add $3$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 2 - λ & 0 & 0 & 3 \\ 3 & - 1 - λ & 0 + 0 & 0 + 0 \\ 8 + 0 & 7 + 0 & \frac{1}{2} - λ & 0 + 0 \\ - 1 + 0 & 9 + 0 & 6 + 0 & 0 - λ \end{array}]$

Step 4.3.5

Add $0$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 2 - λ & 0 & 0 & 3 \\ 3 & - 1 - λ & 0 & 0 + 0 \\ 8 + 0 & 7 + 0 & \frac{1}{2} - λ & 0 + 0 \\ - 1 + 0 & 9 + 0 & 6 + 0 & 0 - λ \end{array}]$

Step 4.3.6

Add $0$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 2 - λ & 0 & 0 & 3 \\ 3 & - 1 - λ & 0 & 0 \\ 8 + 0 & 7 + 0 & \frac{1}{2} - λ & 0 + 0 \\ - 1 + 0 & 9 + 0 & 6 + 0 & 0 - λ \end{array}]$

Step 4.3.7

Add $8$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 2 - λ & 0 & 0 & 3 \\ 3 & - 1 - λ & 0 & 0 \\ 8 & 7 + 0 & \frac{1}{2} - λ & 0 + 0 \\ - 1 + 0 & 9 + 0 & 6 + 0 & 0 - λ \end{array}]$

Step 4.3.8

Add $7$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 2 - λ & 0 & 0 & 3 \\ 3 & - 1 - λ & 0 & 0 \\ 8 & 7 & \frac{1}{2} - λ & 0 + 0 \\ - 1 + 0 & 9 + 0 & 6 + 0 & 0 - λ \end{array}]$

Step 4.3.9

Add $0$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 2 - λ & 0 & 0 & 3 \\ 3 & - 1 - λ & 0 & 0 \\ 8 & 7 & \frac{1}{2} - λ & 0 \\ - 1 + 0 & 9 + 0 & 6 + 0 & 0 - λ \end{array}]$

Step 4.3.10

Add $- 1$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 2 - λ & 0 & 0 & 3 \\ 3 & - 1 - λ & 0 & 0 \\ 8 & 7 & \frac{1}{2} - λ & 0 \\ - 1 & 9 + 0 & 6 + 0 & 0 - λ \end{array}]$

Step 4.3.11

Add $9$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 2 - λ & 0 & 0 & 3 \\ 3 & - 1 - λ & 0 & 0 \\ 8 & 7 & \frac{1}{2} - λ & 0 \\ - 1 & 9 & 6 + 0 & 0 - λ \end{array}]$

Step 4.3.12

Add $6$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 2 - λ & 0 & 0 & 3 \\ 3 & - 1 - λ & 0 & 0 \\ 8 & 7 & \frac{1}{2} - λ & 0 \\ - 1 & 9 & 6 & 0 - λ \end{array}]$

Step 4.3.13

Subtract $λ$ from $0$ .

$p (λ) = determinant [\begin{array}{c} 2 - λ & 0 & 0 & 3 \\ 3 & - 1 - λ & 0 & 0 \\ 8 & 7 & \frac{1}{2} - λ & 0 \\ - 1 & 9 & 6 & - λ \end{array}]$

Step 5

Find the determinant.

Tap for more steps...

Step 5.1

Choose the row or column with the most $0$ elements. If there are no $0$ elements choose any row or column. Multiply every element in row $1$ by its cofactor and add.

Tap for more steps...

Step 5.1.1

Consider the corresponding sign chart.

$| \begin{array}{c} + & - & + & - \\ - & + & - & + \\ + & - & + & - \\ - & + & - & + \end{array} |$

Step 5.1.2

The cofactor is the minor with the sign changed if the indices match a $-$ position on the sign chart.

Step 5.1.3

The minor for $a_{11}$ is the determinant with row $1$ and column $1$ deleted.

$| \begin{array}{c} - 1 - λ & 0 & 0 \\ 7 & \frac{1}{2} - λ & 0 \\ 9 & 6 & - λ \end{array} |$

Step 5.1.4

Multiply element $a_{11}$ by its cofactor.

$(2 - λ) | \begin{array}{c} - 1 - λ & 0 & 0 \\ 7 & \frac{1}{2} - λ & 0 \\ 9 & 6 & - λ \end{array} |$

Step 5.1.5

The minor for $a_{12}$ is the determinant with row $1$ and column $2$ deleted.

$| \begin{array}{c} 3 & 0 & 0 \\ 8 & \frac{1}{2} - λ & 0 \\ - 1 & 6 & - λ \end{array} |$

Step 5.1.6

Multiply element $a_{12}$ by its cofactor.

$0 | \begin{array}{c} 3 & 0 & 0 \\ 8 & \frac{1}{2} - λ & 0 \\ - 1 & 6 & - λ \end{array} |$

Step 5.1.7

The minor for $a_{13}$ is the determinant with row $1$ and column $3$ deleted.

$| \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & 0 \\ - 1 & 9 & - λ \end{array} |$

Step 5.1.8

Multiply element $a_{13}$ by its cofactor.

$0 | \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & 0 \\ - 1 & 9 & - λ \end{array} |$

Step 5.1.9

The minor for $a_{14}$ is the determinant with row $1$ and column $4$ deleted.

$| \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & \frac{1}{2} - λ \\ - 1 & 9 & 6 \end{array} |$

Step 5.1.10

Multiply element $a_{14}$ by its cofactor.

$- 3 | \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & \frac{1}{2} - λ \\ - 1 & 9 & 6 \end{array} |$

Step 5.1.11

Add the terms together.

$p (λ) = (2 - λ) | \begin{array}{c} - 1 - λ & 0 & 0 \\ 7 & \frac{1}{2} - λ & 0 \\ 9 & 6 & - λ \end{array} | + 0 | \begin{array}{c} 3 & 0 & 0 \\ 8 & \frac{1}{2} - λ & 0 \\ - 1 & 6 & - λ \end{array} | + 0 | \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & 0 \\ - 1 & 9 & - λ \end{array} | - 3 | \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & \frac{1}{2} - λ \\ - 1 & 9 & 6 \end{array} |$

Step 5.2

Multiply $0$ by $| \begin{array}{c} 3 & 0 & 0 \\ 8 & \frac{1}{2} - λ & 0 \\ - 1 & 6 & - λ \end{array} |$ .

$p (λ) = (2 - λ) | \begin{array}{c} - 1 - λ & 0 & 0 \\ 7 & \frac{1}{2} - λ & 0 \\ 9 & 6 & - λ \end{array} | + 0 + 0 | \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & 0 \\ - 1 & 9 & - λ \end{array} | - 3 | \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & \frac{1}{2} - λ \\ - 1 & 9 & 6 \end{array} |$

Step 5.3

Multiply $0$ by $| \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & 0 \\ - 1 & 9 & - λ \end{array} |$ .

$p (λ) = (2 - λ) | \begin{array}{c} - 1 - λ & 0 & 0 \\ 7 & \frac{1}{2} - λ & 0 \\ 9 & 6 & - λ \end{array} | + 0 + 0 - 3 | \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & \frac{1}{2} - λ \\ - 1 & 9 & 6 \end{array} |$

Step 5.4

Evaluate $| \begin{array}{c} - 1 - λ & 0 & 0 \\ 7 & \frac{1}{2} - λ & 0 \\ 9 & 6 & - λ \end{array} |$ .

Tap for more steps...

Step 5.4.1

Choose the row or column with the most $0$ elements. If there are no $0$ elements choose any row or column. Multiply every element in row $1$ by its cofactor and add.

Tap for more steps...

Step 5.4.1.1

Consider the corresponding sign chart.

$| \begin{array}{c} + & - & + \\ - & + & - \\ + & - & + \end{array} |$

Step 5.4.1.2

The cofactor is the minor with the sign changed if the indices match a $-$ position on the sign chart.

Step 5.4.1.3

The minor for $a_{11}$ is the determinant with row $1$ and column $1$ deleted.

$| \begin{array}{c} \frac{1}{2} - λ & 0 \\ 6 & - λ \end{array} |$

Step 5.4.1.4

Multiply element $a_{11}$ by its cofactor.

$(- 1 - λ) | \begin{array}{c} \frac{1}{2} - λ & 0 \\ 6 & - λ \end{array} |$

Step 5.4.1.5

The minor for $a_{12}$ is the determinant with row $1$ and column $2$ deleted.

$| \begin{array}{c} 7 & 0 \\ 9 & - λ \end{array} |$

Step 5.4.1.6

Multiply element $a_{12}$ by its cofactor.

$0 | \begin{array}{c} 7 & 0 \\ 9 & - λ \end{array} |$

Step 5.4.1.7

The minor for $a_{13}$ is the determinant with row $1$ and column $3$ deleted.

$| \begin{array}{c} 7 & \frac{1}{2} - λ \\ 9 & 6 \end{array} |$

Step 5.4.1.8

Multiply element $a_{13}$ by its cofactor.

$0 | \begin{array}{c} 7 & \frac{1}{2} - λ \\ 9 & 6 \end{array} |$

Step 5.4.1.9

Add the terms together.

$p (λ) = (2 - λ) ((- 1 - λ) | \begin{array}{c} \frac{1}{2} - λ & 0 \\ 6 & - λ \end{array} | + 0 | \begin{array}{c} 7 & 0 \\ 9 & - λ \end{array} | + 0 | \begin{array}{c} 7 & \frac{1}{2} - λ \\ 9 & 6 \end{array} |) + 0 + 0 - 3 | \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & \frac{1}{2} - λ \\ - 1 & 9 & 6 \end{array} |$

Step 5.4.2

Multiply $0$ by $| \begin{array}{c} 7 & 0 \\ 9 & - λ \end{array} |$ .

$p (λ) = (2 - λ) ((- 1 - λ) | \begin{array}{c} \frac{1}{2} - λ & 0 \\ 6 & - λ \end{array} | + 0 + 0 | \begin{array}{c} 7 & \frac{1}{2} - λ \\ 9 & 6 \end{array} |) + 0 + 0 - 3 | \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & \frac{1}{2} - λ \\ - 1 & 9 & 6 \end{array} |$

Step 5.4.3

Multiply $0$ by $| \begin{array}{c} 7 & \frac{1}{2} - λ \\ 9 & 6 \end{array} |$ .

$p (λ) = (2 - λ) ((- 1 - λ) | \begin{array}{c} \frac{1}{2} - λ & 0 \\ 6 & - λ \end{array} | + 0 + 0) + 0 + 0 - 3 | \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & \frac{1}{2} - λ \\ - 1 & 9 & 6 \end{array} |$

Step 5.4.4

Evaluate $| \begin{array}{c} \frac{1}{2} - λ & 0 \\ 6 & - λ \end{array} |$ .

Tap for more steps...

Step 5.4.4.1

The determinant of a $2 \times 2$ matrix can be found using the formula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$p (λ) = (2 - λ) ((- 1 - λ) ((\frac{1}{2} - λ) (- λ) - 6 \cdot 0) + 0 + 0) + 0 + 0 - 3 | \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & \frac{1}{2} - λ \\ - 1 & 9 & 6 \end{array} |$

Step 5.4.4.2

Simplify the determinant.

Tap for more steps...

Step 5.4.4.2.1

Simplify each term.

Tap for more steps...

Step 5.4.4.2.1.1

Apply the distributive property.

$p (λ) = (2 - λ) ((- 1 - λ) (\frac{1}{2} (- λ) - λ (- λ) - 6 \cdot 0) + 0 + 0) + 0 + 0 - 3 | \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & \frac{1}{2} - λ \\ - 1 & 9 & 6 \end{array} |$

Step 5.4.4.2.1.2

Combine $\frac{1}{2}$ and $λ$ .

$p (λ) = (2 - λ) ((- 1 - λ) (- \frac{λ}{2} - λ (- λ) - 6 \cdot 0) + 0 + 0) + 0 + 0 - 3 | \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & \frac{1}{2} - λ \\ - 1 & 9 & 6 \end{array} |$

Step 5.4.4.2.1.3

Rewrite using the commutative property of multiplication.

$p (λ) = (2 - λ) ((- 1 - λ) (- \frac{λ}{2} - 1 \cdot - 1 λ \cdot λ - 6 \cdot 0) + 0 + 0) + 0 + 0 - 3 | \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & \frac{1}{2} - λ \\ - 1 & 9 & 6 \end{array} |$

Step 5.4.4.2.1.4

Simplify each term.

Tap for more steps...

Step 5.4.4.2.1.4.1

Multiply $λ$ by $λ$ by adding the exponents.

Tap for more steps...

Step 5.4.4.2.1.4.1.1

Move $λ$ .

