Linear Algebra Examples

$[\begin{array}{c} - 6 & - 4 & 10 \\ 2 & \frac{4}{3} & - \frac{10}{3} \\ 6 & 4 & 10 \end{array}]$

Step 1

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

$p (λ) = determinant (A - λ I_{3})$

Step 2

The identity matrix or unit matrix of size $3$ is the $3 \times 3$ square matrix with ones on the main diagonal and zeros elsewhere.

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

Step 3

Substitute the known values into $p (λ) = determinant (A - λ I_{3})$ .

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Step 3.1

Substitute $[\begin{array}{c} - 6 & - 4 & 10 \\ 2 & \frac{4}{3} & - \frac{10}{3} \\ 6 & 4 & 10 \end{array}]$ for $A$ .

$p (λ) = determinant ([\begin{array}{c} - 6 & - 4 & 10 \\ 2 & \frac{4}{3} & - \frac{10}{3} \\ 6 & 4 & 10 \end{array}] - λ I_{3})$

Step 3.2

Substitute $[\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}]$ for $I_{3}$ .

$p (λ) = determinant ([\begin{array}{c} - 6 & - 4 & 10 \\ 2 & \frac{4}{3} & - \frac{10}{3} \\ 6 & 4 & 10 \end{array}] - λ [\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}])$

Step 4

Simplify.

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Step 4.1

Simplify each term.

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Step 4.1.1

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

$p (λ) = determinant ([\begin{array}{c} - 6 & - 4 & 10 \\ 2 & \frac{4}{3} & - \frac{10}{3} \\ 6 & 4 & 10 \end{array}] + [\begin{array}{c} - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2

Simplify each element in the matrix.

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Step 4.1.2.1

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} - 6 & - 4 & 10 \\ 2 & \frac{4}{3} & - \frac{10}{3} \\ 6 & 4 & 10 \end{array}] + [\begin{array}{c} - λ & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.2

Multiply $- λ \cdot 0$ .

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Step 4.1.2.2.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} - 6 & - 4 & 10 \\ 2 & \frac{4}{3} & - \frac{10}{3} \\ 6 & 4 & 10 \end{array}] + [\begin{array}{c} - λ & 0 λ & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.2.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} - 6 & - 4 & 10 \\ 2 & \frac{4}{3} & - \frac{10}{3} \\ 6 & 4 & 10 \end{array}] + [\begin{array}{c} - λ & 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.3

Multiply $- λ \cdot 0$ .

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Step 4.1.2.3.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} - 6 & - 4 & 10 \\ 2 & \frac{4}{3} & - \frac{10}{3} \\ 6 & 4 & 10 \end{array}] + [\begin{array}{c} - λ & 0 & 0 λ \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.3.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} - 6 & - 4 & 10 \\ 2 & \frac{4}{3} & - \frac{10}{3} \\ 6 & 4 & 10 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.4

Multiply $- λ \cdot 0$ .

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Step 4.1.2.4.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} - 6 & - 4 & 10 \\ 2 & \frac{4}{3} & - \frac{10}{3} \\ 6 & 4 & 10 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 λ & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.4.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} - 6 & - 4 & 10 \\ 2 & \frac{4}{3} & - \frac{10}{3} \\ 6 & 4 & 10 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.5

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} - 6 & - 4 & 10 \\ 2 & \frac{4}{3} & - \frac{10}{3} \\ 6 & 4 & 10 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.6

Multiply $- λ \cdot 0$ .

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Step 4.1.2.6.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} - 6 & - 4 & 10 \\ 2 & \frac{4}{3} & - \frac{10}{3} \\ 6 & 4 & 10 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 λ \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.6.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} - 6 & - 4 & 10 \\ 2 & \frac{4}{3} & - \frac{10}{3} \\ 6 & 4 & 10 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.7

Multiply $- λ \cdot 0$ .

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Step 4.1.2.7.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} - 6 & - 4 & 10 \\ 2 & \frac{4}{3} & - \frac{10}{3} \\ 6 & 4 & 10 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 λ & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.7.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} - 6 & - 4 & 10 \\ 2 & \frac{4}{3} & - \frac{10}{3} \\ 6 & 4 & 10 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.8

Multiply $- λ \cdot 0$ .

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Step 4.1.2.8.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} - 6 & - 4 & 10 \\ 2 & \frac{4}{3} & - \frac{10}{3} \\ 6 & 4 & 10 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & 0 λ & - λ \cdot 1 \end{array}])$

Step 4.1.2.8.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} - 6 & - 4 & 10 \\ 2 & \frac{4}{3} & - \frac{10}{3} \\ 6 & 4 & 10 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.9

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} - 6 & - 4 & 10 \\ 2 & \frac{4}{3} & - \frac{10}{3} \\ 6 & 4 & 10 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & 0 & - λ \end{array}])$

Step 4.2

Add the corresponding elements.

$p (λ) = determinant [\begin{array}{c} - 6 - λ & - 4 + 0 & 10 + 0 \\ 2 + 0 & \frac{4}{3} - λ & - \frac{10}{3} + 0 \\ 6 + 0 & 4 + 0 & 10 - λ \end{array}]$

Step 4.3

Simplify each element.

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Step 4.3.1

Add $- 4$ and $0$ .

$p (λ) = determinant [\begin{array}{c} - 6 - λ & - 4 & 10 + 0 \\ 2 + 0 & \frac{4}{3} - λ & - \frac{10}{3} + 0 \\ 6 + 0 & 4 + 0 & 10 - λ \end{array}]$

Step 4.3.2

Add $10$ and $0$ .

$p (λ) = determinant [\begin{array}{c} - 6 - λ & - 4 & 10 \\ 2 + 0 & \frac{4}{3} - λ & - \frac{10}{3} + 0 \\ 6 + 0 & 4 + 0 & 10 - λ \end{array}]$

Step 4.3.3

Add $2$ and $0$ .

$p (λ) = determinant [\begin{array}{c} - 6 - λ & - 4 & 10 \\ 2 & \frac{4}{3} - λ & - \frac{10}{3} + 0 \\ 6 + 0 & 4 + 0 & 10 - λ \end{array}]$

Step 4.3.4

Add $- \frac{10}{3}$ and $0$ .

$p (λ) = determinant [\begin{array}{c} - 6 - λ & - 4 & 10 \\ 2 & \frac{4}{3} - λ & - \frac{10}{3} \\ 6 + 0 & 4 + 0 & 10 - λ \end{array}]$

Step 4.3.5

Add $6$ and $0$ .

$p (λ) = determinant [\begin{array}{c} - 6 - λ & - 4 & 10 \\ 2 & \frac{4}{3} - λ & - \frac{10}{3} \\ 6 & 4 + 0 & 10 - λ \end{array}]$

Step 4.3.6

Add $4$ and $0$ .

$p (λ) = determinant [\begin{array}{c} - 6 - λ & - 4 & 10 \\ 2 & \frac{4}{3} - λ & - \frac{10}{3} \\ 6 & 4 & 10 - λ \end{array}]$

Step 5

Find the determinant.

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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.

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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.

$| \begin{array}{c} \frac{4}{3} - λ & - \frac{10}{3} \\ 4 & 10 - λ \end{array} |$

Step 5.1.4

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

$(- 6 - λ) | \begin{array}{c} \frac{4}{3} - λ & - \frac{10}{3} \\ 4 & 10 - λ \end{array} |$

Step 5.1.5

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

$| \begin{array}{c} 2 & - \frac{10}{3} \\ 6 & 10 - λ \end{array} |$

Step 5.1.6

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

$4 | \begin{array}{c} 2 & - \frac{10}{3} \\ 6 & 10 - λ \end{array} |$

Step 5.1.7

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

$| \begin{array}{c} 2 & \frac{4}{3} - λ \\ 6 & 4 \end{array} |$

Step 5.1.8

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

$10 | \begin{array}{c} 2 & \frac{4}{3} - λ \\ 6 & 4 \end{array} |$

Step 5.1.9

Add the terms together.

$p (λ) = (- 6 - λ) | \begin{array}{c} \frac{4}{3} - λ & - \frac{10}{3} \\ 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 2 & - \frac{10}{3} \\ 6 & 10 - λ \end{array} | + 10 | \begin{array}{c} 2 & \frac{4}{3} - λ \\ 6 & 4 \end{array} |$

Step 5.2

Evaluate $| \begin{array}{c} \frac{4}{3} - λ & - \frac{10}{3} \\ 4 & 10 - λ \end{array} |$ .

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Step 5.2.1

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

$p (λ) = (- 6 - λ) ((\frac{4}{3} - λ) (10 - λ) - 4 (- \frac{10}{3})) + 4 | \begin{array}{c} 2 & - \frac{10}{3} \\ 6 & 10 - λ \end{array} | + 10 | \begin{array}{c} 2 & \frac{4}{3} - λ \\ 6 & 4 \end{array} |$

Step 5.2.2

Simplify the determinant.

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Step 5.2.2.1

Simplify each term.

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Step 5.2.2.1.1

Expand $(\frac{4}{3} - λ) (10 - λ)$ using the FOIL Method.

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Step 5.2.2.1.1.1

Apply the distributive property.

$p (λ) = (- 6 - λ) (\frac{4}{3} (10 - λ) - λ (10 - λ) - 4 (- \frac{10}{3})) + 4 | \begin{array}{c} 2 & - \frac{10}{3} \\ 6 & 10 - λ \end{array} | + 10 | \begin{array}{c} 2 & \frac{4}{3} - λ \\ 6 & 4 \end{array} |$

Step 5.2.2.1.1.2

Apply the distributive property.

$p (λ) = (- 6 - λ) (\frac{4}{3} \cdot 10 + \frac{4}{3} (- λ) - λ (10 - λ) - 4 (- \frac{10}{3})) + 4 | \begin{array}{c} 2 & - \frac{10}{3} \\ 6 & 10 - λ \end{array} | + 10 | \begin{array}{c} 2 & \frac{4}{3} - λ \\ 6 & 4 \end{array} |$

Step 5.2.2.1.1.3

Apply the distributive property.

$p (λ) = (- 6 - λ) (\frac{4}{3} \cdot 10 + \frac{4}{3} (- λ) - λ \cdot 10 - λ (- λ) - 4 (- \frac{10}{3})) + 4 | \begin{array}{c} 2 & - \frac{10}{3} \\ 6 & 10 - λ \end{array} | + 10 | \begin{array}{c} 2 & \frac{4}{3} - λ \\ 6 & 4 \end{array} |$

Step 5.2.2.1.2

Simplify and combine like terms.

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Step 5.2.2.1.2.1

Simplify each term.

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Step 5.2.2.1.2.1.1

Multiply $\frac{4}{3} \cdot 10$ .

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Step 5.2.2.1.2.1.1.1

Combine $\frac{4}{3}$ and $10$ .

$p (λ) = (- 6 - λ) (\frac{4 \cdot 10}{3} + \frac{4}{3} (- λ) - λ \cdot 10 - λ (- λ) - 4 (- \frac{10}{3})) + 4 | \begin{array}{c} 2 & - \frac{10}{3} \\ 6 & 10 - λ \end{array} | + 10 | \begin{array}{c} 2 & \frac{4}{3} - λ \\ 6 & 4 \end{array} |$

Step 5.2.2.1.2.1.1.2

Multiply $4$ by $10$ .

