Algebra Examples

$B = [\begin{array}{c} 9 & 8 & 7 \\ 3 & 4 & 5 \\ 2 & 1 & 0 \end{array}]$

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 $[\begin{array}{c} 9 & 8 & 7 \\ 3 & 4 & 5 \\ 2 & 1 & 0 \end{array}]$ for $A$ .

$p (λ) = determinant ([\begin{array}{c} 9 & 8 & 7 \\ 3 & 4 & 5 \\ 2 & 1 & 0 \end{array}] - λ I_{3})$

Step 3.2

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

$p (λ) = determinant ([\begin{array}{c} 9 & 8 & 7 \\ 3 & 4 & 5 \\ 2 & 1 & 0 \end{array}] - λ [\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}])$

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.

$p (λ) = determinant ([\begin{array}{c} 9 & 8 & 7 \\ 3 & 4 & 5 \\ 2 & 1 & 0 \end{array}] + [\begin{array}{c} - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2

Simplify each element in the matrix.

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Step 4.1.2.1

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} 9 & 8 & 7 \\ 3 & 4 & 5 \\ 2 & 1 & 0 \end{array}] + [\begin{array}{c} - λ & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.2

Multiply $- λ \cdot 0$ .

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Step 4.1.2.2.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 9 & 8 & 7 \\ 3 & 4 & 5 \\ 2 & 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 λ & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.2.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 9 & 8 & 7 \\ 3 & 4 & 5 \\ 2 & 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.3

Multiply $- λ \cdot 0$ .

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Step 4.1.2.3.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 9 & 8 & 7 \\ 3 & 4 & 5 \\ 2 & 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 λ \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.3.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 9 & 8 & 7 \\ 3 & 4 & 5 \\ 2 & 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.4

Multiply $- λ \cdot 0$ .

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Step 4.1.2.4.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 9 & 8 & 7 \\ 3 & 4 & 5 \\ 2 & 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 λ & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.4.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 9 & 8 & 7 \\ 3 & 4 & 5 \\ 2 & 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.5

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} 9 & 8 & 7 \\ 3 & 4 & 5 \\ 2 & 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.6

Multiply $- λ \cdot 0$ .

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Step 4.1.2.6.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 9 & 8 & 7 \\ 3 & 4 & 5 \\ 2 & 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 λ \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.6.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 9 & 8 & 7 \\ 3 & 4 & 5 \\ 2 & 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.7

Multiply $- λ \cdot 0$ .

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Step 4.1.2.7.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 9 & 8 & 7 \\ 3 & 4 & 5 \\ 2 & 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 λ & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.7.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 9 & 8 & 7 \\ 3 & 4 & 5 \\ 2 & 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.8

Multiply $- λ \cdot 0$ .

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Step 4.1.2.8.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 9 & 8 & 7 \\ 3 & 4 & 5 \\ 2 & 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & 0 λ & - λ \cdot 1 \end{array}])$

Step 4.1.2.8.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 9 & 8 & 7 \\ 3 & 4 & 5 \\ 2 & 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.9

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} 9 & 8 & 7 \\ 3 & 4 & 5 \\ 2 & 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & 0 & - λ \end{array}])$

Step 4.2

Add the corresponding elements.

$p (λ) = determinant [\begin{array}{c} 9 - λ & 8 + 0 & 7 + 0 \\ 3 + 0 & 4 - λ & 5 + 0 \\ 2 + 0 & 1 + 0 & 0 - λ \end{array}]$

Step 4.3

Simplify each element.

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Step 4.3.1

Add $8$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 9 - λ & 8 & 7 + 0 \\ 3 + 0 & 4 - λ & 5 + 0 \\ 2 + 0 & 1 + 0 & 0 - λ \end{array}]$

Step 4.3.2

Add $7$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 9 - λ & 8 & 7 \\ 3 + 0 & 4 - λ & 5 + 0 \\ 2 + 0 & 1 + 0 & 0 - λ \end{array}]$

Step 4.3.3

Add $3$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 9 - λ & 8 & 7 \\ 3 & 4 - λ & 5 + 0 \\ 2 + 0 & 1 + 0 & 0 - λ \end{array}]$

Step 4.3.4

Add $5$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 9 - λ & 8 & 7 \\ 3 & 4 - λ & 5 \\ 2 + 0 & 1 + 0 & 0 - λ \end{array}]$

Step 4.3.5

Add $2$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 9 - λ & 8 & 7 \\ 3 & 4 - λ & 5 \\ 2 & 1 + 0 & 0 - λ \end{array}]$

Step 4.3.6

Add $1$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 9 - λ & 8 & 7 \\ 3 & 4 - λ & 5 \\ 2 & 1 & 0 - λ \end{array}]$

Step 4.3.7

Subtract $λ$ from $0$ .