$p (λ) = (2 - λ) ((- 1 - λ) (- \frac{λ}{2} - 1 \cdot - 1 (λ \cdot λ) - 6 \cdot 0) + 0 + 0) + 0 + 0 - 3 | \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & \frac{1}{2} - λ \\ - 1 & 9 & 6 \end{array} |$

Step 5.4.4.2.1.4.1.2

Multiply $λ$ by $λ$ .

$p (λ) = (2 - λ) ((- 1 - λ) (- \frac{λ}{2} - 1 \cdot - 1 λ^{2} - 6 \cdot 0) + 0 + 0) + 0 + 0 - 3 | \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & \frac{1}{2} - λ \\ - 1 & 9 & 6 \end{array} |$

Step 5.4.4.2.1.4.2

Multiply $- 1$ by $- 1$ .

$p (λ) = (2 - λ) ((- 1 - λ) (- \frac{λ}{2} + 1 λ^{2} - 6 \cdot 0) + 0 + 0) + 0 + 0 - 3 | \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & \frac{1}{2} - λ \\ - 1 & 9 & 6 \end{array} |$

Step 5.4.4.2.1.4.3

Multiply $λ^{2}$ by $1$ .

$p (λ) = (2 - λ) ((- 1 - λ) (- \frac{λ}{2} + λ^{2} - 6 \cdot 0) + 0 + 0) + 0 + 0 - 3 | \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & \frac{1}{2} - λ \\ - 1 & 9 & 6 \end{array} |$

Step 5.4.4.2.1.5

Multiply $- 6$ by $0$ .

$p (λ) = (2 - λ) ((- 1 - λ) (- \frac{λ}{2} + λ^{2} + 0) + 0 + 0) + 0 + 0 - 3 | \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & \frac{1}{2} - λ \\ - 1 & 9 & 6 \end{array} |$

Step 5.4.4.2.2

Add $- \frac{λ}{2} + λ^{2}$ and $0$ .

$p (λ) = (2 - λ) ((- 1 - λ) (- \frac{λ}{2} + λ^{2}) + 0 + 0) + 0 + 0 - 3 | \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & \frac{1}{2} - λ \\ - 1 & 9 & 6 \end{array} |$

Step 5.4.4.2.3

Reorder $- \frac{λ}{2}$ and $λ^{2}$ .

$p (λ) = (2 - λ) ((- 1 - λ) (λ^{2} - \frac{λ}{2}) + 0 + 0) + 0 + 0 - 3 | \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & \frac{1}{2} - λ \\ - 1 & 9 & 6 \end{array} |$

Step 5.4.5

Simplify the determinant.

Tap for more steps...

Step 5.4.5.1

Combine the opposite terms in $(- 1 - λ) (λ^{2} - \frac{λ}{2}) + 0 + 0$ .

Tap for more steps...

Step 5.4.5.1.1

Add $(- 1 - λ) (λ^{2} - \frac{λ}{2})$ and $0$ .

$p (λ) = (2 - λ) ((- 1 - λ) (λ^{2} - \frac{λ}{2}) + 0) + 0 + 0 - 3 | \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & \frac{1}{2} - λ \\ - 1 & 9 & 6 \end{array} |$

Step 5.4.5.1.2

Add $(- 1 - λ) (λ^{2} - \frac{λ}{2})$ and $0$ .

$p (λ) = (2 - λ) ((- 1 - λ) (λ^{2} - \frac{λ}{2})) + 0 + 0 - 3 | \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & \frac{1}{2} - λ \\ - 1 & 9 & 6 \end{array} |$

Step 5.4.5.2

Expand $(- 1 - λ) (λ^{2} - \frac{λ}{2})$ using the FOIL Method.

Tap for more steps...

Step 5.4.5.2.1

Apply the distributive property.

$p (λ) = (2 - λ) (- 1 (λ^{2} - \frac{λ}{2}) - λ (λ^{2} - \frac{λ}{2})) + 0 + 0 - 3 | \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & \frac{1}{2} - λ \\ - 1 & 9 & 6 \end{array} |$

Step 5.4.5.2.2

Apply the distributive property.

$p (λ) = (2 - λ) (- 1 λ^{2} - 1 (- \frac{λ}{2}) - λ (λ^{2} - \frac{λ}{2})) + 0 + 0 - 3 | \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & \frac{1}{2} - λ \\ - 1 & 9 & 6 \end{array} |$

Step 5.4.5.2.3

Apply the distributive property.

$p (λ) = (2 - λ) (- 1 λ^{2} - 1 (- \frac{λ}{2}) - λ \cdot λ^{2} - λ (- \frac{λ}{2})) + 0 + 0 - 3 | \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & \frac{1}{2} - λ \\ - 1 & 9 & 6 \end{array} |$

Step 5.4.5.3

Simplify and combine like terms.

Tap for more steps...

Step 5.4.5.3.1

Simplify each term.

Tap for more steps...

Step 5.4.5.3.1.1

Rewrite $- 1 λ^{2}$ as $- λ^{2}$ .

$p (λ) = (2 - λ) (- λ^{2} - 1 (- \frac{λ}{2}) - λ \cdot λ^{2} - λ (- \frac{λ}{2})) + 0 + 0 - 3 | \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & \frac{1}{2} - λ \\ - 1 & 9 & 6 \end{array} |$

Step 5.4.5.3.1.2

Multiply $- 1 (- \frac{λ}{2})$ .

Tap for more steps...

Step 5.4.5.3.1.2.1

Multiply $- 1$ by $- 1$ .

$p (λ) = (2 - λ) (- λ^{2} + 1 \frac{λ}{2} - λ \cdot λ^{2} - λ (- \frac{λ}{2})) + 0 + 0 - 3 | \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & \frac{1}{2} - λ \\ - 1 & 9 & 6 \end{array} |$

Step 5.4.5.3.1.2.2

Multiply $\frac{λ}{2}$ by $1$ .

$p (λ) = (2 - λ) (- λ^{2} + \frac{λ}{2} - λ \cdot λ^{2} - λ (- \frac{λ}{2})) + 0 + 0 - 3 | \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & \frac{1}{2} - λ \\ - 1 & 9 & 6 \end{array} |$

Step 5.4.5.3.1.3

Multiply $λ$ by $λ^{2}$ by adding the exponents.

Tap for more steps...

Step 5.4.5.3.1.3.1

Move $λ^{2}$ .

$p (λ) = (2 - λ) (- λ^{2} + \frac{λ}{2} - (λ^{2} λ) - λ (- \frac{λ}{2})) + 0 + 0 - 3 | \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & \frac{1}{2} - λ \\ - 1 & 9 & 6 \end{array} |$

Step 5.4.5.3.1.3.2

Multiply $λ^{2}$ by $λ$ .

Tap for more steps...

Step 5.4.5.3.1.3.2.1

Raise $λ$ to the power of $1$ .

$p (λ) = (2 - λ) (- λ^{2} + \frac{λ}{2} - (λ^{2} λ^{1}) - λ (- \frac{λ}{2})) + 0 + 0 - 3 | \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & \frac{1}{2} - λ \\ - 1 & 9 & 6 \end{array} |$

Step 5.4.5.3.1.3.2.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$p (λ) = (2 - λ) (- λ^{2} + \frac{λ}{2} - λ^{2 + 1} - λ (- \frac{λ}{2})) + 0 + 0 - 3 | \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & \frac{1}{2} - λ \\ - 1 & 9 & 6 \end{array} |$

Step 5.4.5.3.1.3.3

Add $2$ and $1$ .

$p (λ) = (2 - λ) (- λ^{2} + \frac{λ}{2} - λ^{3} - λ (- \frac{λ}{2})) + 0 + 0 - 3 | \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & \frac{1}{2} - λ \\ - 1 & 9 & 6 \end{array} |$

Step 5.4.5.3.1.4

Multiply $- λ (- \frac{λ}{2})$ .

Tap for more steps...

Step 5.4.5.3.1.4.1

Multiply $- 1$ by $- 1$ .

$p (λ) = (2 - λ) (- λ^{2} + \frac{λ}{2} - λ^{3} + 1 λ \frac{λ}{2}) + 0 + 0 - 3 | \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & \frac{1}{2} - λ \\ - 1 & 9 & 6 \end{array} |$

Step 5.4.5.3.1.4.2

Multiply $λ$ by $1$ .

$p (λ) = (2 - λ) (- λ^{2} + \frac{λ}{2} - λ^{3} + λ \frac{λ}{2}) + 0 + 0 - 3 | \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & \frac{1}{2} - λ \\ - 1 & 9 & 6 \end{array} |$

Step 5.4.5.3.1.4.3

Combine $λ$ and $\frac{λ}{2}$ .

$p (λ) = (2 - λ) (- λ^{2} + \frac{λ}{2} - λ^{3} + \frac{λ \cdot λ}{2}) + 0 + 0 - 3 | \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & \frac{1}{2} - λ \\ - 1 & 9 & 6 \end{array} |$

Step 5.4.5.3.1.4.4

Raise $λ$ to the power of $1$ .

$p (λ) = (2 - λ) (- λ^{2} + \frac{λ}{2} - λ^{3} + \frac{λ^{1} λ}{2}) + 0 + 0 - 3 | \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & \frac{1}{2} - λ \\ - 1 & 9 & 6 \end{array} |$

Step 5.4.5.3.1.4.5

Raise $λ$ to the power of $1$ .

$p (λ) = (2 - λ) (- λ^{2} + \frac{λ}{2} - λ^{3} + \frac{λ^{1} λ^{1}}{2}) + 0 + 0 - 3 | \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & \frac{1}{2} - λ \\ - 1 & 9 & 6 \end{array} |$

Step 5.4.5.3.1.4.6

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$p (λ) = (2 - λ) (- λ^{2} + \frac{λ}{2} - λ^{3} + \frac{λ^{1 + 1}}{2}) + 0 + 0 - 3 | \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & \frac{1}{2} - λ \\ - 1 & 9 & 6 \end{array} |$

Step 5.4.5.3.1.4.7

Add $1$ and $1$ .

$p (λ) = (2 - λ) (- λ^{2} + \frac{λ}{2} - λ^{3} + \frac{λ^{2}}{2}) + 0 + 0 - 3 | \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & \frac{1}{2} - λ \\ - 1 & 9 & 6 \end{array} |$

Step 5.4.5.3.2

To write $- λ^{2}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$p (λ) = (2 - λ) (\frac{λ}{2} - λ^{3} - λ^{2} \cdot \frac{2}{2} + \frac{λ^{2}}{2}) + 0 + 0 - 3 | \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & \frac{1}{2} - λ \\ - 1 & 9 & 6 \end{array} |$

Step 5.4.5.3.3

Combine $- λ^{2}$ and $\frac{2}{2}$ .

$p (λ) = (2 - λ) (\frac{λ}{2} - λ^{3} + \frac{- λ^{2} \cdot 2}{2} + \frac{λ^{2}}{2}) + 0 + 0 - 3 | \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & \frac{1}{2} - λ \\ - 1 & 9 & 6 \end{array} |$

Step 5.4.5.3.4

Combine the numerators over the common denominator.

$p (λ) = (2 - λ) (\frac{λ}{2} - λ^{3} + \frac{- λ^{2} \cdot 2 + λ^{2}}{2}) + 0 + 0 - 3 | \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & \frac{1}{2} - λ \\ - 1 & 9 & 6 \end{array} |$

Step 5.4.5.3.5

Combine the numerators over the common denominator.

$p (λ) = (2 - λ) (- λ^{3} + \frac{λ - λ^{2} \cdot 2 + λ^{2}}{2}) + 0 + 0 - 3 | \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & \frac{1}{2} - λ \\ - 1 & 9 & 6 \end{array} |$

Step 5.4.5.3.6

To write $- λ^{3}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$p (λ) = (2 - λ) (- λ^{3} \cdot \frac{2}{2} + \frac{λ - λ^{2} \cdot 2 + λ^{2}}{2}) + 0 + 0 - 3 | \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & \frac{1}{2} - λ \\ - 1 & 9 & 6 \end{array} |$

Step 5.4.5.3.7

Combine $- λ^{3}$ and $\frac{2}{2}$ .

$p (λ) = (2 - λ) (\frac{- λ^{3} \cdot 2}{2} + \frac{λ - λ^{2} \cdot 2 + λ^{2}}{2}) + 0 + 0 - 3 | \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & \frac{1}{2} - λ \\ - 1 & 9 & 6 \end{array} |$

Step 5.4.5.3.8

Combine the numerators over the common denominator.

$p (λ) = (2 - λ) \frac{- λ^{3} \cdot 2 + λ - λ^{2} \cdot 2 + λ^{2}}{2} + 0 + 0 - 3 | \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & \frac{1}{2} - λ \\ - 1 & 9 & 6 \end{array} |$

Step 5.4.5.4

Simplify the numerator.

Tap for more steps...