$p (λ) = (- 6 - λ) (\frac{40}{3} + \frac{4}{3} (- λ) - λ \cdot 10 - λ (- λ) - 4 (- \frac{10}{3})) + 4 | \begin{array}{c} 2 & - \frac{10}{3} \\ 6 & 10 - λ \end{array} | + 10 | \begin{array}{c} 2 & \frac{4}{3} - λ \\ 6 & 4 \end{array} |$

Step 5.2.2.1.2.1.2

Combine $\frac{4}{3}$ and $λ$ .

$p (λ) = (- 6 - λ) (\frac{40}{3} - \frac{4 λ}{3} - λ \cdot 10 - λ (- λ) - 4 (- \frac{10}{3})) + 4 | \begin{array}{c} 2 & - \frac{10}{3} \\ 6 & 10 - λ \end{array} | + 10 | \begin{array}{c} 2 & \frac{4}{3} - λ \\ 6 & 4 \end{array} |$

Step 5.2.2.1.2.1.3

Multiply $10$ by $- 1$ .

$p (λ) = (- 6 - λ) (\frac{40}{3} - \frac{4 λ}{3} - 10 λ - λ (- λ) - 4 (- \frac{10}{3})) + 4 | \begin{array}{c} 2 & - \frac{10}{3} \\ 6 & 10 - λ \end{array} | + 10 | \begin{array}{c} 2 & \frac{4}{3} - λ \\ 6 & 4 \end{array} |$

Step 5.2.2.1.2.1.4

Rewrite using the commutative property of multiplication.

$p (λ) = (- 6 - λ) (\frac{40}{3} - \frac{4 λ}{3} - 10 λ - 1 \cdot - 1 λ \cdot λ - 4 (- \frac{10}{3})) + 4 | \begin{array}{c} 2 & - \frac{10}{3} \\ 6 & 10 - λ \end{array} | + 10 | \begin{array}{c} 2 & \frac{4}{3} - λ \\ 6 & 4 \end{array} |$

Step 5.2.2.1.2.1.5

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

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Step 5.2.2.1.2.1.5.1

Move $λ$ .

$p (λ) = (- 6 - λ) (\frac{40}{3} - \frac{4 λ}{3} - 10 λ - 1 \cdot - 1 (λ \cdot λ) - 4 (- \frac{10}{3})) + 4 | \begin{array}{c} 2 & - \frac{10}{3} \\ 6 & 10 - λ \end{array} | + 10 | \begin{array}{c} 2 & \frac{4}{3} - λ \\ 6 & 4 \end{array} |$

Step 5.2.2.1.2.1.5.2

Multiply $λ$ by $λ$ .

$p (λ) = (- 6 - λ) (\frac{40}{3} - \frac{4 λ}{3} - 10 λ - 1 \cdot - 1 λ^{2} - 4 (- \frac{10}{3})) + 4 | \begin{array}{c} 2 & - \frac{10}{3} \\ 6 & 10 - λ \end{array} | + 10 | \begin{array}{c} 2 & \frac{4}{3} - λ \\ 6 & 4 \end{array} |$

Step 5.2.2.1.2.1.6

Multiply $- 1$ by $- 1$ .

$p (λ) = (- 6 - λ) (\frac{40}{3} - \frac{4 λ}{3} - 10 λ + 1 λ^{2} - 4 (- \frac{10}{3})) + 4 | \begin{array}{c} 2 & - \frac{10}{3} \\ 6 & 10 - λ \end{array} | + 10 | \begin{array}{c} 2 & \frac{4}{3} - λ \\ 6 & 4 \end{array} |$

Step 5.2.2.1.2.1.7

Multiply $λ^{2}$ by $1$ .

$p (λ) = (- 6 - λ) (\frac{40}{3} - \frac{4 λ}{3} - 10 λ + λ^{2} - 4 (- \frac{10}{3})) + 4 | \begin{array}{c} 2 & - \frac{10}{3} \\ 6 & 10 - λ \end{array} | + 10 | \begin{array}{c} 2 & \frac{4}{3} - λ \\ 6 & 4 \end{array} |$

Step 5.2.2.1.2.2

To write $- 10 λ$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$p (λ) = (- 6 - λ) (\frac{40}{3} - \frac{4 λ}{3} - 10 λ \cdot \frac{3}{3} + λ^{2} - 4 (- \frac{10}{3})) + 4 | \begin{array}{c} 2 & - \frac{10}{3} \\ 6 & 10 - λ \end{array} | + 10 | \begin{array}{c} 2 & \frac{4}{3} - λ \\ 6 & 4 \end{array} |$

Step 5.2.2.1.2.3

Combine $- 10 λ$ and $\frac{3}{3}$ .

$p (λ) = (- 6 - λ) (\frac{40}{3} - \frac{4 λ}{3} + \frac{- 10 λ \cdot 3}{3} + λ^{2} - 4 (- \frac{10}{3})) + 4 | \begin{array}{c} 2 & - \frac{10}{3} \\ 6 & 10 - λ \end{array} | + 10 | \begin{array}{c} 2 & \frac{4}{3} - λ \\ 6 & 4 \end{array} |$

Step 5.2.2.1.2.4

Combine the numerators over the common denominator.

$p (λ) = (- 6 - λ) (\frac{40}{3} + \frac{- 4 λ - 10 λ \cdot 3}{3} + λ^{2} - 4 (- \frac{10}{3})) + 4 | \begin{array}{c} 2 & - \frac{10}{3} \\ 6 & 10 - λ \end{array} | + 10 | \begin{array}{c} 2 & \frac{4}{3} - λ \\ 6 & 4 \end{array} |$

Step 5.2.2.1.2.5

Combine the numerators over the common denominator.

$p (λ) = (- 6 - λ) (\frac{40 - 4 λ - 10 λ \cdot 3}{3} + λ^{2} - 4 (- \frac{10}{3})) + 4 | \begin{array}{c} 2 & - \frac{10}{3} \\ 6 & 10 - λ \end{array} | + 10 | \begin{array}{c} 2 & \frac{4}{3} - λ \\ 6 & 4 \end{array} |$

Step 5.2.2.1.2.6

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

$p (λ) = (- 6 - λ) (\frac{40 - 4 λ - 10 λ \cdot 3}{3} + λ^{2} \cdot \frac{3}{3} - 4 (- \frac{10}{3})) + 4 | \begin{array}{c} 2 & - \frac{10}{3} \\ 6 & 10 - λ \end{array} | + 10 | \begin{array}{c} 2 & \frac{4}{3} - λ \\ 6 & 4 \end{array} |$

Step 5.2.2.1.2.7

Combine $λ^{2}$ and $\frac{3}{3}$ .

$p (λ) = (- 6 - λ) (\frac{40 - 4 λ - 10 λ \cdot 3}{3} + \frac{λ^{2} \cdot 3}{3} - 4 (- \frac{10}{3})) + 4 | \begin{array}{c} 2 & - \frac{10}{3} \\ 6 & 10 - λ \end{array} | + 10 | \begin{array}{c} 2 & \frac{4}{3} - λ \\ 6 & 4 \end{array} |$

Step 5.2.2.1.2.8

Combine the numerators over the common denominator.

$p (λ) = (- 6 - λ) (\frac{40 - 4 λ - 10 λ \cdot 3 + λ^{2} \cdot 3}{3} - 4 (- \frac{10}{3})) + 4 | \begin{array}{c} 2 & - \frac{10}{3} \\ 6 & 10 - λ \end{array} | + 10 | \begin{array}{c} 2 & \frac{4}{3} - λ \\ 6 & 4 \end{array} |$

Step 5.2.2.1.3

Simplify the numerator.

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Step 5.2.2.1.3.1

Multiply $3$ by $- 10$ .

$p (λ) = (- 6 - λ) (\frac{40 - 4 λ - 30 λ + λ^{2} \cdot 3}{3} - 4 (- \frac{10}{3})) + 4 | \begin{array}{c} 2 & - \frac{10}{3} \\ 6 & 10 - λ \end{array} | + 10 | \begin{array}{c} 2 & \frac{4}{3} - λ \\ 6 & 4 \end{array} |$

Step 5.2.2.1.3.2

Move $3$ to the left of $λ^{2}$ .

$p (λ) = (- 6 - λ) (\frac{40 - 4 λ - 30 λ + 3 \cdot λ^{2}}{3} - 4 (- \frac{10}{3})) + 4 | \begin{array}{c} 2 & - \frac{10}{3} \\ 6 & 10 - λ \end{array} | + 10 | \begin{array}{c} 2 & \frac{4}{3} - λ \\ 6 & 4 \end{array} |$

Step 5.2.2.1.3.3

Subtract $30 λ$ from $- 4 λ$ .

$p (λ) = (- 6 - λ) (\frac{40 - 34 λ + 3 λ^{2}}{3} - 4 (- \frac{10}{3})) + 4 | \begin{array}{c} 2 & - \frac{10}{3} \\ 6 & 10 - λ \end{array} | + 10 | \begin{array}{c} 2 & \frac{4}{3} - λ \\ 6 & 4 \end{array} |$

Step 5.2.2.1.3.4

Factor by grouping.

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Step 5.2.2.1.3.4.1

Reorder terms.

$p (λ) = (- 6 - λ) (\frac{3 λ^{2} - 34 λ + 40}{3} - 4 (- \frac{10}{3})) + 4 | \begin{array}{c} 2 & - \frac{10}{3} \\ 6 & 10 - λ \end{array} | + 10 | \begin{array}{c} 2 & \frac{4}{3} - λ \\ 6 & 4 \end{array} |$

Step 5.2.2.1.3.4.2

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 $a \cdot c = 3 \cdot 40 = 120$ and whose sum is $b = - 34$ .

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Step 5.2.2.1.3.4.2.1

Factor $- 34$ out of $- 34 λ$ .

$p (λ) = (- 6 - λ) (\frac{3 λ^{2} - 34 (λ) + 40}{3} - 4 (- \frac{10}{3})) + 4 | \begin{array}{c} 2 & - \frac{10}{3} \\ 6 & 10 - λ \end{array} | + 10 | \begin{array}{c} 2 & \frac{4}{3} - λ \\ 6 & 4 \end{array} |$

Step 5.2.2.1.3.4.2.2

Rewrite $- 34$ as $- 4$ plus $- 30$

$p (λ) = (- 6 - λ) (\frac{3 λ^{2} + (- 4 - 30) λ + 40}{3} - 4 (- \frac{10}{3})) + 4 | \begin{array}{c} 2 & - \frac{10}{3} \\ 6 & 10 - λ \end{array} | + 10 | \begin{array}{c} 2 & \frac{4}{3} - λ \\ 6 & 4 \end{array} |$

Step 5.2.2.1.3.4.2.3

Apply the distributive property.

$p (λ) = (- 6 - λ) (\frac{3 λ^{2} - 4 λ - 30 λ + 40}{3} - 4 (- \frac{10}{3})) + 4 | \begin{array}{c} 2 & - \frac{10}{3} \\ 6 & 10 - λ \end{array} | + 10 | \begin{array}{c} 2 & \frac{4}{3} - λ \\ 6 & 4 \end{array} |$

Step 5.2.2.1.3.4.3

Factor out the greatest common factor from each group.