$p (λ) = determinant [\begin{array}{c} 9 - λ & 8 & 7 \\ 3 & 4 - λ & 5 \\ 2 & 1 & - λ \end{array}]$

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.

$| \begin{array}{c} 4 - λ & 5 \\ 1 & - λ \end{array} |$

Step 5.1.4

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

$(9 - λ) | \begin{array}{c} 4 - λ & 5 \\ 1 & - λ \end{array} |$

Step 5.1.5

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

$| \begin{array}{c} 3 & 5 \\ 2 & - λ \end{array} |$

Step 5.1.6

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

$- 8 | \begin{array}{c} 3 & 5 \\ 2 & - λ \end{array} |$

Step 5.1.7

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

$| \begin{array}{c} 3 & 4 - λ \\ 2 & 1 \end{array} |$

Step 5.1.8

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

$7 | \begin{array}{c} 3 & 4 - λ \\ 2 & 1 \end{array} |$

Step 5.1.9

Add the terms together.

$p (λ) = (9 - λ) | \begin{array}{c} 4 - λ & 5 \\ 1 & - λ \end{array} | - 8 | \begin{array}{c} 3 & 5 \\ 2 & - λ \end{array} | + 7 | \begin{array}{c} 3 & 4 - λ \\ 2 & 1 \end{array} |$

Step 5.2

Evaluate $| \begin{array}{c} 4 - λ & 5 \\ 1 & - λ \end{array} |$ .

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Step 5.2.1

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

$p (λ) = (9 - λ) ((4 - λ) (- λ) - 1 \cdot 5) - 8 | \begin{array}{c} 3 & 5 \\ 2 & - λ \end{array} | + 7 | \begin{array}{c} 3 & 4 - λ \\ 2 & 1 \end{array} |$

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

Apply the distributive property.

$p (λ) = (9 - λ) (4 (- λ) - λ (- λ) - 1 \cdot 5) - 8 | \begin{array}{c} 3 & 5 \\ 2 & - λ \end{array} | + 7 | \begin{array}{c} 3 & 4 - λ \\ 2 & 1 \end{array} |$

Step 5.2.2.1.2

Multiply $- 1$ by $4$ .

$p (λ) = (9 - λ) (- 4 λ - λ (- λ) - 1 \cdot 5) - 8 | \begin{array}{c} 3 & 5 \\ 2 & - λ \end{array} | + 7 | \begin{array}{c} 3 & 4 - λ \\ 2 & 1 \end{array} |$

Step 5.2.2.1.3

Rewrite using the commutative property of multiplication.

$p (λ) = (9 - λ) (- 4 λ - 1 \cdot - 1 λ \cdot λ - 1 \cdot 5) - 8 | \begin{array}{c} 3 & 5 \\ 2 & - λ \end{array} | + 7 | \begin{array}{c} 3 & 4 - λ \\ 2 & 1 \end{array} |$

Step 5.2.2.1.4

Simplify each term.

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Step 5.2.2.1.4.1

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

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Step 5.2.2.1.4.1.1

Move $λ$ .

$p (λ) = (9 - λ) (- 4 λ - 1 \cdot - 1 (λ \cdot λ) - 1 \cdot 5) - 8 | \begin{array}{c} 3 & 5 \\ 2 & - λ \end{array} | + 7 | \begin{array}{c} 3 & 4 - λ \\ 2 & 1 \end{array} |$

Step 5.2.2.1.4.1.2

Multiply $λ$ by $λ$ .

$p (λ) = (9 - λ) (- 4 λ - 1 \cdot - 1 λ^{2} - 1 \cdot 5) - 8 | \begin{array}{c} 3 & 5 \\ 2 & - λ \end{array} | + 7 | \begin{array}{c} 3 & 4 - λ \\ 2 & 1 \end{array} |$

Step 5.2.2.1.4.2

Multiply $- 1$ by $- 1$ .

$p (λ) = (9 - λ) (- 4 λ + 1 λ^{2} - 1 \cdot 5) - 8 | \begin{array}{c} 3 & 5 \\ 2 & - λ \end{array} | + 7 | \begin{array}{c} 3 & 4 - λ \\ 2 & 1 \end{array} |$

Step 5.2.2.1.4.3

Multiply $λ^{2}$ by $1$ .

$p (λ) = (9 - λ) (- 4 λ + λ^{2} - 1 \cdot 5) - 8 | \begin{array}{c} 3 & 5 \\ 2 & - λ \end{array} | + 7 | \begin{array}{c} 3 & 4 - λ \\ 2 & 1 \end{array} |$

Step 5.2.2.1.5

Multiply $- 1$ by $5$ .

$p (λ) = (9 - λ) (- 4 λ + λ^{2} - 5) - 8 | \begin{array}{c} 3 & 5 \\ 2 & - λ \end{array} | + 7 | \begin{array}{c} 3 & 4 - λ \\ 2 & 1 \end{array} |$

Step 5.2.2.2

Reorder $- 4 λ$ and $λ^{2}$ .