Step 5.4.5.4.1

Factor $λ$ out of $- λ^{3} \cdot 2 + λ - λ^{2} \cdot 2 + λ^{2}$ .

Tap for more steps...

Step 5.4.5.4.1.1

Factor $λ$ out of $- λ^{3} \cdot 2$ .

$p (λ) = (2 - λ) \frac{λ (- λ^{2} \cdot 2) + λ - λ^{2} \cdot 2 + λ^{2}}{2} + 0 + 0 - 3 | \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & \frac{1}{2} - λ \\ - 1 & 9 & 6 \end{array} |$

Step 5.4.5.4.1.2

Raise $λ$ to the power of $1$ .

$p (λ) = (2 - λ) \frac{λ (- λ^{2} \cdot 2) + λ^{1} - λ^{2} \cdot 2 + λ^{2}}{2} + 0 + 0 - 3 | \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & \frac{1}{2} - λ \\ - 1 & 9 & 6 \end{array} |$

Step 5.4.5.4.1.3

Factor $λ$ out of $λ^{1}$ .

$p (λ) = (2 - λ) \frac{λ (- λ^{2} \cdot 2) + λ \cdot 1 - λ^{2} \cdot 2 + λ^{2}}{2} + 0 + 0 - 3 | \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & \frac{1}{2} - λ \\ - 1 & 9 & 6 \end{array} |$

Step 5.4.5.4.1.4

Factor $λ$ out of $- λ^{2} \cdot 2$ .

$p (λ) = (2 - λ) \frac{λ (- λ^{2} \cdot 2) + λ \cdot 1 + λ (- λ \cdot 2) + λ^{2}}{2} + 0 + 0 - 3 | \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & \frac{1}{2} - λ \\ - 1 & 9 & 6 \end{array} |$

Step 5.4.5.4.1.5

Factor $λ$ out of $λ^{2}$ .

$p (λ) = (2 - λ) \frac{λ (- λ^{2} \cdot 2) + λ \cdot 1 + λ (- λ \cdot 2) + λ \cdot λ}{2} + 0 + 0 - 3 | \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & \frac{1}{2} - λ \\ - 1 & 9 & 6 \end{array} |$

Step 5.4.5.4.1.6

Factor $λ$ out of $λ (- λ^{2} \cdot 2) + λ \cdot 1$ .

$p (λ) = (2 - λ) \frac{λ (- λ^{2} \cdot 2 + 1) + λ (- λ \cdot 2) + λ \cdot λ}{2} + 0 + 0 - 3 | \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & \frac{1}{2} - λ \\ - 1 & 9 & 6 \end{array} |$

Step 5.4.5.4.1.7

Factor $λ$ out of $λ (- λ^{2} \cdot 2 + 1) + λ (- λ \cdot 2)$ .

$p (λ) = (2 - λ) \frac{λ (- λ^{2} \cdot 2 + 1 - λ \cdot 2) + λ \cdot λ}{2} + 0 + 0 - 3 | \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & \frac{1}{2} - λ \\ - 1 & 9 & 6 \end{array} |$

Step 5.4.5.4.1.8

Factor $λ$ out of $λ (- λ^{2} \cdot 2 + 1 - λ \cdot 2) + λ \cdot λ$ .

$p (λ) = (2 - λ) \frac{λ (- λ^{2} \cdot 2 + 1 - λ \cdot 2 + λ)}{2} + 0 + 0 - 3 | \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & \frac{1}{2} - λ \\ - 1 & 9 & 6 \end{array} |$

Step 5.4.5.4.2

Reorder terms.

$p (λ) = (2 - λ) \frac{λ (- 1 \cdot 2 λ^{2} - 1 \cdot 2 λ + λ + 1)}{2} + 0 + 0 - 3 | \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & \frac{1}{2} - λ \\ - 1 & 9 & 6 \end{array} |$

Step 5.4.5.4.3

Factor out the greatest common factor from each group.

Tap for more steps...

Step 5.4.5.4.3.1

Group the first two terms and the last two terms.

$p (λ) = (2 - λ) \frac{λ ((- 1 \cdot 2 λ^{2} - 1 \cdot 2 λ) + λ + 1)}{2} + 0 + 0 - 3 | \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & \frac{1}{2} - λ \\ - 1 & 9 & 6 \end{array} |$

Step 5.4.5.4.3.2

Factor out the greatest common factor (GCF) from each group.

$p (λ) = (2 - λ) \frac{λ (2 λ (- λ - 1) - 1 (- λ - 1))}{2} + 0 + 0 - 3 | \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & \frac{1}{2} - λ \\ - 1 & 9 & 6 \end{array} |$

Step 5.4.5.4.4

Factor the polynomial by factoring out the greatest common factor, $- λ - 1$ .

$p (λ) = (2 - λ) \frac{λ (- λ - 1) (2 λ - 1)}{2} + 0 + 0 - 3 | \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & \frac{1}{2} - λ \\ - 1 & 9 & 6 \end{array} |$

Step 5.4.5.5

Factor $- 1$ out of $- λ$ .

$p (λ) = (2 - λ) \frac{λ (- (λ) - 1) (2 λ - 1)}{2} + 0 + 0 - 3 | \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & \frac{1}{2} - λ \\ - 1 & 9 & 6 \end{array} |$

Step 5.4.5.6

Rewrite $- 1$ as $- 1 (1)$ .

$p (λ) = (2 - λ) \frac{λ (- (λ) - 1 (1)) (2 λ - 1)}{2} + 0 + 0 - 3 | \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & \frac{1}{2} - λ \\ - 1 & 9 & 6 \end{array} |$

Step 5.4.5.7

Factor $- 1$ out of $- (λ) - 1 (1)$ .

$p (λ) = (2 - λ) \frac{λ (- (λ + 1)) (2 λ - 1)}{2} + 0 + 0 - 3 | \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & \frac{1}{2} - λ \\ - 1 & 9 & 6 \end{array} |$

Step 5.4.5.8

Rewrite $- (λ + 1)$ as $- 1 (λ + 1)$ .

$p (λ) = (2 - λ) \frac{λ (- 1 (λ + 1)) (2 λ - 1)}{2} + 0 + 0 - 3 | \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & \frac{1}{2} - λ \\ - 1 & 9 & 6 \end{array} |$

Step 5.4.5.9

Move the negative in front of the fraction.

$p (λ) = (2 - λ) (- \frac{(λ (λ + 1)) (2 λ - 1)}{2}) + 0 + 0 - 3 | \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & \frac{1}{2} - λ \\ - 1 & 9 & 6 \end{array} |$

Step 5.4.5.10

Reorder factors in $- \frac{(λ (λ + 1)) (2 λ - 1)}{2}$ .

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 | \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & \frac{1}{2} - λ \\ - 1 & 9 & 6 \end{array} |$

Step 5.5

Evaluate $| \begin{array}{c} 3 & - 1 - λ & 0 \\ 8 & 7 & \frac{1}{2} - λ \\ - 1 & 9 & 6 \end{array} |$ .

Tap for more steps...

Step 5.5.1

Choose the row or column with the most $0$ elements. If there are no $0$ elements choose any row or column. Multiply every element in row $1$ by its cofactor and add.

Tap for more steps...

Step 5.5.1.1

Consider the corresponding sign chart.

$| \begin{array}{c} + & - & + \\ - & + & - \\ + & - & + \end{array} |$

Step 5.5.1.2

The cofactor is the minor with the sign changed if the indices match a $-$ position on the sign chart.

Step 5.5.1.3

The minor for $a_{11}$ is the determinant with row $1$ and column $1$ deleted.

$| \begin{array}{c} 7 & \frac{1}{2} - λ \\ 9 & 6 \end{array} |$

Step 5.5.1.4

Multiply element $a_{11}$ by its cofactor.

$3 | \begin{array}{c} 7 & \frac{1}{2} - λ \\ 9 & 6 \end{array} |$

Step 5.5.1.5

The minor for $a_{12}$ is the determinant with row $1$ and column $2$ deleted.

$| \begin{array}{c} 8 & \frac{1}{2} - λ \\ - 1 & 6 \end{array} |$

Step 5.5.1.6

Multiply element $a_{12}$ by its cofactor.

$- (- 1 - λ) | \begin{array}{c} 8 & \frac{1}{2} - λ \\ - 1 & 6 \end{array} |$

Step 5.5.1.7

The minor for $a_{13}$ is the determinant with row $1$ and column $3$ deleted.

$| \begin{array}{c} 8 & 7 \\ - 1 & 9 \end{array} |$

Step 5.5.1.8

Multiply element $a_{13}$ by its cofactor.

$0 | \begin{array}{c} 8 & 7 \\ - 1 & 9 \end{array} |$

Step 5.5.1.9

Add the terms together.

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (3 | \begin{array}{c} 7 & \frac{1}{2} - λ \\ 9 & 6 \end{array} | - (- 1 - λ) | \begin{array}{c} 8 & \frac{1}{2} - λ \\ - 1 & 6 \end{array} | + 0 | \begin{array}{c} 8 & 7 \\ - 1 & 9 \end{array} |)$

Step 5.5.2

Multiply $0$ by $| \begin{array}{c} 8 & 7 \\ - 1 & 9 \end{array} |$ .

Step 5.5.3

Evaluate $| \begin{array}{c} 7 & \frac{1}{2} - λ \\ 9 & 6 \end{array} |$ .

Tap for more steps...

Step 5.5.3.1

The determinant of a $2 \times 2$ matrix can be found using the formula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (3 (7 \cdot 6 - 9 (\frac{1}{2} - λ)) - (- 1 - λ) | \begin{array}{c} 8 & \frac{1}{2} - λ \\ - 1 & 6 \end{array} | + 0)$

Step 5.5.3.2

Simplify the determinant.

Tap for more steps...

Step 5.5.3.2.1

Simplify each term.

Tap for more steps...

Step 5.5.3.2.1.1

Multiply $7$ by $6$ .

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (3 (42 - 9 (\frac{1}{2} - λ)) - (- 1 - λ) | \begin{array}{c} 8 & \frac{1}{2} - λ \\ - 1 & 6 \end{array} | + 0)$

Step 5.5.3.2.1.2

Apply the distributive property.

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (3 (42 - 9 (\frac{1}{2}) - 9 (- λ)) - (- 1 - λ) | \begin{array}{c} 8 & \frac{1}{2} - λ \\ - 1 & 6 \end{array} | + 0)$

Step 5.5.3.2.1.3

Combine $- 9$ and $\frac{1}{2}$ .

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (3 (42 + \frac{- 9}{2} - 9 (- λ)) - (- 1 - λ) | \begin{array}{c} 8 & \frac{1}{2} - λ \\ - 1 & 6 \end{array} | + 0)$

Step 5.5.3.2.1.4

Multiply $- 1$ by $- 9$ .

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (3 (42 + \frac{- 9}{2} + 9 λ) - (- 1 - λ) | \begin{array}{c} 8 & \frac{1}{2} - λ \\ - 1 & 6 \end{array} | + 0)$

Step 5.5.3.2.1.5

Move the negative in front of the fraction.

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (3 (42 - \frac{9}{2} + 9 λ) - (- 1 - λ) | \begin{array}{c} 8 & \frac{1}{2} - λ \\ - 1 & 6 \end{array} | + 0)$

Step 5.5.3.2.2

To write $42$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (3 (42 \cdot \frac{2}{2} - \frac{9}{2} + 9 λ) - (- 1 - λ) | \begin{array}{c} 8 & \frac{1}{2} - λ \\ - 1 & 6 \end{array} | + 0)$

Step 5.5.3.2.3

Combine $42$ and $\frac{2}{2}$ .

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (3 (\frac{42 \cdot 2}{2} - \frac{9}{2} + 9 λ) - (- 1 - λ) | \begin{array}{c} 8 & \frac{1}{2} - λ \\ - 1 & 6 \end{array} | + 0)$

Step 5.5.3.2.4

Combine the numerators over the common denominator.

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (3 (\frac{42 \cdot 2 - 9}{2} + 9 λ) - (- 1 - λ) | \begin{array}{c} 8 & \frac{1}{2} - λ \\ - 1 & 6 \end{array} | + 0)$

Step 5.5.3.2.5

Simplify the numerator.

Tap for more steps...

Step 5.5.3.2.5.1

Multiply $42$ by $2$ .

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (3 (\frac{84 - 9}{2} + 9 λ) - (- 1 - λ) | \begin{array}{c} 8 & \frac{1}{2} - λ \\ - 1 & 6 \end{array} | + 0)$

Step 5.5.3.2.5.2

Subtract $9$ from $84$ .

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (3 (\frac{75}{2} + 9 λ) - (- 1 - λ) | \begin{array}{c} 8 & \frac{1}{2} - λ \\ - 1 & 6 \end{array} | + 0)$

Step 5.5.3.2.6

Reorder $\frac{75}{2}$ and $9 λ$ .