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Step 5.2.2.1.3.4.3.1

Group the first two terms and the last two terms.

$p (λ) = (- 6 - λ) (\frac{(3 λ^{2} - 4 λ) - 30 λ + 40}{3} - 4 (- \frac{10}{3})) + 4 | \begin{array}{c} 2 & - \frac{10}{3} \\ 6 & 10 - λ \end{array} | + 10 | \begin{array}{c} 2 & \frac{4}{3} - λ \\ 6 & 4 \end{array} |$

Step 5.2.2.1.3.4.3.2

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

$p (λ) = (- 6 - λ) (\frac{λ (3 λ - 4) - 10 (3 λ - 4)}{3} - 4 (- \frac{10}{3})) + 4 | \begin{array}{c} 2 & - \frac{10}{3} \\ 6 & 10 - λ \end{array} | + 10 | \begin{array}{c} 2 & \frac{4}{3} - λ \\ 6 & 4 \end{array} |$

Step 5.2.2.1.3.4.4

Factor the polynomial by factoring out the greatest common factor, $3 λ - 4$ .

$p (λ) = (- 6 - λ) (\frac{(3 λ - 4) (λ - 10)}{3} - 4 (- \frac{10}{3})) + 4 | \begin{array}{c} 2 & - \frac{10}{3} \\ 6 & 10 - λ \end{array} | + 10 | \begin{array}{c} 2 & \frac{4}{3} - λ \\ 6 & 4 \end{array} |$

Step 5.2.2.1.4

Multiply $- 4 (- \frac{10}{3})$ .

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Step 5.2.2.1.4.1

Multiply $- 1$ by $- 4$ .

$p (λ) = (- 6 - λ) (\frac{(3 λ - 4) (λ - 10)}{3} + 4 (\frac{10}{3})) + 4 | \begin{array}{c} 2 & - \frac{10}{3} \\ 6 & 10 - λ \end{array} | + 10 | \begin{array}{c} 2 & \frac{4}{3} - λ \\ 6 & 4 \end{array} |$

Step 5.2.2.1.4.2

Combine $4$ and $\frac{10}{3}$ .

$p (λ) = (- 6 - λ) (\frac{(3 λ - 4) (λ - 10)}{3} + \frac{4 \cdot 10}{3}) + 4 | \begin{array}{c} 2 & - \frac{10}{3} \\ 6 & 10 - λ \end{array} | + 10 | \begin{array}{c} 2 & \frac{4}{3} - λ \\ 6 & 4 \end{array} |$

Step 5.2.2.1.4.3

Multiply $4$ by $10$ .

$p (λ) = (- 6 - λ) (\frac{(3 λ - 4) (λ - 10)}{3} + \frac{40}{3}) + 4 | \begin{array}{c} 2 & - \frac{10}{3} \\ 6 & 10 - λ \end{array} | + 10 | \begin{array}{c} 2 & \frac{4}{3} - λ \\ 6 & 4 \end{array} |$

Step 5.2.2.2

Combine the numerators over the common denominator.

$p (λ) = (- 6 - λ) \frac{(3 λ - 4) (λ - 10) + 40}{3} + 4 | \begin{array}{c} 2 & - \frac{10}{3} \\ 6 & 10 - λ \end{array} | + 10 | \begin{array}{c} 2 & \frac{4}{3} - λ \\ 6 & 4 \end{array} |$

Step 5.2.2.3

Simplify each term.

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Step 5.2.2.3.1

Expand $(3 λ - 4) (λ - 10)$ using the FOIL Method.

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Step 5.2.2.3.1.1

Apply the distributive property.

$p (λ) = (- 6 - λ) \frac{3 λ (λ - 10) - 4 (λ - 10) + 40}{3} + 4 | \begin{array}{c} 2 & - \frac{10}{3} \\ 6 & 10 - λ \end{array} | + 10 | \begin{array}{c} 2 & \frac{4}{3} - λ \\ 6 & 4 \end{array} |$

Step 5.2.2.3.1.2

Apply the distributive property.

$p (λ) = (- 6 - λ) \frac{3 λ \cdot λ + 3 λ \cdot - 10 - 4 (λ - 10) + 40}{3} + 4 | \begin{array}{c} 2 & - \frac{10}{3} \\ 6 & 10 - λ \end{array} | + 10 | \begin{array}{c} 2 & \frac{4}{3} - λ \\ 6 & 4 \end{array} |$

Step 5.2.2.3.1.3

Apply the distributive property.

$p (λ) = (- 6 - λ) \frac{3 λ \cdot λ + 3 λ \cdot - 10 - 4 λ - 4 \cdot - 10 + 40}{3} + 4 | \begin{array}{c} 2 & - \frac{10}{3} \\ 6 & 10 - λ \end{array} | + 10 | \begin{array}{c} 2 & \frac{4}{3} - λ \\ 6 & 4 \end{array} |$

Step 5.2.2.3.2

Simplify and combine like terms.

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Step 5.2.2.3.2.1

Simplify each term.

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Step 5.2.2.3.2.1.1

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

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Step 5.2.2.3.2.1.1.1

Move $λ$ .

$p (λ) = (- 6 - λ) \frac{3 (λ \cdot λ) + 3 λ \cdot - 10 - 4 λ - 4 \cdot - 10 + 40}{3} + 4 | \begin{array}{c} 2 & - \frac{10}{3} \\ 6 & 10 - λ \end{array} | + 10 | \begin{array}{c} 2 & \frac{4}{3} - λ \\ 6 & 4 \end{array} |$

Step 5.2.2.3.2.1.1.2

Multiply $λ$ by $λ$ .

$p (λ) = (- 6 - λ) \frac{3 λ^{2} + 3 λ \cdot - 10 - 4 λ - 4 \cdot - 10 + 40}{3} + 4 | \begin{array}{c} 2 & - \frac{10}{3} \\ 6 & 10 - λ \end{array} | + 10 | \begin{array}{c} 2 & \frac{4}{3} - λ \\ 6 & 4 \end{array} |$

Step 5.2.2.3.2.1.2

Multiply $- 10$ by $3$ .

$p (λ) = (- 6 - λ) \frac{3 λ^{2} - 30 λ - 4 λ - 4 \cdot - 10 + 40}{3} + 4 | \begin{array}{c} 2 & - \frac{10}{3} \\ 6 & 10 - λ \end{array} | + 10 | \begin{array}{c} 2 & \frac{4}{3} - λ \\ 6 & 4 \end{array} |$

Step 5.2.2.3.2.1.3

Multiply $- 4$ by $- 10$ .

$p (λ) = (- 6 - λ) \frac{3 λ^{2} - 30 λ - 4 λ + 40 + 40}{3} + 4 | \begin{array}{c} 2 & - \frac{10}{3} \\ 6 & 10 - λ \end{array} | + 10 | \begin{array}{c} 2 & \frac{4}{3} - λ \\ 6 & 4 \end{array} |$

Step 5.2.2.3.2.2

Subtract $4 λ$ from $- 30 λ$ .

$p (λ) = (- 6 - λ) \frac{3 λ^{2} - 34 λ + 40 + 40}{3} + 4 | \begin{array}{c} 2 & - \frac{10}{3} \\ 6 & 10 - λ \end{array} | + 10 | \begin{array}{c} 2 & \frac{4}{3} - λ \\ 6 & 4 \end{array} |$

Step 5.2.2.4

Add $40$ and $40$ .

$p (λ) = (- 6 - λ) \frac{3 λ^{2} - 34 λ + 80}{3} + 4 | \begin{array}{c} 2 & - \frac{10}{3} \\ 6 & 10 - λ \end{array} | + 10 | \begin{array}{c} 2 & \frac{4}{3} - λ \\ 6 & 4 \end{array} |$

Step 5.2.2.5

Factor by grouping.

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Step 5.2.2.5.1

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 $a \cdot c = 3 \cdot 80 = 240$ and whose sum is $b = - 34$ .

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Step 5.2.2.5.1.1

Factor $- 34$ out of $- 34 λ$ .

$p (λ) = (- 6 - λ) \frac{3 λ^{2} - 34 (λ) + 80}{3} + 4 | \begin{array}{c} 2 & - \frac{10}{3} \\ 6 & 10 - λ \end{array} | + 10 | \begin{array}{c} 2 & \frac{4}{3} - λ \\ 6 & 4 \end{array} |$

Step 5.2.2.5.1.2

Rewrite $- 34$ as $- 10$ plus $- 24$

$p (λ) = (- 6 - λ) \frac{3 λ^{2} + (- 10 - 24) λ + 80}{3} + 4 | \begin{array}{c} 2 & - \frac{10}{3} \\ 6 & 10 - λ \end{array} | + 10 | \begin{array}{c} 2 & \frac{4}{3} - λ \\ 6 & 4 \end{array} |$

Step 5.2.2.5.1.3

Apply the distributive property.

$p (λ) = (- 6 - λ) \frac{3 λ^{2} - 10 λ - 24 λ + 80}{3} + 4 | \begin{array}{c} 2 & - \frac{10}{3} \\ 6 & 10 - λ \end{array} | + 10 | \begin{array}{c} 2 & \frac{4}{3} - λ \\ 6 & 4 \end{array} |$

Step 5.2.2.5.2

Factor out the greatest common factor from each group.

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Step 5.2.2.5.2.1

Group the first two terms and the last two terms.

$p (λ) = (- 6 - λ) \frac{(3 λ^{2} - 10 λ) - 24 λ + 80}{3} + 4 | \begin{array}{c} 2 & - \frac{10}{3} \\ 6 & 10 - λ \end{array} | + 10 | \begin{array}{c} 2 & \frac{4}{3} - λ \\ 6 & 4 \end{array} |$

Step 5.2.2.5.2.2

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

$p (λ) = (- 6 - λ) \frac{λ (3 λ - 10) - 8 (3 λ - 10)}{3} + 4 | \begin{array}{c} 2 & - \frac{10}{3} \\ 6 & 10 - λ \end{array} | + 10 | \begin{array}{c} 2 & \frac{4}{3} - λ \\ 6 & 4 \end{array} |$

Step 5.2.2.5.3

Factor the polynomial by factoring out the greatest common factor, $3 λ - 10$ .

$p (λ) = (- 6 - λ) \frac{(3 λ - 10) (λ - 8)}{3} + 4 | \begin{array}{c} 2 & - \frac{10}{3} \\ 6 & 10 - λ \end{array} | + 10 | \begin{array}{c} 2 & \frac{4}{3} - λ \\ 6 & 4 \end{array} |$

Step 5.3

Evaluate $| \begin{array}{c} 2 & - \frac{10}{3} \\ 6 & 10 - λ \end{array} |$ .

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Step 5.3.1

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

$p (λ) = (- 6 - λ) \frac{(3 λ - 10) (λ - 8)}{3} + 4 (2 (10 - λ) - 6 (- \frac{10}{3})) + 10 | \begin{array}{c} 2 & \frac{4}{3} - λ \\ 6 & 4 \end{array} |$

Step 5.3.2

Simplify the determinant.

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Step 5.3.2.1

Simplify each term.