$p (λ) = (9 - λ) (λ^{2} - 4 λ - 5) - 8 | \begin{array}{c} 3 & 5 \\ 2 & - λ \end{array} | + 7 | \begin{array}{c} 3 & 4 - λ \\ 2 & 1 \end{array} |$

Step 5.3

Evaluate $| \begin{array}{c} 3 & 5 \\ 2 & - λ \end{array} |$ .

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Step 5.3.1

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

$p (λ) = (9 - λ) (λ^{2} - 4 λ - 5) - 8 (3 (- λ) - 2 \cdot 5) + 7 | \begin{array}{c} 3 & 4 - λ \\ 2 & 1 \end{array} |$

Step 5.3.2

Simplify each term.

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Step 5.3.2.1

Multiply $- 1$ by $3$ .

$p (λ) = (9 - λ) (λ^{2} - 4 λ - 5) - 8 (- 3 λ - 2 \cdot 5) + 7 | \begin{array}{c} 3 & 4 - λ \\ 2 & 1 \end{array} |$

Step 5.3.2.2

Multiply $- 2$ by $5$ .

$p (λ) = (9 - λ) (λ^{2} - 4 λ - 5) - 8 (- 3 λ - 10) + 7 | \begin{array}{c} 3 & 4 - λ \\ 2 & 1 \end{array} |$

Step 5.4

Evaluate $| \begin{array}{c} 3 & 4 - λ \\ 2 & 1 \end{array} |$ .

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Step 5.4.1

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

$p (λ) = (9 - λ) (λ^{2} - 4 λ - 5) - 8 (- 3 λ - 10) + 7 (3 \cdot 1 - 2 (4 - λ))$

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 $3$ by $1$ .

$p (λ) = (9 - λ) (λ^{2} - 4 λ - 5) - 8 (- 3 λ - 10) + 7 (3 - 2 (4 - λ))$

Step 5.4.2.1.2

Apply the distributive property.

$p (λ) = (9 - λ) (λ^{2} - 4 λ - 5) - 8 (- 3 λ - 10) + 7 (3 - 2 \cdot 4 - 2 (- λ))$

Step 5.4.2.1.3

Multiply $- 2$ by $4$ .

$p (λ) = (9 - λ) (λ^{2} - 4 λ - 5) - 8 (- 3 λ - 10) + 7 (3 - 8 - 2 (- λ))$

Step 5.4.2.1.4

Multiply $- 1$ by $- 2$ .

$p (λ) = (9 - λ) (λ^{2} - 4 λ - 5) - 8 (- 3 λ - 10) + 7 (3 - 8 + 2 λ)$

Step 5.4.2.2

Subtract $8$ from $3$ .

$p (λ) = (9 - λ) (λ^{2} - 4 λ - 5) - 8 (- 3 λ - 10) + 7 (- 5 + 2 λ)$

Step 5.4.2.3

Reorder $- 5$ and $2 λ$ .

$p (λ) = (9 - λ) (λ^{2} - 4 λ - 5) - 8 (- 3 λ - 10) + 7 (2 λ - 5)$

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

Expand $(9 - λ) (λ^{2} - 4 λ - 5)$ by multiplying each term in the first expression by each term in the second expression.

$p (λ) = 9 λ^{2} + 9 (- 4 λ) + 9 \cdot - 5 - λ \cdot λ^{2} - λ (- 4 λ) - λ \cdot - 5 - 8 (- 3 λ - 10) + 7 (2 λ - 5)$

Step 5.5.1.2

Simplify each term.

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Step 5.5.1.2.1

Multiply $- 4$ by $9$ .

$p (λ) = 9 λ^{2} - 36 λ + 9 \cdot - 5 - λ \cdot λ^{2} - λ (- 4 λ) - λ \cdot - 5 - 8 (- 3 λ - 10) + 7 (2 λ - 5)$

Step 5.5.1.2.2

Multiply $9$ by $- 5$ .

$p (λ) = 9 λ^{2} - 36 λ - 45 - λ \cdot λ^{2} - λ (- 4 λ) - λ \cdot - 5 - 8 (- 3 λ - 10) + 7 (2 λ - 5)$

Step 5.5.1.2.3

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

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Step 5.5.1.2.3.1

Move $λ^{2}$ .

$p (λ) = 9 λ^{2} - 36 λ - 45 - (λ^{2} λ) - λ (- 4 λ) - λ \cdot - 5 - 8 (- 3 λ - 10) + 7 (2 λ - 5)$

Step 5.5.1.2.3.2

Multiply $λ^{2}$ by $λ$ .

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Step 5.5.1.2.3.2.1

Raise $λ$ to the power of $1$ .