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (3 (9 λ + \frac{75}{2}) - (- 1 - λ) | \begin{array}{c} 8 & \frac{1}{2} - λ \\ - 1 & 6 \end{array} | + 0)$

Step 5.5.4

Evaluate $| \begin{array}{c} 8 & \frac{1}{2} - λ \\ - 1 & 6 \end{array} |$ .

Tap for more steps...

Step 5.5.4.1

The determinant of a $2 \times 2$ matrix can be found using the formula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (3 (9 λ + \frac{75}{2}) - (- 1 - λ) (8 \cdot 6 - - (\frac{1}{2} - λ)) + 0)$

Step 5.5.4.2

Simplify the determinant.

Tap for more steps...

Step 5.5.4.2.1

Simplify each term.

Tap for more steps...

Step 5.5.4.2.1.1

Multiply $8$ by $6$ .

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (3 (9 λ + \frac{75}{2}) - (- 1 - λ) (48 - - (\frac{1}{2} - λ)) + 0)$

Step 5.5.4.2.1.2

Apply the distributive property.

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (3 (9 λ + \frac{75}{2}) - (- 1 - λ) (48 - (- \frac{1}{2} - - λ)) + 0)$

Step 5.5.4.2.1.3

Multiply $- - λ$ .

Tap for more steps...

Step 5.5.4.2.1.3.1

Multiply $- 1$ by $- 1$ .

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (3 (9 λ + \frac{75}{2}) - (- 1 - λ) (48 - (- \frac{1}{2} + 1 λ)) + 0)$

Step 5.5.4.2.1.3.2

Multiply $λ$ by $1$ .

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (3 (9 λ + \frac{75}{2}) - (- 1 - λ) (48 - (- \frac{1}{2} + λ)) + 0)$

Step 5.5.4.2.1.4

Apply the distributive property.

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (3 (9 λ + \frac{75}{2}) - (- 1 - λ) (48 - - \frac{1}{2} - λ) + 0)$

Step 5.5.4.2.1.5

Multiply $- - \frac{1}{2}$ .

Tap for more steps...

Step 5.5.4.2.1.5.1

Multiply $- 1$ by $- 1$ .

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (3 (9 λ + \frac{75}{2}) - (- 1 - λ) (48 + 1 (\frac{1}{2}) - λ) + 0)$

Step 5.5.4.2.1.5.2

Multiply $\frac{1}{2}$ by $1$ .

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (3 (9 λ + \frac{75}{2}) - (- 1 - λ) (48 + \frac{1}{2} - λ) + 0)$

Step 5.5.4.2.2

To write $48$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (3 (9 λ + \frac{75}{2}) - (- 1 - λ) (48 \cdot \frac{2}{2} + \frac{1}{2} - λ) + 0)$

Step 5.5.4.2.3

Combine $48$ and $\frac{2}{2}$ .

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (3 (9 λ + \frac{75}{2}) - (- 1 - λ) (\frac{48 \cdot 2}{2} + \frac{1}{2} - λ) + 0)$

Step 5.5.4.2.4

Combine the numerators over the common denominator.

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (3 (9 λ + \frac{75}{2}) - (- 1 - λ) (\frac{48 \cdot 2 + 1}{2} - λ) + 0)$

Step 5.5.4.2.5

Simplify the numerator.

Tap for more steps...

Step 5.5.4.2.5.1

Multiply $48$ by $2$ .

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (3 (9 λ + \frac{75}{2}) - (- 1 - λ) (\frac{96 + 1}{2} - λ) + 0)$

Step 5.5.4.2.5.2

Add $96$ and $1$ .

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (3 (9 λ + \frac{75}{2}) - (- 1 - λ) (\frac{97}{2} - λ) + 0)$

Step 5.5.4.2.6

Reorder $\frac{97}{2}$ and $- λ$ .

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (3 (9 λ + \frac{75}{2}) - (- 1 - λ) (- λ + \frac{97}{2}) + 0)$

Step 5.5.5

Simplify the determinant.

Tap for more steps...

Step 5.5.5.1

Add $3 (9 λ + \frac{75}{2}) - (- 1 - λ) (- λ + \frac{97}{2})$ and $0$ .

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (3 (9 λ + \frac{75}{2}) - (- 1 - λ) (- λ + \frac{97}{2}))$

Step 5.5.5.2

Simplify each term.

Tap for more steps...

Step 5.5.5.2.1

Apply the distributive property.

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (3 (9 λ) + 3 (\frac{75}{2}) - (- 1 - λ) (- λ + \frac{97}{2}))$

Step 5.5.5.2.2

Multiply $9$ by $3$ .

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (27 λ + 3 (\frac{75}{2}) - (- 1 - λ) (- λ + \frac{97}{2}))$

Step 5.5.5.2.3

Multiply $3 (\frac{75}{2})$ .

Tap for more steps...

Step 5.5.5.2.3.1

Combine $3$ and $\frac{75}{2}$ .

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (27 λ + \frac{3 \cdot 75}{2} - (- 1 - λ) (- λ + \frac{97}{2}))$

Step 5.5.5.2.3.2

Multiply $3$ by $75$ .

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (27 λ + \frac{225}{2} - (- 1 - λ) (- λ + \frac{97}{2}))$

Step 5.5.5.2.4

Apply the distributive property.

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (27 λ + \frac{225}{2} + (- - 1 - - λ) (- λ + \frac{97}{2}))$

Step 5.5.5.2.5

Multiply $- 1$ by $- 1$ .

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (27 λ + \frac{225}{2} + (1 - - λ) (- λ + \frac{97}{2}))$

Step 5.5.5.2.6

Multiply $- - λ$ .

Tap for more steps...

Step 5.5.5.2.6.1

Multiply $- 1$ by $- 1$ .

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (27 λ + \frac{225}{2} + (1 + 1 λ) (- λ + \frac{97}{2}))$

Step 5.5.5.2.6.2

Multiply $λ$ by $1$ .

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (27 λ + \frac{225}{2} + (1 + λ) (- λ + \frac{97}{2}))$

Step 5.5.5.2.7

Expand $(1 + λ) (- λ + \frac{97}{2})$ using the FOIL Method.

Tap for more steps...

Step 5.5.5.2.7.1

Apply the distributive property.

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (27 λ + \frac{225}{2} + 1 (- λ + \frac{97}{2}) + λ (- λ + \frac{97}{2}))$

Step 5.5.5.2.7.2

Apply the distributive property.

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (27 λ + \frac{225}{2} + 1 (- λ) + 1 (\frac{97}{2}) + λ (- λ + \frac{97}{2}))$

Step 5.5.5.2.7.3

Apply the distributive property.

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (27 λ + \frac{225}{2} + 1 (- λ) + 1 (\frac{97}{2}) + λ (- λ) + λ \frac{97}{2})$

Step 5.5.5.2.8

Simplify and combine like terms.

Tap for more steps...

Step 5.5.5.2.8.1

Simplify each term.

Tap for more steps...

Step 5.5.5.2.8.1.1

Multiply $- λ$ by $1$ .

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (27 λ + \frac{225}{2} - λ + 1 (\frac{97}{2}) + λ (- λ) + λ \frac{97}{2})$

Step 5.5.5.2.8.1.2

Multiply $\frac{97}{2}$ by $1$ .

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (27 λ + \frac{225}{2} - λ + \frac{97}{2} + λ (- λ) + λ \frac{97}{2})$

Step 5.5.5.2.8.1.3

Rewrite using the commutative property of multiplication.

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (27 λ + \frac{225}{2} - λ + \frac{97}{2} - λ \cdot λ + λ \frac{97}{2})$

Step 5.5.5.2.8.1.4

Multiply $λ$ by $λ$ by adding the exponents.

Tap for more steps...

Step 5.5.5.2.8.1.4.1

Move $λ$ .

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (27 λ + \frac{225}{2} - λ + \frac{97}{2} - (λ \cdot λ) + λ \frac{97}{2})$

Step 5.5.5.2.8.1.4.2

Multiply $λ$ by $λ$ .

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (27 λ + \frac{225}{2} - λ + \frac{97}{2} - λ^{2} + λ \frac{97}{2})$

Step 5.5.5.2.8.1.5

Combine $λ$ and $\frac{97}{2}$ .

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (27 λ + \frac{225}{2} - λ + \frac{97}{2} - λ^{2} + \frac{λ \cdot 97}{2})$

Step 5.5.5.2.8.1.6

Move $97$ to the left of $λ$ .

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (27 λ + \frac{225}{2} - λ + \frac{97}{2} - λ^{2} + \frac{97 λ}{2})$

Step 5.5.5.2.8.2

To write $- λ$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (27 λ + \frac{225}{2} + \frac{97}{2} - λ^{2} - λ \cdot \frac{2}{2} + \frac{97 λ}{2})$

Step 5.5.5.2.8.3

Combine $- λ$ and $\frac{2}{2}$ .

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (27 λ + \frac{225}{2} + \frac{97}{2} - λ^{2} + \frac{- λ \cdot 2}{2} + \frac{97 λ}{2})$

Step 5.5.5.2.8.4

Combine the numerators over the common denominator.

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (27 λ + \frac{225}{2} + \frac{97}{2} - λ^{2} + \frac{- λ \cdot 2 + 97 λ}{2})$

Step 5.5.5.2.8.5

Combine the numerators over the common denominator.

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (27 λ + \frac{225}{2} - λ^{2} + \frac{97 - λ \cdot 2 + 97 λ}{2})$

Step 5.5.5.2.8.6

To write $- λ^{2}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (27 λ + \frac{225}{2} - λ^{2} \cdot \frac{2}{2} + \frac{97 - λ \cdot 2 + 97 λ}{2})$

Step 5.5.5.2.8.7

Combine $- λ^{2}$ and $\frac{2}{2}$ .

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (27 λ + \frac{225}{2} + \frac{- λ^{2} \cdot 2}{2} + \frac{97 - λ \cdot 2 + 97 λ}{2})$

Step 5.5.5.2.8.8

Combine the numerators over the common denominator.

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (27 λ + \frac{225}{2} + \frac{- λ^{2} \cdot 2 + 97 - λ \cdot 2 + 97 λ}{2})$

Step 5.5.5.2.9

Simplify the numerator.

Tap for more steps...

Step 5.5.5.2.9.1

Reorder terms.

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (27 λ + \frac{225}{2} + \frac{- 1 \cdot 2 λ^{2} - 1 \cdot 2 λ + 97 λ + 97}{2})$

Step 5.5.5.2.9.2

Factor out the greatest common factor from each group.

Tap for more steps...

Step 5.5.5.2.9.2.1

Group the first two terms and the last two terms.

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (27 λ + \frac{225}{2} + \frac{(- 1 \cdot 2 λ^{2} - 1 \cdot 2 λ) + 97 λ + 97}{2})$

Step 5.5.5.2.9.2.2

Factor out the greatest common factor (GCF) from each group.

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (27 λ + \frac{225}{2} + \frac{2 λ (- λ - 1) - 97 (- λ - 1)}{2})$

Step 5.5.5.2.9.3

Factor the polynomial by factoring out the greatest common factor, $- λ - 1$ .

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (27 λ + \frac{225}{2} + \frac{(- λ - 1) (2 λ - 97)}{2})$

Step 5.5.5.3

Combine the numerators over the common denominator.

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (27 λ + \frac{225 + (- λ - 1) (2 λ - 97)}{2})$

Step 5.5.5.4

Simplify each term.

Tap for more steps...

Step 5.5.5.4.1

Expand $(- λ - 1) (2 λ - 97)$ using the FOIL Method.

Tap for more steps...

Step 5.5.5.4.1.1

Apply the distributive property.

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (27 λ + \frac{225 - λ (2 λ - 97) - 1 (2 λ - 97)}{2})$

Step 5.5.5.4.1.2

Apply the distributive property.

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (27 λ + \frac{225 - λ (2 λ) - λ \cdot - 97 - 1 (2 λ - 97)}{2})$

Step 5.5.5.4.1.3

Apply the distributive property.

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (27 λ + \frac{225 - λ (2 λ) - λ \cdot - 97 - 1 (2 λ) - 1 \cdot - 97}{2})$

Step 5.5.5.4.2

Simplify and combine like terms.

Tap for more steps...

Step 5.5.5.4.2.1

Simplify each term.

Tap for more steps...

Step 5.5.5.4.2.1.1

Rewrite using the commutative property of multiplication.

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (27 λ + \frac{225 - 1 \cdot 2 λ \cdot λ - λ \cdot - 97 - 1 (2 λ) - 1 \cdot - 97}{2})$

Step 5.5.5.4.2.1.2

Multiply $λ$ by $λ$ by adding the exponents.

Tap for more steps...

Step 5.5.5.4.2.1.2.1

Move $λ$ .