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Step 5.3.2.1.1

Apply the distributive property.

$p (λ) = (- 6 - λ) \frac{(3 λ - 10) (λ - 8)}{3} + 4 (2 \cdot 10 + 2 (- λ) - 6 (- \frac{10}{3})) + 10 | \begin{array}{c} 2 & \frac{4}{3} - λ \\ 6 & 4 \end{array} |$

Step 5.3.2.1.2

Multiply $2$ by $10$ .

$p (λ) = (- 6 - λ) \frac{(3 λ - 10) (λ - 8)}{3} + 4 (20 + 2 (- λ) - 6 (- \frac{10}{3})) + 10 | \begin{array}{c} 2 & \frac{4}{3} - λ \\ 6 & 4 \end{array} |$

Step 5.3.2.1.3

Multiply $- 1$ by $2$ .

$p (λ) = (- 6 - λ) \frac{(3 λ - 10) (λ - 8)}{3} + 4 (20 - 2 λ - 6 (- \frac{10}{3})) + 10 | \begin{array}{c} 2 & \frac{4}{3} - λ \\ 6 & 4 \end{array} |$

Step 5.3.2.1.4

Cancel the common factor of $3$ .

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Step 5.3.2.1.4.1

Move the leading negative in $- \frac{10}{3}$ into the numerator.

$p (λ) = (- 6 - λ) \frac{(3 λ - 10) (λ - 8)}{3} + 4 (20 - 2 λ - 6 (\frac{- 10}{3})) + 10 | \begin{array}{c} 2 & \frac{4}{3} - λ \\ 6 & 4 \end{array} |$

Step 5.3.2.1.4.2

Factor $3$ out of $- 6$ .

$p (λ) = (- 6 - λ) \frac{(3 λ - 10) (λ - 8)}{3} + 4 (20 - 2 λ + 3 (- 2) \frac{- 10}{3}) + 10 | \begin{array}{c} 2 & \frac{4}{3} - λ \\ 6 & 4 \end{array} |$

Step 5.3.2.1.4.3

Cancel the common factor.

$p (λ) = (- 6 - λ) \frac{(3 λ - 10) (λ - 8)}{3} + 4 (20 - 2 λ + 3 \cdot - 2 \frac{- 10}{3}) + 10 | \begin{array}{c} 2 & \frac{4}{3} - λ \\ 6 & 4 \end{array} |$

Step 5.3.2.1.4.4

Rewrite the expression.

$p (λ) = (- 6 - λ) \frac{(3 λ - 10) (λ - 8)}{3} + 4 (20 - 2 λ - 2 \cdot - 10) + 10 | \begin{array}{c} 2 & \frac{4}{3} - λ \\ 6 & 4 \end{array} |$

Step 5.3.2.1.5

Multiply $- 2$ by $- 10$ .

$p (λ) = (- 6 - λ) \frac{(3 λ - 10) (λ - 8)}{3} + 4 (20 - 2 λ + 20) + 10 | \begin{array}{c} 2 & \frac{4}{3} - λ \\ 6 & 4 \end{array} |$

Step 5.3.2.2

Add $20$ and $20$ .

$p (λ) = (- 6 - λ) \frac{(3 λ - 10) (λ - 8)}{3} + 4 (- 2 λ + 40) + 10 | \begin{array}{c} 2 & \frac{4}{3} - λ \\ 6 & 4 \end{array} |$

Step 5.4

Evaluate $| \begin{array}{c} 2 & \frac{4}{3} - λ \\ 6 & 4 \end{array} |$ .

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Step 5.4.1

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

$p (λ) = (- 6 - λ) \frac{(3 λ - 10) (λ - 8)}{3} + 4 (- 2 λ + 40) + 10 (2 \cdot 4 - 6 (\frac{4}{3} - λ))$

Step 5.4.2

Simplify the determinant.

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Step 5.4.2.1

Simplify each term.

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Step 5.4.2.1.1

Multiply $2$ by $4$ .

$p (λ) = (- 6 - λ) \frac{(3 λ - 10) (λ - 8)}{3} + 4 (- 2 λ + 40) + 10 (8 - 6 (\frac{4}{3} - λ))$

Step 5.4.2.1.2

Apply the distributive property.

$p (λ) = (- 6 - λ) \frac{(3 λ - 10) (λ - 8)}{3} + 4 (- 2 λ + 40) + 10 (8 - 6 (\frac{4}{3}) - 6 (- λ))$

Step 5.4.2.1.3

Cancel the common factor of $3$ .

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Step 5.4.2.1.3.1

Factor $3$ out of $- 6$ .

$p (λ) = (- 6 - λ) \frac{(3 λ - 10) (λ - 8)}{3} + 4 (- 2 λ + 40) + 10 (8 + 3 (- 2) \frac{4}{3} - 6 (- λ))$

Step 5.4.2.1.3.2

Cancel the common factor.

$p (λ) = (- 6 - λ) \frac{(3 λ - 10) (λ - 8)}{3} + 4 (- 2 λ + 40) + 10 (8 + 3 \cdot - 2 \frac{4}{3} - 6 (- λ))$

Step 5.4.2.1.3.3

Rewrite the expression.

$p (λ) = (- 6 - λ) \frac{(3 λ - 10) (λ - 8)}{3} + 4 (- 2 λ + 40) + 10 (8 - 2 \cdot 4 - 6 (- λ))$

Step 5.4.2.1.4

Multiply $- 2$ by $4$ .

$p (λ) = (- 6 - λ) \frac{(3 λ - 10) (λ - 8)}{3} + 4 (- 2 λ + 40) + 10 (8 - 8 - 6 (- λ))$

Step 5.4.2.1.5

Multiply $- 1$ by $- 6$ .

$p (λ) = (- 6 - λ) \frac{(3 λ - 10) (λ - 8)}{3} + 4 (- 2 λ + 40) + 10 (8 - 8 + 6 λ)$

Step 5.4.2.2

Subtract $8$ from $8$ .

$p (λ) = (- 6 - λ) \frac{(3 λ - 10) (λ - 8)}{3} + 4 (- 2 λ + 40) + 10 (0 + 6 λ)$

Step 5.4.2.3

Add $0$ and $6 λ$ .

$p (λ) = (- 6 - λ) \frac{(3 λ - 10) (λ - 8)}{3} + 4 (- 2 λ + 40) + 10 (6 λ)$

Step 5.5

Simplify the determinant.

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Step 5.5.1

Simplify each term.

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Step 5.5.1.1

Apply the distributive property.

$p (λ) = - 6 \frac{(3 λ - 10) (λ - 8)}{3} - λ \frac{(3 λ - 10) (λ - 8)}{3} + 4 (- 2 λ + 40) + 10 (6 λ)$

Step 5.5.1.2

Cancel the common factor of $3$ .

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Step 5.5.1.2.1

Factor $3$ out of $- 6$ .

$p (λ) = 3 (- 2) \frac{(3 λ - 10) (λ - 8)}{3} - λ \frac{(3 λ - 10) (λ - 8)}{3} + 4 (- 2 λ + 40) + 10 (6 λ)$

Step 5.5.1.2.2

Cancel the common factor.

$p (λ) = 3 \cdot - 2 \frac{(3 λ - 10) (λ - 8)}{3} - λ \frac{(3 λ - 10) (λ - 8)}{3} + 4 (- 2 λ + 40) + 10 (6 λ)$

Step 5.5.1.2.3

Rewrite the expression.

$p (λ) = - 2 ((3 λ - 10) (λ - 8)) - λ \frac{(3 λ - 10) (λ - 8)}{3} + 4 (- 2 λ + 40) + 10 (6 λ)$

Step 5.5.1.3

Combine $\frac{(3 λ - 10) (λ - 8)}{3}$ and $λ$ .

$p (λ) = - 2 ((3 λ - 10) (λ - 8)) - \frac{(3 λ - 10) (λ - 8) λ}{3} + 4 (- 2 λ + 40) + 10 (6 λ)$

Step 5.5.1.4

Simplify each term.

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Step 5.5.1.4.1

Expand $(3 λ - 10) (λ - 8)$ using the FOIL Method.

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Step 5.5.1.4.1.1

Apply the distributive property.

$p (λ) = - 2 (3 λ (λ - 8) - 10 (λ - 8)) - \frac{(3 λ - 10) (λ - 8) λ}{3} + 4 (- 2 λ + 40) + 10 (6 λ)$

Step 5.5.1.4.1.2

Apply the distributive property.

$p (λ) = - 2 (3 λ \cdot λ + 3 λ \cdot - 8 - 10 (λ - 8)) - \frac{(3 λ - 10) (λ - 8) λ}{3} + 4 (- 2 λ + 40) + 10 (6 λ)$

Step 5.5.1.4.1.3

Apply the distributive property.

$p (λ) = - 2 (3 λ \cdot λ + 3 λ \cdot - 8 - 10 λ - 10 \cdot - 8) - \frac{(3 λ - 10) (λ - 8) λ}{3} + 4 (- 2 λ + 40) + 10 (6 λ)$

Step 5.5.1.4.2

Simplify and combine like terms.

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Step 5.5.1.4.2.1

Simplify each term.

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Step 5.5.1.4.2.1.1

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

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Step 5.5.1.4.2.1.1.1

Move $λ$ .

$p (λ) = - 2 (3 (λ \cdot λ) + 3 λ \cdot - 8 - 10 λ - 10 \cdot - 8) - \frac{(3 λ - 10) (λ - 8) λ}{3} + 4 (- 2 λ + 40) + 10 (6 λ)$

Step 5.5.1.4.2.1.1.2

Multiply $λ$ by $λ$ .

$p (λ) = - 2 (3 λ^{2} + 3 λ \cdot - 8 - 10 λ - 10 \cdot - 8) - \frac{(3 λ - 10) (λ - 8) λ}{3} + 4 (- 2 λ + 40) + 10 (6 λ)$

Step 5.5.1.4.2.1.2

Multiply $- 8$ by $3$ .

$p (λ) = - 2 (3 λ^{2} - 24 λ - 10 λ - 10 \cdot - 8) - \frac{(3 λ - 10) (λ - 8) λ}{3} + 4 (- 2 λ + 40) + 10 (6 λ)$

Step 5.5.1.4.2.1.3

Multiply $- 10$ by $- 8$ .

$p (λ) = - 2 (3 λ^{2} - 24 λ - 10 λ + 80) - \frac{(3 λ - 10) (λ - 8) λ}{3} + 4 (- 2 λ + 40) + 10 (6 λ)$

Step 5.5.1.4.2.2

Subtract $10 λ$ from $- 24 λ$ .

$p (λ) = - 2 (3 λ^{2} - 34 λ + 80) - \frac{(3 λ - 10) (λ - 8) λ}{3} + 4 (- 2 λ + 40) + 10 (6 λ)$

Step 5.5.1.4.3

Apply the distributive property.

$p (λ) = - 2 (3 λ^{2}) - 2 (- 34 λ) - 2 \cdot 80 - \frac{(3 λ - 10) (λ - 8) λ}{3} + 4 (- 2 λ + 40) + 10 (6 λ)$

Step 5.5.1.4.4

Simplify.

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Step 5.5.1.4.4.1

Multiply $3$ by $- 2$ .

$p (λ) = - 6 λ^{2} - 2 (- 34 λ) - 2 \cdot 80 - \frac{(3 λ - 10) (λ - 8) λ}{3} + 4 (- 2 λ + 40) + 10 (6 λ)$

Step 5.5.1.4.4.2

Multiply $- 34$ by $- 2$ .

$p (λ) = - 6 λ^{2} + 68 λ - 2 \cdot 80 - \frac{(3 λ - 10) (λ - 8) λ}{3} + 4 (- 2 λ + 40) + 10 (6 λ)$

Step 5.5.1.4.4.3

Multiply $- 2$ by $80$ .

$p (λ) = - 6 λ^{2} + 68 λ - 160 - \frac{(3 λ - 10) (λ - 8) λ}{3} + 4 (- 2 λ + 40) + 10 (6 λ)$

Step 5.5.1.5

To write $- 6 λ^{2}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$p (λ) = 68 λ - 160 - 6 λ^{2} \cdot \frac{3}{3} - \frac{(3 λ - 10) (λ - 8) λ}{3} + 4 (- 2 λ + 40) + 10 (6 λ)$

Step 5.5.1.6

Combine $- 6 λ^{2}$ and $\frac{3}{3}$ .

$p (λ) = 68 λ - 160 + \frac{- 6 λ^{2} \cdot 3}{3} - \frac{(3 λ - 10) (λ - 8) λ}{3} + 4 (- 2 λ + 40) + 10 (6 λ)$

Step 5.5.1.7

Combine the numerators over the common denominator.

$p (λ) = 68 λ - 160 + \frac{- 6 λ^{2} \cdot 3 - (3 λ - 10) (λ - 8) λ}{3} + 4 (- 2 λ + 40) + 10 (6 λ)$

Step 5.5.1.8

Simplify the numerator.

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Step 5.5.1.8.1

Factor $λ$ out of $- 6 λ^{2} \cdot 3 - (3 λ - 10) (λ - 8) λ$ .

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Step 5.5.1.8.1.1

Factor $λ$ out of $- 6 λ^{2} \cdot 3$ .

$p (λ) = 68 λ - 160 + \frac{λ (- 6 λ \cdot 3) - (3 λ - 10) (λ - 8) λ}{3} + 4 (- 2 λ + 40) + 10 (6 λ)$

Step 5.5.1.8.1.2

Factor $λ$ out of $- (3 λ - 10) (λ - 8) λ$ .

$p (λ) = 68 λ - 160 + \frac{λ (- 6 λ \cdot 3) + λ (- (3 λ - 10) (λ - 8))}{3} + 4 (- 2 λ + 40) + 10 (6 λ)$

Step 5.5.1.8.1.3

Factor $λ$ out of $λ (- 6 λ \cdot 3) + λ (- (3 λ - 10) (λ - 8))$ .

$p (λ) = 68 λ - 160 + \frac{λ (- 6 λ \cdot 3 - (3 λ - 10) (λ - 8))}{3} + 4 (- 2 λ + 40) + 10 (6 λ)$

Step 5.5.1.8.2

Multiply $3$ by $- 6$ .

$p (λ) = 68 λ - 160 + \frac{λ (- 18 λ - (3 λ - 10) (λ - 8))}{3} + 4 (- 2 λ + 40) + 10 (6 λ)$

Step 5.5.1.8.3

Apply the distributive property.

$p (λ) = 68 λ - 160 + \frac{λ (- 18 λ + (- (3 λ) - - 10) (λ - 8))}{3} + 4 (- 2 λ + 40) + 10 (6 λ)$

Step 5.5.1.8.4

Multiply $3$ by $- 1$ .

$p (λ) = 68 λ - 160 + \frac{λ (- 18 λ + (- 3 λ - - 10) (λ - 8))}{3} + 4 (- 2 λ + 40) + 10 (6 λ)$

Step 5.5.1.8.5

Multiply $- 1$ by $- 10$ .

$p (λ) = 68 λ - 160 + \frac{λ (- 18 λ + (- 3 λ + 10) (λ - 8))}{3} + 4 (- 2 λ + 40) + 10 (6 λ)$

Step 5.5.1.8.6

Expand $(- 3 λ + 10) (λ - 8)$ using the FOIL Method.

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Step 5.5.1.8.6.1

Apply the distributive property.

$p (λ) = 68 λ - 160 + \frac{λ (- 18 λ - 3 λ (λ - 8) + 10 (λ - 8))}{3} + 4 (- 2 λ + 40) + 10 (6 λ)$

Step 5.5.1.8.6.2

Apply the distributive property.

$p (λ) = 68 λ - 160 + \frac{λ (- 18 λ - 3 λ \cdot λ - 3 λ \cdot - 8 + 10 (λ - 8))}{3} + 4 (- 2 λ + 40) + 10 (6 λ)$

Step 5.5.1.8.6.3

Apply the distributive property.

$p (λ) = 68 λ - 160 + \frac{λ (- 18 λ - 3 λ \cdot λ - 3 λ \cdot - 8 + 10 λ + 10 \cdot - 8)}{3} + 4 (- 2 λ + 40) + 10 (6 λ)$

Step 5.5.1.8.7

Simplify and combine like terms.

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Step 5.5.1.8.7.1

Simplify each term.

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Step 5.5.1.8.7.1.1

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

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Step 5.5.1.8.7.1.1.1

Move $λ$ .

$p (λ) = 68 λ - 160 + \frac{λ (- 18 λ - 3 (λ \cdot λ) - 3 λ \cdot - 8 + 10 λ + 10 \cdot - 8)}{3} + 4 (- 2 λ + 40) + 10 (6 λ)$

Step 5.5.1.8.7.1.1.2

Multiply $λ$ by $λ$ .

$p (λ) = 68 λ - 160 + \frac{λ (- 18 λ - 3 λ^{2} - 3 λ \cdot - 8 + 10 λ + 10 \cdot - 8)}{3} + 4 (- 2 λ + 40) + 10 (6 λ)$

Step 5.5.1.8.7.1.2

Multiply $- 8$ by $- 3$ .

$p (λ) = 68 λ - 160 + \frac{λ (- 18 λ - 3 λ^{2} + 24 λ + 10 λ + 10 \cdot - 8)}{3} + 4 (- 2 λ + 40) + 10 (6 λ)$

Step 5.5.1.8.7.1.3

Multiply $10$ by $- 8$ .

$p (λ) = 68 λ - 160 + \frac{λ (- 18 λ - 3 λ^{2} + 24 λ + 10 λ - 80)}{3} + 4 (- 2 λ + 40) + 10 (6 λ)$

Step 5.5.1.8.7.2

Add $24 λ$ and $10 λ$ .

$p (λ) = 68 λ - 160 + \frac{λ (- 18 λ - 3 λ^{2} + 34 λ - 80)}{3} + 4 (- 2 λ + 40) + 10 (6 λ)$

Step 5.5.1.8.8

Add $- 18 λ$ and $34 λ$ .

$p (λ) = 68 λ - 160 + \frac{λ (- 3 λ^{2} + 16 λ - 80)}{3} + 4 (- 2 λ + 40) + 10 (6 λ)$

Step 5.5.1.9

To write $68 λ$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$p (λ) = 68 λ \cdot \frac{3}{3} + \frac{λ (- 3 λ^{2} + 16 λ - 80)}{3} - 160 + 4 (- 2 λ + 40) + 10 (6 λ)$

Step 5.5.1.10

Combine $68 λ$ and $\frac{3}{3}$ .

$p (λ) = \frac{68 λ \cdot 3}{3} + \frac{λ (- 3 λ^{2} + 16 λ - 80)}{3} - 160 + 4 (- 2 λ + 40) + 10 (6 λ)$

Step 5.5.1.11

Combine the numerators over the common denominator.

$p (λ) = \frac{68 λ \cdot 3 + λ (- 3 λ^{2} + 16 λ - 80)}{3} - 160 + 4 (- 2 λ + 40) + 10 (6 λ)$

Step 5.5.1.12

Simplify the numerator.

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Step 5.5.1.12.1

Factor $λ$ out of $68 λ \cdot 3 + λ (- 3 λ^{2} + 16 λ - 80)$ .

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Step 5.5.1.12.1.1

Factor $λ$ out of $68 λ \cdot 3$ .

$p (λ) = \frac{λ (68 \cdot 3) + λ (- 3 λ^{2} + 16 λ - 80)}{3} - 160 + 4 (- 2 λ + 40) + 10 (6 λ)$

Step 5.5.1.12.1.2

Factor $λ$ out of $λ (68 \cdot 3) + λ (- 3 λ^{2} + 16 λ - 80)$ .

$p (λ) = \frac{λ (68 \cdot 3 - 3 λ^{2} + 16 λ - 80)}{3} - 160 + 4 (- 2 λ + 40) + 10 (6 λ)$

Step 5.5.1.12.2

Multiply $68$ by $3$ .

$p (λ) = \frac{λ (204 - 3 λ^{2} + 16 λ - 80)}{3} - 160 + 4 (- 2 λ + 40) + 10 (6 λ)$

Step 5.5.1.12.3

Subtract $80$ from $204$ .

$p (λ) = \frac{λ (- 3 λ^{2} + 16 λ + 124)}{3} - 160 + 4 (- 2 λ + 40) + 10 (6 λ)$

Step 5.5.1.13

To write $- 160$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$p (λ) = \frac{λ (- 3 λ^{2} + 16 λ + 124)}{3} - 160 \cdot \frac{3}{3} + 4 (- 2 λ + 40) + 10 (6 λ)$

Step 5.5.1.14

Combine $- 160$ and $\frac{3}{3}$ .

$p (λ) = \frac{λ (- 3 λ^{2} + 16 λ + 124)}{3} + \frac{- 160 \cdot 3}{3} + 4 (- 2 λ + 40) + 10 (6 λ)$

Step 5.5.1.15

Combine the numerators over the common denominator.

$p (λ) = \frac{λ (- 3 λ^{2} + 16 λ + 124) - 160 \cdot 3}{3} + 4 (- 2 λ + 40) + 10 (6 λ)$

Step 5.5.1.16

Simplify the numerator.

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Step 5.5.1.16.1

Apply the distributive property.

$p (λ) = \frac{λ (- 3 λ^{2}) + λ (16 λ) + λ \cdot 124 - 160 \cdot 3}{3} + 4 (- 2 λ + 40) + 10 (6 λ)$

Step 5.5.1.16.2

Simplify.

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Step 5.5.1.16.2.1

Rewrite using the commutative property of multiplication.

$p (λ) = \frac{- 3 λ \cdot λ^{2} + λ (16 λ) + λ \cdot 124 - 160 \cdot 3}{3} + 4 (- 2 λ + 40) + 10 (6 λ)$

Step 5.5.1.16.2.2

Rewrite using the commutative property of multiplication.

$p (λ) = \frac{- 3 λ \cdot λ^{2} + 16 λ \cdot λ + λ \cdot 124 - 160 \cdot 3}{3} + 4 (- 2 λ + 40) + 10 (6 λ)$

Step 5.5.1.16.2.3

Move $124$ to the left of $λ$ .