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (27 λ + \frac{225 - 1 \cdot 2 (λ \cdot λ) - λ \cdot - 97 - 1 (2 λ) - 1 \cdot - 97}{2})$

Step 5.5.5.4.2.1.2.2

Multiply $λ$ by $λ$ .

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (27 λ + \frac{225 - 1 \cdot 2 λ^{2} - λ \cdot - 97 - 1 (2 λ) - 1 \cdot - 97}{2})$

Step 5.5.5.4.2.1.3

Multiply $- 1$ by $2$ .

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (27 λ + \frac{225 - 2 λ^{2} - λ \cdot - 97 - 1 (2 λ) - 1 \cdot - 97}{2})$

Step 5.5.5.4.2.1.4

Multiply $- 97$ by $- 1$ .

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (27 λ + \frac{225 - 2 λ^{2} + 97 λ - 1 (2 λ) - 1 \cdot - 97}{2})$

Step 5.5.5.4.2.1.5

Multiply $2$ by $- 1$ .

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (27 λ + \frac{225 - 2 λ^{2} + 97 λ - 2 λ - 1 \cdot - 97}{2})$

Step 5.5.5.4.2.1.6

Multiply $- 1$ by $- 97$ .

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (27 λ + \frac{225 - 2 λ^{2} + 97 λ - 2 λ + 97}{2})$

Step 5.5.5.4.2.2

Subtract $2 λ$ from $97 λ$ .

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (27 λ + \frac{225 - 2 λ^{2} + 95 λ + 97}{2})$

Step 5.5.5.5

Add $225$ and $97$ .

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (27 λ + \frac{- 2 λ^{2} + 95 λ + 322}{2})$

Step 5.5.5.6

To write $27 λ$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (27 λ \cdot \frac{2}{2} + \frac{- 2 λ^{2} + 95 λ + 322}{2})$

Step 5.5.5.7

Combine $27 λ$ and $\frac{2}{2}$ .

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (\frac{27 λ \cdot 2}{2} + \frac{- 2 λ^{2} + 95 λ + 322}{2})$

Step 5.5.5.8

Combine the numerators over the common denominator.

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 \frac{27 λ \cdot 2 - 2 λ^{2} + 95 λ + 322}{2}$

Step 5.5.5.9

Simplify the numerator.

Tap for more steps...

Step 5.5.5.9.1

Multiply $2$ by $27$ .

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 \frac{54 λ - 2 λ^{2} + 95 λ + 322}{2}$

Step 5.5.5.9.2

Add $54 λ$ and $95 λ$ .

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 \frac{- 2 λ^{2} + 149 λ + 322}{2}$

Step 5.5.5.10

Factor $- 1$ out of $- 2 λ^{2}$ .

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 \frac{- (2 λ^{2}) + 149 λ + 322}{2}$

Step 5.5.5.11

Factor $- 1$ out of $149 λ$ .

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 \frac{- (2 λ^{2}) - (- 149 λ) + 322}{2}$

Step 5.5.5.12

Factor $- 1$ out of $- (2 λ^{2}) - (- 149 λ)$ .

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 \frac{- (2 λ^{2} - 149 λ) + 322}{2}$

Step 5.5.5.13

Rewrite $322$ as $- 1 (- 322)$ .

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 \frac{- (2 λ^{2} - 149 λ) - 1 (- 322)}{2}$

Step 5.5.5.14

Factor $- 1$ out of $- (2 λ^{2} - 149 λ) - 1 (- 322)$ .

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 \frac{- (2 λ^{2} - 149 λ - 322)}{2}$

Step 5.5.5.15

Rewrite $- (2 λ^{2} - 149 λ - 322)$ as $- 1 (2 λ^{2} - 149 λ - 322)$ .

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 \frac{- 1 (2 λ^{2} - 149 λ - 322)}{2}$

Step 5.5.5.16

Move the negative in front of the fraction.

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6

Simplify the determinant.

Tap for more steps...

Step 5.6.1

Combine the opposite terms in $(2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 + 0 - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$ .

Tap for more steps...

Step 5.6.1.1

Add $(2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2})$ and $0$ .

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) + 0 - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.1.2

Add $(2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2})$ and $0$ .

$p (λ) = (2 - λ) (- \frac{λ (λ + 1) (2 λ - 1)}{2}) - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2

Simplify each term.

Tap for more steps...

Step 5.6.2.1

Apply the distributive property.

$p (λ) = 2 (- \frac{λ (λ + 1) (2 λ - 1)}{2}) - λ (- \frac{λ (λ + 1) (2 λ - 1)}{2}) - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.2

Cancel the common factor of $2$ .

Tap for more steps...

Step 5.6.2.2.1

Move the leading negative in $- \frac{λ (λ + 1) (2 λ - 1)}{2}$ into the numerator.

$p (λ) = 2 \frac{- λ (λ + 1) (2 λ - 1)}{2} - λ (- \frac{λ (λ + 1) (2 λ - 1)}{2}) - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.2.2

Cancel the common factor.

$p (λ) = 2 \frac{- λ (λ + 1) (2 λ - 1)}{2} - λ (- \frac{λ (λ + 1) (2 λ - 1)}{2}) - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.2.3

Rewrite the expression.

$p (λ) = - λ (λ + 1) (2 λ - 1) - λ (- \frac{λ (λ + 1) (2 λ - 1)}{2}) - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.3

Multiply $- λ (- \frac{λ (λ + 1) (2 λ - 1)}{2})$ .

Tap for more steps...

Step 5.6.2.3.1

Multiply $- 1$ by $- 1$ .

$p (λ) = - λ (λ + 1) (2 λ - 1) + 1 λ \frac{λ (λ + 1) (2 λ - 1)}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.3.2

Multiply $λ$ by $1$ .

$p (λ) = - λ (λ + 1) (2 λ - 1) + λ \frac{λ (λ + 1) (2 λ - 1)}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.3.3

Combine $λ$ and $\frac{λ (λ + 1) (2 λ - 1)}{2}$ .

$p (λ) = - λ (λ + 1) (2 λ - 1) + \frac{λ (λ (λ + 1) (2 λ - 1))}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.3.4

Raise $λ$ to the power of $1$ .

$p (λ) = - λ (λ + 1) (2 λ - 1) + \frac{λ^{1} λ ((λ + 1) (2 λ - 1))}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.3.5

Raise $λ$ to the power of $1$ .

$p (λ) = - λ (λ + 1) (2 λ - 1) + \frac{λ^{1} λ^{1} ((λ + 1) (2 λ - 1))}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.3.6

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$p (λ) = - λ (λ + 1) (2 λ - 1) + \frac{λ^{1 + 1} ((λ + 1) (2 λ - 1))}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.3.7

Add $1$ and $1$ .

$p (λ) = - λ (λ + 1) (2 λ - 1) + \frac{λ^{2} ((λ + 1) (2 λ - 1))}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.4

Simplify each term.

Tap for more steps...

Step 5.6.2.4.1

Apply the distributive property.

$p (λ) = (- λ \cdot λ - λ \cdot 1) (2 λ - 1) + \frac{λ^{2} ((λ + 1) (2 λ - 1))}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.4.2

Multiply $λ$ by $λ$ by adding the exponents.

Tap for more steps...

Step 5.6.2.4.2.1

Move $λ$ .

$p (λ) = (- (λ \cdot λ) - λ \cdot 1) (2 λ - 1) + \frac{λ^{2} ((λ + 1) (2 λ - 1))}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.4.2.2

Multiply $λ$ by $λ$ .

$p (λ) = (- λ^{2} - λ \cdot 1) (2 λ - 1) + \frac{λ^{2} ((λ + 1) (2 λ - 1))}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.4.3

Multiply $- 1$ by $1$ .

$p (λ) = (- λ^{2} - λ) (2 λ - 1) + \frac{λ^{2} ((λ + 1) (2 λ - 1))}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.4.4

Expand $(- λ^{2} - λ) (2 λ - 1)$ using the FOIL Method.

Tap for more steps...

Step 5.6.2.4.4.1

Apply the distributive property.

$p (λ) = - λ^{2} (2 λ - 1) - λ (2 λ - 1) + \frac{λ^{2} ((λ + 1) (2 λ - 1))}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.4.4.2

Apply the distributive property.

$p (λ) = - λ^{2} (2 λ) - λ^{2} \cdot - 1 - λ (2 λ - 1) + \frac{λ^{2} ((λ + 1) (2 λ - 1))}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.4.4.3

Apply the distributive property.

$p (λ) = - λ^{2} (2 λ) - λ^{2} \cdot - 1 - λ (2 λ) - λ \cdot - 1 + \frac{λ^{2} ((λ + 1) (2 λ - 1))}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.4.5

Simplify and combine like terms.

Tap for more steps...

Step 5.6.2.4.5.1

Simplify each term.

Tap for more steps...

Step 5.6.2.4.5.1.1

Rewrite using the commutative property of multiplication.

$p (λ) = - 1 \cdot 2 λ^{2} λ - λ^{2} \cdot - 1 - λ (2 λ) - λ \cdot - 1 + \frac{λ^{2} ((λ + 1) (2 λ - 1))}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.4.5.1.2

Multiply $λ^{2}$ by $λ$ by adding the exponents.

Tap for more steps...

Step 5.6.2.4.5.1.2.1

Move $λ$ .

$p (λ) = - 1 \cdot 2 (λ \cdot λ^{2}) - λ^{2} \cdot - 1 - λ (2 λ) - λ \cdot - 1 + \frac{λ^{2} ((λ + 1) (2 λ - 1))}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.4.5.1.2.2

Multiply $λ$ by $λ^{2}$ .

Tap for more steps...

Step 5.6.2.4.5.1.2.2.1

Raise $λ$ to the power of $1$ .

$p (λ) = - 1 \cdot 2 (λ^{1} λ^{2}) - λ^{2} \cdot - 1 - λ (2 λ) - λ \cdot - 1 + \frac{λ^{2} ((λ + 1) (2 λ - 1))}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.4.5.1.2.2.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$p (λ) = - 1 \cdot 2 λ^{1 + 2} - λ^{2} \cdot - 1 - λ (2 λ) - λ \cdot - 1 + \frac{λ^{2} ((λ + 1) (2 λ - 1))}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.4.5.1.2.3

Add $1$ and $2$ .

$p (λ) = - 1 \cdot 2 λ^{3} - λ^{2} \cdot - 1 - λ (2 λ) - λ \cdot - 1 + \frac{λ^{2} ((λ + 1) (2 λ - 1))}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.4.5.1.3

Multiply $- 1$ by $2$ .

$p (λ) = - 2 λ^{3} - λ^{2} \cdot - 1 - λ (2 λ) - λ \cdot - 1 + \frac{λ^{2} ((λ + 1) (2 λ - 1))}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.4.5.1.4

Multiply $- λ^{2} \cdot - 1$ .

Tap for more steps...

Step 5.6.2.4.5.1.4.1

Multiply $- 1$ by $- 1$ .

$p (λ) = - 2 λ^{3} + 1 λ^{2} - λ (2 λ) - λ \cdot - 1 + \frac{λ^{2} ((λ + 1) (2 λ - 1))}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.4.5.1.4.2

Multiply $λ^{2}$ by $1$ .

$p (λ) = - 2 λ^{3} + λ^{2} - λ (2 λ) - λ \cdot - 1 + \frac{λ^{2} ((λ + 1) (2 λ - 1))}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.4.5.1.5

Rewrite using the commutative property of multiplication.

$p (λ) = - 2 λ^{3} + λ^{2} - 1 \cdot 2 λ \cdot λ - λ \cdot - 1 + \frac{λ^{2} ((λ + 1) (2 λ - 1))}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.4.5.1.6

Multiply $λ$ by $λ$ by adding the exponents.

Tap for more steps...

Step 5.6.2.4.5.1.6.1

Move $λ$ .

$p (λ) = - 2 λ^{3} + λ^{2} - 1 \cdot 2 (λ \cdot λ) - λ \cdot - 1 + \frac{λ^{2} ((λ + 1) (2 λ - 1))}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.4.5.1.6.2

Multiply $λ$ by $λ$ .

$p (λ) = - 2 λ^{3} + λ^{2} - 1 \cdot 2 λ^{2} - λ \cdot - 1 + \frac{λ^{2} ((λ + 1) (2 λ - 1))}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.4.5.1.7

Multiply $- 1$ by $2$ .

$p (λ) = - 2 λ^{3} + λ^{2} - 2 λ^{2} - λ \cdot - 1 + \frac{λ^{2} ((λ + 1) (2 λ - 1))}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.4.5.1.8

Multiply $- λ \cdot - 1$ .

Tap for more steps...

Step 5.6.2.4.5.1.8.1

Multiply $- 1$ by $- 1$ .