$p (λ) = \frac{- 3 λ \cdot λ^{2} + 16 λ \cdot λ + 124 \cdot λ - 160 \cdot 3}{3} + 4 (- 2 λ + 40) + 10 (6 λ)$

Step 5.5.1.16.3

Simplify each term.

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Step 5.5.1.16.3.1

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

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Step 5.5.1.16.3.1.1

Move $λ^{2}$ .

$p (λ) = \frac{- 3 (λ^{2} λ) + 16 λ \cdot λ + 124 \cdot λ - 160 \cdot 3}{3} + 4 (- 2 λ + 40) + 10 (6 λ)$

Step 5.5.1.16.3.1.2

Multiply $λ^{2}$ by $λ$ .

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Step 5.5.1.16.3.1.2.1

Raise $λ$ to the power of $1$ .

$p (λ) = \frac{- 3 (λ^{2} λ^{1}) + 16 λ \cdot λ + 124 \cdot λ - 160 \cdot 3}{3} + 4 (- 2 λ + 40) + 10 (6 λ)$

Step 5.5.1.16.3.1.2.2

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

$p (λ) = \frac{- 3 λ^{2 + 1} + 16 λ \cdot λ + 124 \cdot λ - 160 \cdot 3}{3} + 4 (- 2 λ + 40) + 10 (6 λ)$

Step 5.5.1.16.3.1.3

Add $2$ and $1$ .

$p (λ) = \frac{- 3 λ^{3} + 16 λ \cdot λ + 124 \cdot λ - 160 \cdot 3}{3} + 4 (- 2 λ + 40) + 10 (6 λ)$

Step 5.5.1.16.3.2

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

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Step 5.5.1.16.3.2.1

Move $λ$ .

$p (λ) = \frac{- 3 λ^{3} + 16 (λ \cdot λ) + 124 \cdot λ - 160 \cdot 3}{3} + 4 (- 2 λ + 40) + 10 (6 λ)$

Step 5.5.1.16.3.2.2

Multiply $λ$ by $λ$ .

$p (λ) = \frac{- 3 λ^{3} + 16 λ^{2} + 124 \cdot λ - 160 \cdot 3}{3} + 4 (- 2 λ + 40) + 10 (6 λ)$

$p (λ) = \frac{- 3 λ^{3} + 16 λ^{2} + 124 λ - 160 \cdot 3}{3} + 4 (- 2 λ + 40) + 10 (6 λ)$

Step 5.5.1.16.4

Multiply $- 160$ by $3$ .

$p (λ) = \frac{- 3 λ^{3} + 16 λ^{2} + 124 λ - 480}{3} + 4 (- 2 λ + 40) + 10 (6 λ)$

Step 5.5.1.16.5

Rewrite $- 3 λ^{3} + 16 λ^{2} + 124 λ - 480$ in a factored form.

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Step 5.5.1.16.5.1

Factor $- 3 λ^{3} + 16 λ^{2} + 124 λ - 480$ using the rational roots test.

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Step 5.5.1.16.5.1.1

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.

$p = \pm 1, \pm 480, \pm 2, \pm 240, \pm 3, \pm 160, \pm 4, \pm 120, \pm 5, \pm 96, \pm 6, \pm 80, \pm 8, \pm 60, \pm 10, \pm 48, \pm 12, \pm 40, \pm 15, \pm 32, \pm 16, \pm 30, \pm 20, \pm 24$

$q = \pm 1, \pm 3$

Step 5.5.1.16.5.1.2

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

$\pm 1, \pm 0.\overline{3}, \pm 480, \pm 160, \pm 2, \pm 0.\overline{6}, \pm 240, \pm 80, \pm 3, \pm 53.\overline{3}, \pm 4, \pm 1.\overline{3}, \pm 120, \pm 40, \pm 5, \pm 1.\overline{6}, \pm 96, \pm 32, \pm 6, \pm 26.\overline{6}, \pm 8, \pm 2.\overline{6}, \pm 60, \pm 20, \pm 10, \pm 3.\overline{3}, \pm 48, \pm 16, \pm 12, \pm 13.\overline{3}, \pm 15, \pm 10.\overline{6}, \pm 5.\overline{3}, \pm 30, \pm 6.\overline{6}, \pm 24$

Step 5.5.1.16.5.1.3

Substitute $- 6$ and simplify the expression. In this case, the expression is equal to $0$ so $- 6$ is a root of the polynomial.

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Step 5.5.1.16.5.1.3.1

Substitute $- 6$ into the polynomial.

$- 3 {(- 6)}^{3} + 16 {(- 6)}^{2} + 124 \cdot - 6 - 480$

Step 5.5.1.16.5.1.3.2

Raise $- 6$ to the power of $3$ .

$- 3 \cdot - 216 + 16 {(- 6)}^{2} + 124 \cdot - 6 - 480$

Step 5.5.1.16.5.1.3.3

Multiply $- 3$ by $- 216$ .

$648 + 16 {(- 6)}^{2} + 124 \cdot - 6 - 480$

Step 5.5.1.16.5.1.3.4

Raise $- 6$ to the power of $2$ .

$648 + 16 \cdot 36 + 124 \cdot - 6 - 480$

Step 5.5.1.16.5.1.3.5

Multiply $16$ by $36$ .

$648 + 576 + 124 \cdot - 6 - 480$

Step 5.5.1.16.5.1.3.6

Add $648$ and $576$ .

$1224 + 124 \cdot - 6 - 480$

Step 5.5.1.16.5.1.3.7

Multiply $124$ by $- 6$ .

$1224 - 744 - 480$

Step 5.5.1.16.5.1.3.8

Subtract $744$ from $1224$ .

$480 - 480$

Step 5.5.1.16.5.1.3.9

Subtract $480$ from $480$ .

$0$

Step 5.5.1.16.5.1.4

Since $- 6$ is a known root, divide the polynomial by $λ + 6$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{- 3 λ^{3} + 16 λ^{2} + 124 λ - 480}{λ + 6}$

Step 5.5.1.16.5.1.5

Divide $- 3 λ^{3} + 16 λ^{2} + 124 λ - 480$ by $λ + 6$ .

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Step 5.5.1.16.5.1.5.1

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

$λ$

$6$

$3 λ^{3}$

$16 λ^{2}$

$124 λ$

$480$

Step 5.5.1.16.5.1.5.2

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

				-	$3 λ^{2}$
	$λ$	+	$6$	-	$3 λ^{3}$	+	$16 λ^{2}$	+	$124 λ$	-	$480$

Step 5.5.1.16.5.1.5.3

Multiply the new quotient term by the divisor.

			-	$3 λ^{2}$
$λ$	+	$6$	-	$3 λ^{3}$	+	$16 λ^{2}$	+	$124 λ$	-	$480$
			-	$3 λ^{3}$	-	$18 λ^{2}$

Step 5.5.1.16.5.1.5.4

The expression needs to be subtracted from the dividend, so change all the signs in $- 3 λ^{3} - 18 λ^{2}$

			-	$3 λ^{2}$
$λ$	+	$6$	-	$3 λ^{3}$	+	$16 λ^{2}$	+	$124 λ$	-	$480$
			+	$3 λ^{3}$	+	$18 λ^{2}$

Step 5.5.1.16.5.1.5.5

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

			-	$3 λ^{2}$
$λ$	+	$6$	-	$3 λ^{3}$	+	$16 λ^{2}$	+	$124 λ$	-	$480$
			+	$3 λ^{3}$	+	$18 λ^{2}$
					+	$34 λ^{2}$

Step 5.5.1.16.5.1.5.6

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

			-	$3 λ^{2}$
$λ$	+	$6$	-	$3 λ^{3}$	+	$16 λ^{2}$	+	$124 λ$	-	$480$
			+	$3 λ^{3}$	+	$18 λ^{2}$
					+	$34 λ^{2}$	+	$124 λ$

Step 5.5.1.16.5.1.5.7

Divide the highest order term in the dividend $34 λ^{2}$ by the highest order term in divisor $λ$ .

			-	$3 λ^{2}$	+	$34 λ$
$λ$	+	$6$	-	$3 λ^{3}$	+	$16 λ^{2}$	+	$124 λ$	-	$480$
			+	$3 λ^{3}$	+	$18 λ^{2}$
					+	$34 λ^{2}$	+	$124 λ$

Step 5.5.1.16.5.1.5.8

Multiply the new quotient term by the divisor.

			-	$3 λ^{2}$	+	$34 λ$
$λ$	+	$6$	-	$3 λ^{3}$	+	$16 λ^{2}$	+	$124 λ$	-	$480$
			+	$3 λ^{3}$	+	$18 λ^{2}$
					+	$34 λ^{2}$	+	$124 λ$
					+	$34 λ^{2}$	+	$204 λ$

Step 5.5.1.16.5.1.5.9

The expression needs to be subtracted from the dividend, so change all the signs in $34 λ^{2} + 204 λ$

			-	$3 λ^{2}$	+	$34 λ$
$λ$	+	$6$	-	$3 λ^{3}$	+	$16 λ^{2}$	+	$124 λ$	-	$480$
			+	$3 λ^{3}$	+	$18 λ^{2}$
					+	$34 λ^{2}$	+	$124 λ$
					-	$34 λ^{2}$	-	$204 λ$

Step 5.5.1.16.5.1.5.10

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

			-	$3 λ^{2}$	+	$34 λ$
$λ$	+	$6$	-	$3 λ^{3}$	+	$16 λ^{2}$	+	$124 λ$	-	$480$
			+	$3 λ^{3}$	+	$18 λ^{2}$
					+	$34 λ^{2}$	+	$124 λ$
					-	$34 λ^{2}$	-	$204 λ$
							-	$80 λ$

Step 5.5.1.16.5.1.5.11

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

			-	$3 λ^{2}$	+	$34 λ$
$λ$	+	$6$	-	$3 λ^{3}$	+	$16 λ^{2}$	+	$124 λ$	-	$480$
			+	$3 λ^{3}$	+	$18 λ^{2}$
					+	$34 λ^{2}$	+	$124 λ$
					-	$34 λ^{2}$	-	$204 λ$
							-	$80 λ$	-	$480$

Step 5.5.1.16.5.1.5.12

Divide the highest order term in the dividend $- 80 λ$ by the highest order term in divisor $λ$ .

			-	$3 λ^{2}$	+	$34 λ$	-	$80$
$λ$	+	$6$	-	$3 λ^{3}$	+	$16 λ^{2}$	+	$124 λ$	-	$480$
			+	$3 λ^{3}$	+	$18 λ^{2}$
					+	$34 λ^{2}$	+	$124 λ$
					-	$34 λ^{2}$	-	$204 λ$
							-	$80 λ$	-	$480$

Step 5.5.1.16.5.1.5.13

Multiply the new quotient term by the divisor.

			-	$3 λ^{2}$	+	$34 λ$	-	$80$
$λ$	+	$6$	-	$3 λ^{3}$	+	$16 λ^{2}$	+	$124 λ$	-	$480$
			+	$3 λ^{3}$	+	$18 λ^{2}$
					+	$34 λ^{2}$	+	$124 λ$
					-	$34 λ^{2}$	-	$204 λ$
							-	$80 λ$	-	$480$
							-	$80 λ$	-	$480$

Step 5.5.1.16.5.1.5.14

The expression needs to be subtracted from the dividend, so change all the signs in $- 80 λ - 480$

			-	$3 λ^{2}$	+	$34 λ$	-	$80$
$λ$	+	$6$	-	$3 λ^{3}$	+	$16 λ^{2}$	+	$124 λ$	-	$480$
			+	$3 λ^{3}$	+	$18 λ^{2}$
					+	$34 λ^{2}$	+	$124 λ$
					-	$34 λ^{2}$	-	$204 λ$
							-	$80 λ$	-	$480$
							+	$80 λ$	+	$480$

Step 5.5.1.16.5.1.5.15

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

			-	$3 λ^{2}$	+	$34 λ$	-	$80$
$λ$	+	$6$	-	$3 λ^{3}$	+	$16 λ^{2}$	+	$124 λ$	-	$480$
			+	$3 λ^{3}$	+	$18 λ^{2}$
					+	$34 λ^{2}$	+	$124 λ$
					-	$34 λ^{2}$	-	$204 λ$
							-	$80 λ$	-	$480$
							+	$80 λ$	+	$480$
										$0$

Step 5.5.1.16.5.1.5.16

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

$- 3 λ^{2} + 34 λ - 80$

Step 5.5.1.16.5.1.6

Write $- 3 λ^{3} + 16 λ^{2} + 124 λ - 480$ as a set of factors.

$p (λ) = \frac{(λ + 6) (- 3 λ^{2} + 34 λ - 80)}{3} + 4 (- 2 λ + 40) + 10 (6 λ)$

Step 5.5.1.16.5.2

Factor by grouping.

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Step 5.5.1.16.5.2.1

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 $a \cdot c = - 3 \cdot - 80 = 240$ and whose sum is $b = 34$ .

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Step 5.5.1.16.5.2.1.1

Factor $34$ out of $34 λ$ .

$p (λ) = \frac{(λ + 6) (- 3 λ^{2} + 34 (λ) - 80)}{3} + 4 (- 2 λ + 40) + 10 (6 λ)$

Step 5.5.1.16.5.2.1.2

Rewrite $34$ as $10$ plus $24$

$p (λ) = \frac{(λ + 6) (- 3 λ^{2} + (10 + 24) λ - 80)}{3} + 4 (- 2 λ + 40) + 10 (6 λ)$

Step 5.5.1.16.5.2.1.3

Apply the distributive property.

$p (λ) = \frac{(λ + 6) (- 3 λ^{2} + 10 λ + 24 λ - 80)}{3} + 4 (- 2 λ + 40) + 10 (6 λ)$

Step 5.5.1.16.5.2.2

Factor out the greatest common factor from each group.

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Step 5.5.1.16.5.2.2.1

Group the first two terms and the last two terms.

$p (λ) = \frac{(λ + 6) ((- 3 λ^{2} + 10 λ) + 24 λ - 80)}{3} + 4 (- 2 λ + 40) + 10 (6 λ)$

Step 5.5.1.16.5.2.2.2

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

$p (λ) = \frac{(λ + 6) (λ (- 3 λ + 10) - 8 (- 3 λ + 10))}{3} + 4 (- 2 λ + 40) + 10 (6 λ)$

Step 5.5.1.16.5.2.3

Factor the polynomial by factoring out the greatest common factor, $- 3 λ + 10$ .

$p (λ) = \frac{(λ + 6) ((- 3 λ + 10) (λ - 8))}{3} + 4 (- 2 λ + 40) + 10 (6 λ)$

$p (λ) = \frac{(λ + 6) (- 3 λ + 10) (λ - 8)}{3} + 4 (- 2 λ + 40) + 10 (6 λ)$

Step 5.5.1.17

Apply the distributive property.

$p (λ) = \frac{(λ + 6) (- 3 λ + 10) (λ - 8)}{3} + 4 (- 2 λ) + 4 \cdot 40 + 10 (6 λ)$

Step 5.5.1.18

Multiply $- 2$ by $4$ .

$p (λ) = \frac{(λ + 6) (- 3 λ + 10) (λ - 8)}{3} - 8 λ + 4 \cdot 40 + 10 (6 λ)$

Step 5.5.1.19

Multiply $4$ by $40$ .

$p (λ) = \frac{(λ + 6) (- 3 λ + 10) (λ - 8)}{3} - 8 λ + 160 + 10 (6 λ)$

Step 5.5.1.20

Multiply $6$ by $10$ .

$p (λ) = \frac{(λ + 6) (- 3 λ + 10) (λ - 8)}{3} - 8 λ + 160 + 60 λ$

Step 5.5.2

To write $- 8 λ$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$p (λ) = \frac{(λ + 6) (- 3 λ + 10) (λ - 8)}{3} - 8 λ \cdot \frac{3}{3} + 160 + 60 λ$

Step 5.5.3

Combine $- 8 λ$ and $\frac{3}{3}$ .

$p (λ) = \frac{(λ + 6) (- 3 λ + 10) (λ - 8)}{3} + \frac{- 8 λ \cdot 3}{3} + 160 + 60 λ$

Step 5.5.4

Combine the numerators over the common denominator.

$p (λ) = \frac{(λ + 6) (- 3 λ + 10) (λ - 8) - 8 λ \cdot 3}{3} + 160 + 60 λ$

Step 5.5.5

Simplify the numerator.

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Step 5.5.5.1

Expand $(λ + 6) (- 3 λ + 10)$ using the FOIL Method.

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Step 5.5.5.1.1

Apply the distributive property.

$p (λ) = \frac{(λ (- 3 λ + 10) + 6 (- 3 λ + 10)) (λ - 8) - 8 λ \cdot 3}{3} + 160 + 60 λ$

Step 5.5.5.1.2

Apply the distributive property.

$p (λ) = \frac{(λ (- 3 λ) + λ \cdot 10 + 6 (- 3 λ + 10)) (λ - 8) - 8 λ \cdot 3}{3} + 160 + 60 λ$

Step 5.5.5.1.3

Apply the distributive property.

$p (λ) = \frac{(λ (- 3 λ) + λ \cdot 10 + 6 (- 3 λ) + 6 \cdot 10) (λ - 8) - 8 λ \cdot 3}{3} + 160 + 60 λ$

Step 5.5.5.2

Simplify and combine like terms.

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Step 5.5.5.2.1

Simplify each term.

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Step 5.5.5.2.1.1

Rewrite using the commutative property of multiplication.

$p (λ) = \frac{(- 3 λ \cdot λ + λ \cdot 10 + 6 (- 3 λ) + 6 \cdot 10) (λ - 8) - 8 λ \cdot 3}{3} + 160 + 60 λ$

Step 5.5.5.2.1.2

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

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Step 5.5.5.2.1.2.1

Move $λ$ .

$p (λ) = \frac{(- 3 (λ \cdot λ) + λ \cdot 10 + 6 (- 3 λ) + 6 \cdot 10) (λ - 8) - 8 λ \cdot 3}{3} + 160 + 60 λ$

Step 5.5.5.2.1.2.2

Multiply $λ$ by $λ$ .

$p (λ) = \frac{(- 3 λ^{2} + λ \cdot 10 + 6 (- 3 λ) + 6 \cdot 10) (λ - 8) - 8 λ \cdot 3}{3} + 160 + 60 λ$

Step 5.5.5.2.1.3

Move $10$ to the left of $λ$ .

$p (λ) = \frac{(- 3 λ^{2} + 10 \cdot λ + 6 (- 3 λ) + 6 \cdot 10) (λ - 8) - 8 λ \cdot 3}{3} + 160 + 60 λ$

Step 5.5.5.2.1.4

Multiply $- 3$ by $6$ .

$p (λ) = \frac{(- 3 λ^{2} + 10 λ - 18 λ + 6 \cdot 10) (λ - 8) - 8 λ \cdot 3}{3} + 160 + 60 λ$

Step 5.5.5.2.1.5

Multiply $6$ by $10$ .

$p (λ) = \frac{(- 3 λ^{2} + 10 λ - 18 λ + 60) (λ - 8) - 8 λ \cdot 3}{3} + 160 + 60 λ$

Step 5.5.5.2.2

Subtract $18 λ$ from $10 λ$ .

$p (λ) = \frac{(- 3 λ^{2} - 8 λ + 60) (λ - 8) - 8 λ \cdot 3}{3} + 160 + 60 λ$

Step 5.5.5.3

Expand $(- 3 λ^{2} - 8 λ + 60) (λ - 8)$ by multiplying each term in the first expression by each term in the second expression.

$p (λ) = \frac{- 3 λ^{2} λ - 3 λ^{2} \cdot - 8 - 8 λ \cdot λ - 8 λ \cdot - 8 + 60 λ + 60 \cdot - 8 - 8 λ \cdot 3}{3} + 160 + 60 λ$

Step 5.5.5.4

Simplify each term.

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Step 5.5.5.4.1

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

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Step 5.5.5.4.1.1

Move $λ$ .

$p (λ) = \frac{- 3 (λ \cdot λ^{2}) - 3 λ^{2} \cdot - 8 - 8 λ \cdot λ - 8 λ \cdot - 8 + 60 λ + 60 \cdot - 8 - 8 λ \cdot 3}{3} + 160 + 60 λ$

Step 5.5.5.4.1.2

Multiply $λ$ by $λ^{2}$ .

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Step 5.5.5.4.1.2.1

Raise $λ$ to the power of $1$ .

$p (λ) = \frac{- 3 (λ^{1} λ^{2}) - 3 λ^{2} \cdot - 8 - 8 λ \cdot λ - 8 λ \cdot - 8 + 60 λ + 60 \cdot - 8 - 8 λ \cdot 3}{3} + 160 + 60 λ$

Step 5.5.5.4.1.2.2

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

$p (λ) = \frac{- 3 λ^{1 + 2} - 3 λ^{2} \cdot - 8 - 8 λ \cdot λ - 8 λ \cdot - 8 + 60 λ + 60 \cdot - 8 - 8 λ \cdot 3}{3} + 160 + 60 λ$

Step 5.5.5.4.1.3

Add $1$ and $2$ .

$p (λ) = \frac{- 3 λ^{3} - 3 λ^{2} \cdot - 8 - 8 λ \cdot λ - 8 λ \cdot - 8 + 60 λ + 60 \cdot - 8 - 8 λ \cdot 3}{3} + 160 + 60 λ$

Step 5.5.5.4.2

Multiply $- 8$ by $- 3$ .

$p (λ) = \frac{- 3 λ^{3} + 24 λ^{2} - 8 λ \cdot λ - 8 λ \cdot - 8 + 60 λ + 60 \cdot - 8 - 8 λ \cdot 3}{3} + 160 + 60 λ$

Step 5.5.5.4.3

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

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Step 5.5.5.4.3.1

Move $λ$ .

$p (λ) = \frac{- 3 λ^{3} + 24 λ^{2} - 8 (λ \cdot λ) - 8 λ \cdot - 8 + 60 λ + 60 \cdot - 8 - 8 λ \cdot 3}{3} + 160 + 60 λ$

Step 5.5.5.4.3.2

Multiply $λ$ by $λ$ .

$p (λ) = \frac{- 3 λ^{3} + 24 λ^{2} - 8 λ^{2} - 8 λ \cdot - 8 + 60 λ + 60 \cdot - 8 - 8 λ \cdot 3}{3} + 160 + 60 λ$

Step 5.5.5.4.4

Multiply $- 8$ by $- 8$ .