$p (λ) = - 2 λ^{3} + λ^{2} - 2 λ^{2} + 1 λ + \frac{λ^{2} ((λ + 1) (2 λ - 1))}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.4.5.1.8.2

Multiply $λ$ by $1$ .

$p (λ) = - 2 λ^{3} + λ^{2} - 2 λ^{2} + λ + \frac{λ^{2} ((λ + 1) (2 λ - 1))}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.4.5.2

Subtract $2 λ^{2}$ from $λ^{2}$ .

$p (λ) = - 2 λ^{3} - λ^{2} + λ + \frac{λ^{2} ((λ + 1) (2 λ - 1))}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.4.6

Simplify the numerator.

Tap for more steps...

Step 5.6.2.4.6.1

Rewrite.

$p (λ) = - 2 λ^{3} - λ^{2} + λ + \frac{λ (λ (λ + 1) (2 λ - 1))}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.4.6.2

Raise $λ$ to the power of $1$ .

$p (λ) = - 2 λ^{3} - λ^{2} + λ + \frac{λ^{1} λ ((λ + 1) (2 λ - 1))}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.4.6.3

Raise $λ$ to the power of $1$ .

$p (λ) = - 2 λ^{3} - λ^{2} + λ + \frac{λ^{1} λ^{1} ((λ + 1) (2 λ - 1))}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.4.6.4

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$p (λ) = - 2 λ^{3} - λ^{2} + λ + \frac{λ^{1 + 1} ((λ + 1) (2 λ - 1))}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.4.6.5

Add $1$ and $1$ .

$p (λ) = - 2 λ^{3} - λ^{2} + λ + \frac{λ^{2} ((λ + 1) (2 λ - 1))}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.4.6.6

Remove unnecessary parentheses.

$p (λ) = - 2 λ^{3} - λ^{2} + λ + \frac{λ^{2} (λ + 1) (2 λ - 1)}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.5

To write $- 2 λ^{3}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$p (λ) = - λ^{2} + λ - 2 λ^{3} \cdot \frac{2}{2} + \frac{λ^{2} (λ + 1) (2 λ - 1)}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.6

Combine $- 2 λ^{3}$ and $\frac{2}{2}$ .

$p (λ) = - λ^{2} + λ + \frac{- 2 λ^{3} \cdot 2}{2} + \frac{λ^{2} (λ + 1) (2 λ - 1)}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.7

Combine the numerators over the common denominator.

$p (λ) = - λ^{2} + λ + \frac{- 2 λ^{3} \cdot 2 + λ^{2} (λ + 1) (2 λ - 1)}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.8

Simplify the numerator.

Tap for more steps...

Step 5.6.2.8.1

Factor $λ^{2}$ out of $- 2 λ^{3} \cdot 2 + λ^{2} (λ + 1) (2 λ - 1)$ .

Tap for more steps...

Step 5.6.2.8.1.1

Factor $λ^{2}$ out of $- 2 λ^{3} \cdot 2$ .

$p (λ) = - λ^{2} + λ + \frac{λ^{2} (- 2 λ \cdot 2) + λ^{2} (λ + 1) (2 λ - 1)}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.8.1.2

Factor $λ^{2}$ out of $λ^{2} (λ + 1) (2 λ - 1)$ .

$p (λ) = - λ^{2} + λ + \frac{λ^{2} (- 2 λ \cdot 2) + λ^{2} ((λ + 1) (2 λ - 1))}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.8.1.3

Factor $λ^{2}$ out of $λ^{2} (- 2 λ \cdot 2) + λ^{2} ((λ + 1) (2 λ - 1))$ .

$p (λ) = - λ^{2} + λ + \frac{λ^{2} (- 2 λ \cdot 2 + (λ + 1) (2 λ - 1))}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.8.2

Multiply $2$ by $- 2$ .

$p (λ) = - λ^{2} + λ + \frac{λ^{2} (- 4 λ + (λ + 1) (2 λ - 1))}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.8.3

Expand $(λ + 1) (2 λ - 1)$ using the FOIL Method.

Tap for more steps...

Step 5.6.2.8.3.1

Apply the distributive property.

$p (λ) = - λ^{2} + λ + \frac{λ^{2} (- 4 λ + λ (2 λ - 1) + 1 (2 λ - 1))}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.8.3.2

Apply the distributive property.

$p (λ) = - λ^{2} + λ + \frac{λ^{2} (- 4 λ + λ (2 λ) + λ \cdot - 1 + 1 (2 λ - 1))}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.8.3.3

Apply the distributive property.

$p (λ) = - λ^{2} + λ + \frac{λ^{2} (- 4 λ + λ (2 λ) + λ \cdot - 1 + 1 (2 λ) + 1 \cdot - 1)}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.8.4

Simplify and combine like terms.

Tap for more steps...

Step 5.6.2.8.4.1

Simplify each term.

Tap for more steps...

Step 5.6.2.8.4.1.1

Rewrite using the commutative property of multiplication.

$p (λ) = - λ^{2} + λ + \frac{λ^{2} (- 4 λ + 2 λ \cdot λ + λ \cdot - 1 + 1 (2 λ) + 1 \cdot - 1)}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.8.4.1.2

Multiply $λ$ by $λ$ by adding the exponents.

Tap for more steps...

Step 5.6.2.8.4.1.2.1

Move $λ$ .

$p (λ) = - λ^{2} + λ + \frac{λ^{2} (- 4 λ + 2 (λ \cdot λ) + λ \cdot - 1 + 1 (2 λ) + 1 \cdot - 1)}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.8.4.1.2.2

Multiply $λ$ by $λ$ .

$p (λ) = - λ^{2} + λ + \frac{λ^{2} (- 4 λ + 2 λ^{2} + λ \cdot - 1 + 1 (2 λ) + 1 \cdot - 1)}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.8.4.1.3

Move $- 1$ to the left of $λ$ .

$p (λ) = - λ^{2} + λ + \frac{λ^{2} (- 4 λ + 2 λ^{2} - 1 \cdot λ + 1 (2 λ) + 1 \cdot - 1)}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.8.4.1.4

Rewrite $- 1 λ$ as $- λ$ .

$p (λ) = - λ^{2} + λ + \frac{λ^{2} (- 4 λ + 2 λ^{2} - λ + 1 (2 λ) + 1 \cdot - 1)}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.8.4.1.5

Multiply $2 λ$ by $1$ .

$p (λ) = - λ^{2} + λ + \frac{λ^{2} (- 4 λ + 2 λ^{2} - λ + 2 λ + 1 \cdot - 1)}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.8.4.1.6

Multiply $- 1$ by $1$ .

$p (λ) = - λ^{2} + λ + \frac{λ^{2} (- 4 λ + 2 λ^{2} - λ + 2 λ - 1)}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.8.4.2

Add $- λ$ and $2 λ$ .

$p (λ) = - λ^{2} + λ + \frac{λ^{2} (- 4 λ + 2 λ^{2} + λ - 1)}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.8.5

Add $- 4 λ$ and $λ$ .

$p (λ) = - λ^{2} + λ + \frac{λ^{2} (2 λ^{2} - 3 λ - 1)}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.9

To write $- λ^{2}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$p (λ) = λ - λ^{2} \cdot \frac{2}{2} + \frac{λ^{2} (2 λ^{2} - 3 λ - 1)}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.10

Combine $- λ^{2}$ and $\frac{2}{2}$ .

$p (λ) = λ + \frac{- λ^{2} \cdot 2}{2} + \frac{λ^{2} (2 λ^{2} - 3 λ - 1)}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.11

Combine the numerators over the common denominator.

$p (λ) = λ + \frac{- λ^{2} \cdot 2 + λ^{2} (2 λ^{2} - 3 λ - 1)}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.12

Simplify the numerator.

Tap for more steps...

Step 5.6.2.12.1

Factor $λ^{2}$ out of $- λ^{2} \cdot 2 + λ^{2} (2 λ^{2} - 3 λ - 1)$ .

Tap for more steps...

Step 5.6.2.12.1.1

Factor $λ^{2}$ out of $- λ^{2} \cdot 2$ .

$p (λ) = λ + \frac{λ^{2} (- 1 \cdot 2) + λ^{2} (2 λ^{2} - 3 λ - 1)}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.12.1.2

Factor $λ^{2}$ out of $λ^{2} (- 1 \cdot 2) + λ^{2} (2 λ^{2} - 3 λ - 1)$ .

$p (λ) = λ + \frac{λ^{2} (- 1 \cdot 2 + 2 λ^{2} - 3 λ - 1)}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.12.2

Multiply $- 1$ by $2$ .

$p (λ) = λ + \frac{λ^{2} (- 2 + 2 λ^{2} - 3 λ - 1)}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.12.3

Subtract $1$ from $- 2$ .

$p (λ) = λ + \frac{λ^{2} (2 λ^{2} - 3 λ - 3)}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.13

To write $λ$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$p (λ) = λ \cdot \frac{2}{2} + \frac{λ^{2} (2 λ^{2} - 3 λ - 3)}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.14

Combine $λ$ and $\frac{2}{2}$ .

$p (λ) = \frac{λ \cdot 2}{2} + \frac{λ^{2} (2 λ^{2} - 3 λ - 3)}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.15

Combine the numerators over the common denominator.

$p (λ) = \frac{λ \cdot 2 + λ^{2} (2 λ^{2} - 3 λ - 3)}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.16

Simplify the numerator.

Tap for more steps...

Step 5.6.2.16.1

Factor $λ$ out of $λ \cdot 2 + λ^{2} (2 λ^{2} - 3 λ - 3)$ .

Tap for more steps...

Step 5.6.2.16.1.1

Factor $λ$ out of $λ \cdot 2$ .

$p (λ) = \frac{λ (2) + λ^{2} (2 λ^{2} - 3 λ - 3)}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.16.1.2

Factor $λ$ out of $λ^{2} (2 λ^{2} - 3 λ - 3)$ .

$p (λ) = \frac{λ (2) + λ (λ (2 λ^{2} - 3 λ - 3))}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.16.1.3

Factor $λ$ out of $λ (2) + λ (λ (2 λ^{2} - 3 λ - 3))$ .

$p (λ) = \frac{λ (2 + λ (2 λ^{2} - 3 λ - 3))}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.16.2

Apply the distributive property.

$p (λ) = \frac{λ (2 + λ (2 λ^{2}) + λ (- 3 λ) + λ \cdot - 3)}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.16.3

Simplify.

Tap for more steps...

Step 5.6.2.16.3.1

Rewrite using the commutative property of multiplication.

$p (λ) = \frac{λ (2 + 2 λ \cdot λ^{2} + λ (- 3 λ) + λ \cdot - 3)}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.16.3.2

Rewrite using the commutative property of multiplication.

$p (λ) = \frac{λ (2 + 2 λ \cdot λ^{2} - 3 λ \cdot λ + λ \cdot - 3)}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.16.3.3

Move $- 3$ to the left of $λ$ .

$p (λ) = \frac{λ (2 + 2 λ \cdot λ^{2} - 3 λ \cdot λ - 3 \cdot λ)}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.16.4

Simplify each term.

Tap for more steps...

Step 5.6.2.16.4.1

Multiply $λ$ by $λ^{2}$ by adding the exponents.

Tap for more steps...

Step 5.6.2.16.4.1.1

Move $λ^{2}$ .

$p (λ) = \frac{λ (2 + 2 (λ^{2} λ) - 3 λ \cdot λ - 3 \cdot λ)}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.16.4.1.2

Multiply $λ^{2}$ by $λ$ .

Tap for more steps...

Step 5.6.2.16.4.1.2.1

Raise $λ$ to the power of $1$ .

$p (λ) = \frac{λ (2 + 2 (λ^{2} λ^{1}) - 3 λ \cdot λ - 3 \cdot λ)}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.16.4.1.2.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$p (λ) = \frac{λ (2 + 2 λ^{2 + 1} - 3 λ \cdot λ - 3 \cdot λ)}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.16.4.1.3

Add $2$ and $1$ .

$p (λ) = \frac{λ (2 + 2 λ^{3} - 3 λ \cdot λ - 3 \cdot λ)}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.16.4.2

Multiply $λ$ by $λ$ by adding the exponents.

Tap for more steps...

Step 5.6.2.16.4.2.1

Move $λ$ .

$p (λ) = \frac{λ (2 + 2 λ^{3} - 3 (λ \cdot λ) - 3 \cdot λ)}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.16.4.2.2

Multiply $λ$ by $λ$ .

$p (λ) = \frac{λ (2 + 2 λ^{3} - 3 λ^{2} - 3 \cdot λ)}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

$p (λ) = \frac{λ (2 + 2 λ^{3} - 3 λ^{2} - 3 λ)}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.16.5

Reorder terms.

$p (λ) = \frac{λ (2 λ^{3} - 3 λ^{2} - 3 λ + 2)}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.16.6

Rewrite $2 λ^{3} - 3 λ^{2} - 3 λ + 2$ in a factored form.