$p (λ) = \frac{- 3 λ^{3} + 24 λ^{2} - 8 λ^{2} + 64 λ + 60 λ + 60 \cdot - 8 - 8 λ \cdot 3}{3} + 160 + 60 λ$

Step 5.5.5.4.5

Multiply $60$ by $- 8$ .

$p (λ) = \frac{- 3 λ^{3} + 24 λ^{2} - 8 λ^{2} + 64 λ + 60 λ - 480 - 8 λ \cdot 3}{3} + 160 + 60 λ$

Step 5.5.5.5

Subtract $8 λ^{2}$ from $24 λ^{2}$ .

$p (λ) = \frac{- 3 λ^{3} + 16 λ^{2} + 64 λ + 60 λ - 480 - 8 λ \cdot 3}{3} + 160 + 60 λ$

Step 5.5.5.6

Add $64 λ$ and $60 λ$ .

$p (λ) = \frac{- 3 λ^{3} + 16 λ^{2} + 124 λ - 480 - 8 λ \cdot 3}{3} + 160 + 60 λ$

Step 5.5.5.7

Multiply $3$ by $- 8$ .

$p (λ) = \frac{- 3 λ^{3} + 16 λ^{2} + 124 λ - 480 - 24 λ}{3} + 160 + 60 λ$

Step 5.5.5.8

Subtract $24 λ$ from $124 λ$ .

$p (λ) = \frac{- 3 λ^{3} + 16 λ^{2} + 100 λ - 480}{3} + 160 + 60 λ$

Step 5.5.6

To write $160$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$p (λ) = \frac{- 3 λ^{3} + 16 λ^{2} + 100 λ - 480}{3} + 160 \cdot \frac{3}{3} + 60 λ$

Step 5.5.7

Combine $160$ and $\frac{3}{3}$ .

$p (λ) = \frac{- 3 λ^{3} + 16 λ^{2} + 100 λ - 480}{3} + \frac{160 \cdot 3}{3} + 60 λ$

Step 5.5.8

Combine the numerators over the common denominator.

$p (λ) = \frac{- 3 λ^{3} + 16 λ^{2} + 100 λ - 480 + 160 \cdot 3}{3} + 60 λ$

Step 5.5.9

Multiply $160$ by $3$ .

$p (λ) = \frac{- 3 λ^{3} + 16 λ^{2} + 100 λ - 480 + 480}{3} + 60 λ$

Step 5.5.10

Add $- 480$ and $480$ .

$p (λ) = \frac{- 3 λ^{3} + 16 λ^{2} + 100 λ + 0}{3} + 60 λ$

Step 5.5.11

Add $- 3 λ^{3}$ and $0$ .

$p (λ) = \frac{- 3 λ^{3} + 16 λ^{2} + 100 λ}{3} + 60 λ$

Step 5.5.12

Factor $λ$ out of $- 3 λ^{3} + 16 λ^{2} + 100 λ$ .

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Step 5.5.12.1

Factor $λ$ out of $- 3 λ^{3}$ .

$p (λ) = \frac{λ (- 3 λ^{2}) + 16 λ^{2} + 100 λ}{3} + 60 λ$

Step 5.5.12.2

Factor $λ$ out of $16 λ^{2}$ .

$p (λ) = \frac{λ (- 3 λ^{2}) + λ (16 λ) + 100 λ}{3} + 60 λ$

Step 5.5.12.3

Factor $λ$ out of $100 λ$ .

$p (λ) = \frac{λ (- 3 λ^{2}) + λ (16 λ) + λ \cdot 100}{3} + 60 λ$

Step 5.5.12.4

Factor $λ$ out of $λ (- 3 λ^{2}) + λ (16 λ)$ .

$p (λ) = \frac{λ (- 3 λ^{2} + 16 λ) + λ \cdot 100}{3} + 60 λ$

Step 5.5.12.5

Factor $λ$ out of $λ (- 3 λ^{2} + 16 λ) + λ \cdot 100$ .

$p (λ) = \frac{λ (- 3 λ^{2} + 16 λ + 100)}{3} + 60 λ$

Step 5.5.13

To write $60 λ$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$p (λ) = \frac{λ (- 3 λ^{2} + 16 λ + 100)}{3} + 60 λ \cdot \frac{3}{3}$

Step 5.5.14

Combine $60 λ$ and $\frac{3}{3}$ .

$p (λ) = \frac{λ (- 3 λ^{2} + 16 λ + 100)}{3} + \frac{60 λ \cdot 3}{3}$

Step 5.5.15

Combine the numerators over the common denominator.

$p (λ) = \frac{λ (- 3 λ^{2} + 16 λ + 100) + 60 λ \cdot 3}{3}$

Step 5.5.16

Simplify the numerator.

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Step 5.5.16.1

Factor $λ$ out of $λ (- 3 λ^{2} + 16 λ + 100) + 60 λ \cdot 3$ .

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Step 5.5.16.1.1

Factor $λ$ out of $60 λ \cdot 3$ .

$p (λ) = \frac{λ (- 3 λ^{2} + 16 λ + 100) + λ (60 \cdot 3)}{3}$

Step 5.5.16.1.2

Factor $λ$ out of $λ (- 3 λ^{2} + 16 λ + 100) + λ (60 \cdot 3)$ .

$p (λ) = \frac{λ (- 3 λ^{2} + 16 λ + 100 + 60 \cdot 3)}{3}$

Step 5.5.16.2

Multiply $60$ by $3$ .

$p (λ) = \frac{λ (- 3 λ^{2} + 16 λ + 100 + 180)}{3}$

Step 5.5.16.3

Add $100$ and $180$ .

$p (λ) = \frac{λ (- 3 λ^{2} + 16 λ + 280)}{3}$

Step 5.5.17

Factor $- 1$ out of $- 3 λ^{2}$ .

$p (λ) = \frac{λ (- (3 λ^{2}) + 16 λ + 280)}{3}$

Step 5.5.18

Factor $- 1$ out of $16 λ$ .

$p (λ) = \frac{λ (- (3 λ^{2}) - (- 16 λ) + 280)}{3}$

Step 5.5.19

Factor $- 1$ out of $- (3 λ^{2}) - (- 16 λ)$ .

$p (λ) = \frac{λ (- (3 λ^{2} - 16 λ) + 280)}{3}$

Step 5.5.20

Rewrite $280$ as $- 1 (- 280)$ .

$p (λ) = \frac{λ (- (3 λ^{2} - 16 λ) - 1 (- 280))}{3}$

Step 5.5.21

Factor $- 1$ out of $- (3 λ^{2} - 16 λ) - 1 (- 280)$ .

$p (λ) = \frac{λ (- (3 λ^{2} - 16 λ - 280))}{3}$

Step 5.5.22

Rewrite $- (3 λ^{2} - 16 λ - 280)$ as $- 1 (3 λ^{2} - 16 λ - 280)$ .

$p (λ) = \frac{λ (- 1 (3 λ^{2} - 16 λ - 280))}{3}$

Step 5.5.23

Move the negative in front of the fraction.

$p (λ) = - \frac{λ (3 λ^{2} - 16 λ - 280)}{3}$

Step 6

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

$- \frac{λ (3 λ^{2} - 16 λ - 280)}{3} = 0$

Step 7

Solve for $λ$ .

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Step 7.1

Set the numerator equal to zero.

$λ (3 λ^{2} - 16 λ - 280) = 0$

Step 7.2

Solve the equation for $λ$ .

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Step 7.2.1

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$λ = 0$

$3 λ^{2} - 16 λ - 280 = 0$

Step 7.2.2

Set $λ$ equal to $0$ .

$λ = 0$

Step 7.2.3

Set $3 λ^{2} - 16 λ - 280$ equal to $0$ and solve for $λ$ .

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Step 7.2.3.1

Set $3 λ^{2} - 16 λ - 280$ equal to $0$ .

$3 λ^{2} - 16 λ - 280 = 0$

Step 7.2.3.2

Solve $3 λ^{2} - 16 λ - 280 = 0$ for $λ$ .

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Step 7.2.3.2.1

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 7.2.3.2.2

Substitute the values $a = 3$ , $b = - 16$ , and $c = - 280$ into the quadratic formula and solve for $λ$ .

$\frac{16 \pm \sqrt{{(- 16)}^{2} - 4 \cdot (3 \cdot - 280)}}{2 \cdot 3}$

Step 7.2.3.2.3

Simplify.

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Step 7.2.3.2.3.1

Simplify the numerator.

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Step 7.2.3.2.3.1.1

Raise $- 16$ to the power of $2$ .

$λ = \frac{16 \pm \sqrt{256 - 4 \cdot 3 \cdot - 280}}{2 \cdot 3}$

Step 7.2.3.2.3.1.2

Multiply $- 4 \cdot 3 \cdot - 280$ .

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Step 7.2.3.2.3.1.2.1

Multiply $- 4$ by $3$ .

$λ = \frac{16 \pm \sqrt{256 - 12 \cdot - 280}}{2 \cdot 3}$

Step 7.2.3.2.3.1.2.2

Multiply $- 12$ by $- 280$ .

$λ = \frac{16 \pm \sqrt{256 + 3360}}{2 \cdot 3}$

Step 7.2.3.2.3.1.3

Add $256$ and $3360$ .

$λ = \frac{16 \pm \sqrt{3616}}{2 \cdot 3}$

Step 7.2.3.2.3.1.4

Rewrite $3616$ as $4^{2} \cdot 226$ .

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Step 7.2.3.2.3.1.4.1

Factor $16$ out of $3616$ .

$λ = \frac{16 \pm \sqrt{16 (226)}}{2 \cdot 3}$

Step 7.2.3.2.3.1.4.2

Rewrite $16$ as $4^{2}$ .

$λ = \frac{16 \pm \sqrt{4^{2} \cdot 226}}{2 \cdot 3}$

Step 7.2.3.2.3.1.5

Pull terms out from under the radical.

$λ = \frac{16 \pm 4 \sqrt{226}}{2 \cdot 3}$

Step 7.2.3.2.3.2

Multiply $2$ by $3$ .

$λ = \frac{16 \pm 4 \sqrt{226}}{6}$

Step 7.2.3.2.3.3

Simplify $\frac{16 \pm 4 \sqrt{226}}{6}$ .

$λ = \frac{8 \pm 2 \sqrt{226}}{3}$

Step 7.2.3.2.4

The final answer is the combination of both solutions.

$λ = \frac{8 + 2 \sqrt{226}}{3}, \frac{8 - 2 \sqrt{226}}{3}$

Step 7.2.4

The final solution is all the values that make $λ (3 λ^{2} - 16 λ - 280) = 0$ true.

$λ = 0, \frac{8 + 2 \sqrt{226}}{3}, \frac{8 - 2 \sqrt{226}}{3}$

Step 8

The result can be shown in multiple forms.

Exact Form:

$λ = 0, \frac{8 + 2 \sqrt{226}}{3}, \frac{8 - 2 \sqrt{226}}{3}$

Decimal Form:

$λ = 0, 12.68886425 \dots, - 7.35553091 \dots$