Tap for more steps...

Step 5.6.2.16.6.1

Factor $2 λ^{3} - 3 λ^{2} - 3 λ + 2$ using the rational roots test.

Tap for more steps...

Step 5.6.2.16.6.1.1

If a polynomial function has integer coefficients, then every rational zero will have the form $\frac{p}{q}$ where $p$ is a factor of the constant and $q$ is a factor of the leading coefficient.

$p = \pm 1, \pm 2$

$q = \pm 1, \pm 2$

Step 5.6.2.16.6.1.2

Find every combination of $\pm \frac{p}{q}$ . These are the possible roots of the polynomial function.

$\pm 1, \pm 0.5, \pm 2$

Step 5.6.2.16.6.1.3

Substitute $- 1$ and simplify the expression. In this case, the expression is equal to $0$ so $- 1$ is a root of the polynomial.

Tap for more steps...

Step 5.6.2.16.6.1.3.1

Substitute $- 1$ into the polynomial.

$2 {(- 1)}^{3} - 3 {(- 1)}^{2} - 3 \cdot - 1 + 2$

Step 5.6.2.16.6.1.3.2

Raise $- 1$ to the power of $3$ .

$2 \cdot - 1 - 3 {(- 1)}^{2} - 3 \cdot - 1 + 2$

Step 5.6.2.16.6.1.3.3

Multiply $2$ by $- 1$ .

$- 2 - 3 {(- 1)}^{2} - 3 \cdot - 1 + 2$

Step 5.6.2.16.6.1.3.4

Raise $- 1$ to the power of $2$ .

$- 2 - 3 \cdot 1 - 3 \cdot - 1 + 2$

Step 5.6.2.16.6.1.3.5

Multiply $- 3$ by $1$ .

$- 2 - 3 - 3 \cdot - 1 + 2$

Step 5.6.2.16.6.1.3.6

Subtract $3$ from $- 2$ .

$- 5 - 3 \cdot - 1 + 2$

Step 5.6.2.16.6.1.3.7

Multiply $- 3$ by $- 1$ .

$- 5 + 3 + 2$

Step 5.6.2.16.6.1.3.8

Add $- 5$ and $3$ .

$- 2 + 2$

Step 5.6.2.16.6.1.3.9

Add $- 2$ and $2$ .

$0$

Step 5.6.2.16.6.1.4

Since $- 1$ is a known root, divide the polynomial by $λ + 1$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{2 λ^{3} - 3 λ^{2} - 3 λ + 2}{λ + 1}$

Step 5.6.2.16.6.1.5

Divide $2 λ^{3} - 3 λ^{2} - 3 λ + 2$ by $λ + 1$ .

Tap for more steps...

Step 5.6.2.16.6.1.5.1

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$λ$

$1$

$2 λ^{3}$

$3 λ^{2}$

$3 λ$

$2$

Step 5.6.2.16.6.1.5.2

Divide the highest order term in the dividend $2 λ^{3}$ by the highest order term in divisor $λ$ .

					$2 λ^{2}$
	$λ$	+	$1$		$2 λ^{3}$	-	$3 λ^{2}$	-	$3 λ$	+	$2$

Step 5.6.2.16.6.1.5.3

Multiply the new quotient term by the divisor.

				$2 λ^{2}$
$λ$	+	$1$		$2 λ^{3}$	-	$3 λ^{2}$	-	$3 λ$	+	$2$
			+	$2 λ^{3}$	+	$2 λ^{2}$

Step 5.6.2.16.6.1.5.4

The expression needs to be subtracted from the dividend, so change all the signs in $2 λ^{3} + 2 λ^{2}$

				$2 λ^{2}$
$λ$	+	$1$		$2 λ^{3}$	-	$3 λ^{2}$	-	$3 λ$	+	$2$
			-	$2 λ^{3}$	-	$2 λ^{2}$

Step 5.6.2.16.6.1.5.5

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$2 λ^{2}$
$λ$	+	$1$		$2 λ^{3}$	-	$3 λ^{2}$	-	$3 λ$	+	$2$
			-	$2 λ^{3}$	-	$2 λ^{2}$
					-	$5 λ^{2}$

Step 5.6.2.16.6.1.5.6

Pull the next terms from the original dividend down into the current dividend.

				$2 λ^{2}$
$λ$	+	$1$		$2 λ^{3}$	-	$3 λ^{2}$	-	$3 λ$	+	$2$
			-	$2 λ^{3}$	-	$2 λ^{2}$
					-	$5 λ^{2}$	-	$3 λ$

Step 5.6.2.16.6.1.5.7

Divide the highest order term in the dividend $- 5 λ^{2}$ by the highest order term in divisor $λ$ .

				$2 λ^{2}$	-	$5 λ$
$λ$	+	$1$		$2 λ^{3}$	-	$3 λ^{2}$	-	$3 λ$	+	$2$
			-	$2 λ^{3}$	-	$2 λ^{2}$
					-	$5 λ^{2}$	-	$3 λ$

Step 5.6.2.16.6.1.5.8

Multiply the new quotient term by the divisor.

				$2 λ^{2}$	-	$5 λ$
$λ$	+	$1$		$2 λ^{3}$	-	$3 λ^{2}$	-	$3 λ$	+	$2$
			-	$2 λ^{3}$	-	$2 λ^{2}$
					-	$5 λ^{2}$	-	$3 λ$
					-	$5 λ^{2}$	-	$5 λ$

Step 5.6.2.16.6.1.5.9

The expression needs to be subtracted from the dividend, so change all the signs in $- 5 λ^{2} - 5 λ$

				$2 λ^{2}$	-	$5 λ$
$λ$	+	$1$		$2 λ^{3}$	-	$3 λ^{2}$	-	$3 λ$	+	$2$
			-	$2 λ^{3}$	-	$2 λ^{2}$
					-	$5 λ^{2}$	-	$3 λ$
					+	$5 λ^{2}$	+	$5 λ$

Step 5.6.2.16.6.1.5.10

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$2 λ^{2}$	-	$5 λ$
$λ$	+	$1$		$2 λ^{3}$	-	$3 λ^{2}$	-	$3 λ$	+	$2$
			-	$2 λ^{3}$	-	$2 λ^{2}$
					-	$5 λ^{2}$	-	$3 λ$
					+	$5 λ^{2}$	+	$5 λ$
							+	$2 λ$

Step 5.6.2.16.6.1.5.11

Pull the next terms from the original dividend down into the current dividend.

				$2 λ^{2}$	-	$5 λ$
$λ$	+	$1$		$2 λ^{3}$	-	$3 λ^{2}$	-	$3 λ$	+	$2$
			-	$2 λ^{3}$	-	$2 λ^{2}$
					-	$5 λ^{2}$	-	$3 λ$
					+	$5 λ^{2}$	+	$5 λ$
							+	$2 λ$	+	$2$

Step 5.6.2.16.6.1.5.12

Divide the highest order term in the dividend $2 λ$ by the highest order term in divisor $λ$ .

				$2 λ^{2}$	-	$5 λ$	+	$2$
$λ$	+	$1$		$2 λ^{3}$	-	$3 λ^{2}$	-	$3 λ$	+	$2$
			-	$2 λ^{3}$	-	$2 λ^{2}$
					-	$5 λ^{2}$	-	$3 λ$
					+	$5 λ^{2}$	+	$5 λ$
							+	$2 λ$	+	$2$

Step 5.6.2.16.6.1.5.13

Multiply the new quotient term by the divisor.

				$2 λ^{2}$	-	$5 λ$	+	$2$
$λ$	+	$1$		$2 λ^{3}$	-	$3 λ^{2}$	-	$3 λ$	+	$2$
			-	$2 λ^{3}$	-	$2 λ^{2}$
					-	$5 λ^{2}$	-	$3 λ$
					+	$5 λ^{2}$	+	$5 λ$
							+	$2 λ$	+	$2$
							+	$2 λ$	+	$2$

Step 5.6.2.16.6.1.5.14

The expression needs to be subtracted from the dividend, so change all the signs in $2 λ + 2$

				$2 λ^{2}$	-	$5 λ$	+	$2$
$λ$	+	$1$		$2 λ^{3}$	-	$3 λ^{2}$	-	$3 λ$	+	$2$
			-	$2 λ^{3}$	-	$2 λ^{2}$
					-	$5 λ^{2}$	-	$3 λ$
					+	$5 λ^{2}$	+	$5 λ$
							+	$2 λ$	+	$2$
							-	$2 λ$	-	$2$

Step 5.6.2.16.6.1.5.15

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$2 λ^{2}$	-	$5 λ$	+	$2$
$λ$	+	$1$		$2 λ^{3}$	-	$3 λ^{2}$	-	$3 λ$	+	$2$
			-	$2 λ^{3}$	-	$2 λ^{2}$
					-	$5 λ^{2}$	-	$3 λ$
					+	$5 λ^{2}$	+	$5 λ$
							+	$2 λ$	+	$2$
							-	$2 λ$	-	$2$
										$0$

Step 5.6.2.16.6.1.5.16

Since the remander is $0$ , the final answer is the quotient.

$2 λ^{2} - 5 λ + 2$

Step 5.6.2.16.6.1.6

Write $2 λ^{3} - 3 λ^{2} - 3 λ + 2$ as a set of factors.

$p (λ) = \frac{λ ((λ + 1) (2 λ^{2} - 5 λ + 2))}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.16.6.2

Factor by grouping.

Tap for more steps...

Step 5.6.2.16.6.2.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot 2 = 4$ and whose sum is $b = - 5$ .

Tap for more steps...

Step 5.6.2.16.6.2.1.1

Factor $- 5$ out of $- 5 λ$ .

$p (λ) = \frac{λ ((λ + 1) (2 λ^{2} - 5 (λ) + 2))}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.16.6.2.1.2

Rewrite $- 5$ as $- 1$ plus $- 4$

$p (λ) = \frac{λ ((λ + 1) (2 λ^{2} + (- 1 - 4) λ + 2))}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.16.6.2.1.3

Apply the distributive property.

$p (λ) = \frac{λ ((λ + 1) (2 λ^{2} - 1 λ - 4 λ + 2))}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.16.6.2.2

Factor out the greatest common factor from each group.

Tap for more steps...

Step 5.6.2.16.6.2.2.1

Group the first two terms and the last two terms.

$p (λ) = \frac{λ ((λ + 1) ((2 λ^{2} - 1 λ) - 4 λ + 2))}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.16.6.2.2.2

Factor out the greatest common factor (GCF) from each group.

$p (λ) = \frac{λ ((λ + 1) (λ (2 λ - 1) - 2 (2 λ - 1)))}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.16.6.2.3

Factor the polynomial by factoring out the greatest common factor, $2 λ - 1$ .

$p (λ) = \frac{λ ((λ + 1) ((2 λ - 1) (λ - 2)))}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

$p (λ) = \frac{λ ((λ + 1) (2 λ - 1) (λ - 2))}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

$p (λ) = \frac{λ (λ + 1) (2 λ - 1) (λ - 2)}{2} - 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$

Step 5.6.2.17

Multiply $- 3 (- \frac{2 λ^{2} - 149 λ - 322}{2})$ .

Tap for more steps...

Step 5.6.2.17.1

Multiply $- 1$ by $- 3$ .

$p (λ) = \frac{λ (λ + 1) (2 λ - 1) (λ - 2)}{2} + 3 \frac{2 λ^{2} - 149 λ - 322}{2}$

Step 5.6.2.17.2

Combine $3$ and $\frac{2 λ^{2} - 149 λ - 322}{2}$ .

$p (λ) = \frac{λ (λ + 1) (2 λ - 1) (λ - 2)}{2} + \frac{3 (2 λ^{2} - 149 λ - 322)}{2}$

Step 5.6.3

Combine the numerators over the common denominator.

$p (λ) = \frac{λ (λ + 1) (2 λ - 1) (λ - 2) + 3 (2 λ^{2} - 149 λ - 322)}{2}$

Step 5.6.4

Simplify each term.

Tap for more steps...

Step 5.6.4.1

Apply the distributive property.

$p (λ) = \frac{(λ \cdot λ + λ \cdot 1) (2 λ - 1) (λ - 2) + 3 (2 λ^{2} - 149 λ - 322)}{2}$

Step 5.6.4.2

Multiply $λ$ by $λ$ .

$p (λ) = \frac{(λ^{2} + λ \cdot 1) (2 λ - 1) (λ - 2) + 3 (2 λ^{2} - 149 λ - 322)}{2}$

Step 5.6.4.3

Multiply $λ$ by $1$ .

$p (λ) = \frac{(λ^{2} + λ) (2 λ - 1) (λ - 2) + 3 (2 λ^{2} - 149 λ - 322)}{2}$

Step 5.6.4.4

Expand $(λ^{2} + λ) (2 λ - 1)$ using the FOIL Method.

Tap for more steps...

Step 5.6.4.4.1

Apply the distributive property.

$p (λ) = \frac{(λ^{2} (2 λ - 1) + λ (2 λ - 1)) (λ - 2) + 3 (2 λ^{2} - 149 λ - 322)}{2}$

Step 5.6.4.4.2

Apply the distributive property.

$p (λ) = \frac{(λ^{2} (2 λ) + λ^{2} \cdot - 1 + λ (2 λ - 1)) (λ - 2) + 3 (2 λ^{2} - 149 λ - 322)}{2}$

Step 5.6.4.4.3

Apply the distributive property.

$p (λ) = \frac{(λ^{2} (2 λ) + λ^{2} \cdot - 1 + λ (2 λ) + λ \cdot - 1) (λ - 2) + 3 (2 λ^{2} - 149 λ - 322)}{2}$

Step 5.6.4.5

Simplify and combine like terms.

Tap for more steps...

Step 5.6.4.5.1

Simplify each term.

Tap for more steps...

Step 5.6.4.5.1.1

Rewrite using the commutative property of multiplication.

$p (λ) = \frac{(2 λ^{2} λ + λ^{2} \cdot - 1 + λ (2 λ) + λ \cdot - 1) (λ - 2) + 3 (2 λ^{2} - 149 λ - 322)}{2}$

Step 5.6.4.5.1.2

Multiply $λ^{2}$ by $λ$ by adding the exponents.

Tap for more steps...

Step 5.6.4.5.1.2.1

Move $λ$ .

$p (λ) = \frac{(2 (λ \cdot λ^{2}) + λ^{2} \cdot - 1 + λ (2 λ) + λ \cdot - 1) (λ - 2) + 3 (2 λ^{2} - 149 λ - 322)}{2}$

Step 5.6.4.5.1.2.2

Multiply $λ$ by $λ^{2}$ .

Tap for more steps...

Step 5.6.4.5.1.2.2.1

Raise $λ$ to the power of $1$ .

$p (λ) = \frac{(2 (λ^{1} λ^{2}) + λ^{2} \cdot - 1 + λ (2 λ) + λ \cdot - 1) (λ - 2) + 3 (2 λ^{2} - 149 λ - 322)}{2}$

Step 5.6.4.5.1.2.2.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$p (λ) = \frac{(2 λ^{1 + 2} + λ^{2} \cdot - 1 + λ (2 λ) + λ \cdot - 1) (λ - 2) + 3 (2 λ^{2} - 149 λ - 322)}{2}$

Step 5.6.4.5.1.2.3

Add $1$ and $2$ .

$p (λ) = \frac{(2 λ^{3} + λ^{2} \cdot - 1 + λ (2 λ) + λ \cdot - 1) (λ - 2) + 3 (2 λ^{2} - 149 λ - 322)}{2}$

Step 5.6.4.5.1.3

Move $- 1$ to the left of $λ^{2}$ .

$p (λ) = \frac{(2 λ^{3} - 1 \cdot λ^{2} + λ (2 λ) + λ \cdot - 1) (λ - 2) + 3 (2 λ^{2} - 149 λ - 322)}{2}$

Step 5.6.4.5.1.4

Rewrite $- 1 λ^{2}$ as $- λ^{2}$ .

$p (λ) = \frac{(2 λ^{3} - λ^{2} + λ (2 λ) + λ \cdot - 1) (λ - 2) + 3 (2 λ^{2} - 149 λ - 322)}{2}$

Step 5.6.4.5.1.5

Rewrite using the commutative property of multiplication.

$p (λ) = \frac{(2 λ^{3} - λ^{2} + 2 λ \cdot λ + λ \cdot - 1) (λ - 2) + 3 (2 λ^{2} - 149 λ - 322)}{2}$

Step 5.6.4.5.1.6

Multiply $λ$ by $λ$ by adding the exponents.

Tap for more steps...

Step 5.6.4.5.1.6.1

Move $λ$ .

$p (λ) = \frac{(2 λ^{3} - λ^{2} + 2 (λ \cdot λ) + λ \cdot - 1) (λ - 2) + 3 (2 λ^{2} - 149 λ - 322)}{2}$

Step 5.6.4.5.1.6.2

Multiply $λ$ by $λ$ .

$p (λ) = \frac{(2 λ^{3} - λ^{2} + 2 λ^{2} + λ \cdot - 1) (λ - 2) + 3 (2 λ^{2} - 149 λ - 322)}{2}$

Step 5.6.4.5.1.7

Move $- 1$ to the left of $λ$ .

$p (λ) = \frac{(2 λ^{3} - λ^{2} + 2 λ^{2} - 1 \cdot λ) (λ - 2) + 3 (2 λ^{2} - 149 λ - 322)}{2}$

Step 5.6.4.5.1.8

Rewrite $- 1 λ$ as $- λ$ .

$p (λ) = \frac{(2 λ^{3} - λ^{2} + 2 λ^{2} - λ) (λ - 2) + 3 (2 λ^{2} - 149 λ - 322)}{2}$

Step 5.6.4.5.2

Add $- λ^{2}$ and $2 λ^{2}$ .

$p (λ) = \frac{(2 λ^{3} + λ^{2} - λ) (λ - 2) + 3 (2 λ^{2} - 149 λ - 322)}{2}$

Step 5.6.4.6

Expand $(2 λ^{3} + λ^{2} - λ) (λ - 2)$ by multiplying each term in the first expression by each term in the second expression.

$p (λ) = \frac{2 λ^{3} λ + 2 λ^{3} \cdot - 2 + λ^{2} λ + λ^{2} \cdot - 2 - λ \cdot λ - λ \cdot - 2 + 3 (2 λ^{2} - 149 λ - 322)}{2}$

Step 5.6.4.7

Simplify each term.

Tap for more steps...

Step 5.6.4.7.1

Multiply $λ^{3}$ by $λ$ by adding the exponents.

Tap for more steps...

Step 5.6.4.7.1.1

Move $λ$ .

$p (λ) = \frac{2 (λ \cdot λ^{3}) + 2 λ^{3} \cdot - 2 + λ^{2} λ + λ^{2} \cdot - 2 - λ \cdot λ - λ \cdot - 2 + 3 (2 λ^{2} - 149 λ - 322)}{2}$

Step 5.6.4.7.1.2

Multiply $λ$ by $λ^{3}$ .

Tap for more steps...

Step 5.6.4.7.1.2.1

Raise $λ$ to the power of $1$ .

$p (λ) = \frac{2 (λ^{1} λ^{3}) + 2 λ^{3} \cdot - 2 + λ^{2} λ + λ^{2} \cdot - 2 - λ \cdot λ - λ \cdot - 2 + 3 (2 λ^{2} - 149 λ - 322)}{2}$

Step 5.6.4.7.1.2.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$p (λ) = \frac{2 λ^{1 + 3} + 2 λ^{3} \cdot - 2 + λ^{2} λ + λ^{2} \cdot - 2 - λ \cdot λ - λ \cdot - 2 + 3 (2 λ^{2} - 149 λ - 322)}{2}$

Step 5.6.4.7.1.3

Add $1$ and $3$ .

$p (λ) = \frac{2 λ^{4} + 2 λ^{3} \cdot - 2 + λ^{2} λ + λ^{2} \cdot - 2 - λ \cdot λ - λ \cdot - 2 + 3 (2 λ^{2} - 149 λ - 322)}{2}$

Step 5.6.4.7.2

Multiply $- 2$ by $2$ .

$p (λ) = \frac{2 λ^{4} - 4 λ^{3} + λ^{2} λ + λ^{2} \cdot - 2 - λ \cdot λ - λ \cdot - 2 + 3 (2 λ^{2} - 149 λ - 322)}{2}$

Step 5.6.4.7.3

Multiply $λ^{2}$ by $λ$ by adding the exponents.

Tap for more steps...

Step 5.6.4.7.3.1

Multiply $λ^{2}$ by $λ$ .

Tap for more steps...

Step 5.6.4.7.3.1.1

Raise $λ$ to the power of $1$ .

$p (λ) = \frac{2 λ^{4} - 4 λ^{3} + λ^{2} λ^{1} + λ^{2} \cdot - 2 - λ \cdot λ - λ \cdot - 2 + 3 (2 λ^{2} - 149 λ - 322)}{2}$

Step 5.6.4.7.3.1.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$p (λ) = \frac{2 λ^{4} - 4 λ^{3} + λ^{2 + 1} + λ^{2} \cdot - 2 - λ \cdot λ - λ \cdot - 2 + 3 (2 λ^{2} - 149 λ - 322)}{2}$

Step 5.6.4.7.3.2

Add $2$ and $1$ .

$p (λ) = \frac{2 λ^{4} - 4 λ^{3} + λ^{3} + λ^{2} \cdot - 2 - λ \cdot λ - λ \cdot - 2 + 3 (2 λ^{2} - 149 λ - 322)}{2}$

Step 5.6.4.7.4

Move $- 2$ to the left of $λ^{2}$ .

$p (λ) = \frac{2 λ^{4} - 4 λ^{3} + λ^{3} - 2 \cdot λ^{2} - λ \cdot λ - λ \cdot - 2 + 3 (2 λ^{2} - 149 λ - 322)}{2}$

Step 5.6.4.7.5

Multiply $λ$ by $λ$ by adding the exponents.

Tap for more steps...

Step 5.6.4.7.5.1

Move $λ$ .

$p (λ) = \frac{2 λ^{4} - 4 λ^{3} + λ^{3} - 2 λ^{2} - (λ \cdot λ) - λ \cdot - 2 + 3 (2 λ^{2} - 149 λ - 322)}{2}$

Step 5.6.4.7.5.2

Multiply $λ$ by $λ$ .

$p (λ) = \frac{2 λ^{4} - 4 λ^{3} + λ^{3} - 2 λ^{2} - λ^{2} - λ \cdot - 2 + 3 (2 λ^{2} - 149 λ - 322)}{2}$

Step 5.6.4.7.6

Multiply $- 2$ by $- 1$ .

$p (λ) = \frac{2 λ^{4} - 4 λ^{3} + λ^{3} - 2 λ^{2} - λ^{2} + 2 λ + 3 (2 λ^{2} - 149 λ - 322)}{2}$

Step 5.6.4.8

Add $- 4 λ^{3}$ and $λ^{3}$ .

$p (λ) = \frac{2 λ^{4} - 3 λ^{3} - 2 λ^{2} - λ^{2} + 2 λ + 3 (2 λ^{2} - 149 λ - 322)}{2}$

Step 5.6.4.9

Subtract $λ^{2}$ from $- 2 λ^{2}$ .

$p (λ) = \frac{2 λ^{4} - 3 λ^{3} - 3 λ^{2} + 2 λ + 3 (2 λ^{2} - 149 λ - 322)}{2}$

Step 5.6.4.10

Apply the distributive property.

$p (λ) = \frac{2 λ^{4} - 3 λ^{3} - 3 λ^{2} + 2 λ + 3 (2 λ^{2}) + 3 (- 149 λ) + 3 \cdot - 322}{2}$

Step 5.6.4.11

Simplify.

Tap for more steps...

Step 5.6.4.11.1

Multiply $2$ by $3$ .

$p (λ) = \frac{2 λ^{4} - 3 λ^{3} - 3 λ^{2} + 2 λ + 6 λ^{2} + 3 (- 149 λ) + 3 \cdot - 322}{2}$

Step 5.6.4.11.2

Multiply $- 149$ by $3$ .

$p (λ) = \frac{2 λ^{4} - 3 λ^{3} - 3 λ^{2} + 2 λ + 6 λ^{2} - 447 λ + 3 \cdot - 322}{2}$

Step 5.6.4.11.3

Multiply $3$ by $- 322$ .

$p (λ) = \frac{2 λ^{4} - 3 λ^{3} - 3 λ^{2} + 2 λ + 6 λ^{2} - 447 λ - 966}{2}$

Step 5.6.5

Add $- 3 λ^{2}$ and $6 λ^{2}$ .

$p (λ) = \frac{2 λ^{4} - 3 λ^{3} + 3 λ^{2} + 2 λ - 447 λ - 966}{2}$

Step 5.6.6

Subtract $447 λ$ from $2 λ$ .

$p (λ) = \frac{2 λ^{4} - 3 λ^{3} + 3 λ^{2} - 445 λ - 966}{2}$

Step 6

Set the characteristic polynomial equal to $0$ to find the eigenvalues $λ$ .

$\frac{2 λ^{4} - 3 λ^{3} + 3 λ^{2} - 445 λ - 966}{2} = 0$

Step 7

Solve for $λ$ .

Tap for more steps...

Step 7.1

Graph each side of the equation. The solution is the x-value of the point of intersection.

$λ \approx - 2.01433709, 7.08281505$