$p (λ) = 9 λ^{2} - 36 λ - 45 - (λ^{2} λ^{1}) - λ (- 4 λ) - λ \cdot - 5 - 8 (- 3 λ - 10) + 7 (2 λ - 5)$

Step 5.5.1.2.3.2.2

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

$p (λ) = 9 λ^{2} - 36 λ - 45 - λ^{2 + 1} - λ (- 4 λ) - λ \cdot - 5 - 8 (- 3 λ - 10) + 7 (2 λ - 5)$

Step 5.5.1.2.3.3

Add $2$ and $1$ .

$p (λ) = 9 λ^{2} - 36 λ - 45 - λ^{3} - λ (- 4 λ) - λ \cdot - 5 - 8 (- 3 λ - 10) + 7 (2 λ - 5)$

Step 5.5.1.2.4

Rewrite using the commutative property of multiplication.

$p (λ) = 9 λ^{2} - 36 λ - 45 - λ^{3} - 1 \cdot - 4 λ \cdot λ - λ \cdot - 5 - 8 (- 3 λ - 10) + 7 (2 λ - 5)$

Step 5.5.1.2.5

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

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Step 5.5.1.2.5.1

Move $λ$ .

$p (λ) = 9 λ^{2} - 36 λ - 45 - λ^{3} - 1 \cdot - 4 (λ \cdot λ) - λ \cdot - 5 - 8 (- 3 λ - 10) + 7 (2 λ - 5)$

Step 5.5.1.2.5.2

Multiply $λ$ by $λ$ .

$p (λ) = 9 λ^{2} - 36 λ - 45 - λ^{3} - 1 \cdot - 4 λ^{2} - λ \cdot - 5 - 8 (- 3 λ - 10) + 7 (2 λ - 5)$

Step 5.5.1.2.6

Multiply $- 1$ by $- 4$ .

$p (λ) = 9 λ^{2} - 36 λ - 45 - λ^{3} + 4 λ^{2} - λ \cdot - 5 - 8 (- 3 λ - 10) + 7 (2 λ - 5)$

Step 5.5.1.2.7

Multiply $- 5$ by $- 1$ .

$p (λ) = 9 λ^{2} - 36 λ - 45 - λ^{3} + 4 λ^{2} + 5 λ - 8 (- 3 λ - 10) + 7 (2 λ - 5)$

Step 5.5.1.3

Add $9 λ^{2}$ and $4 λ^{2}$ .

$p (λ) = 13 λ^{2} - 36 λ - 45 - λ^{3} + 5 λ - 8 (- 3 λ - 10) + 7 (2 λ - 5)$

Step 5.5.1.4

Add $- 36 λ$ and $5 λ$ .

$p (λ) = 13 λ^{2} - 31 λ - 45 - λ^{3} - 8 (- 3 λ - 10) + 7 (2 λ - 5)$

Step 5.5.1.5

Apply the distributive property.

$p (λ) = 13 λ^{2} - 31 λ - 45 - λ^{3} - 8 (- 3 λ) - 8 \cdot - 10 + 7 (2 λ - 5)$

Step 5.5.1.6

Multiply $- 3$ by $- 8$ .

$p (λ) = 13 λ^{2} - 31 λ - 45 - λ^{3} + 24 λ - 8 \cdot - 10 + 7 (2 λ - 5)$

Step 5.5.1.7

Multiply $- 8$ by $- 10$ .

$p (λ) = 13 λ^{2} - 31 λ - 45 - λ^{3} + 24 λ + 80 + 7 (2 λ - 5)$

Step 5.5.1.8

Apply the distributive property.

$p (λ) = 13 λ^{2} - 31 λ - 45 - λ^{3} + 24 λ + 80 + 7 (2 λ) + 7 \cdot - 5$

Step 5.5.1.9

Multiply $2$ by $7$ .

$p (λ) = 13 λ^{2} - 31 λ - 45 - λ^{3} + 24 λ + 80 + 14 λ + 7 \cdot - 5$

Step 5.5.1.10

Multiply $7$ by $- 5$ .

$p (λ) = 13 λ^{2} - 31 λ - 45 - λ^{3} + 24 λ + 80 + 14 λ - 35$

Step 5.5.2

Add $- 31 λ$ and $24 λ$ .

$p (λ) = 13 λ^{2} - 7 λ - 45 - λ^{3} + 80 + 14 λ - 35$

Step 5.5.3

Add $- 7 λ$ and $14 λ$ .

$p (λ) = 13 λ^{2} + 7 λ - 45 - λ^{3} + 80 - 35$

Step 5.5.4

Add $- 45$ and $80$ .

$p (λ) = 13 λ^{2} + 7 λ - λ^{3} + 35 - 35$

Step 5.5.5

Combine the opposite terms in $13 λ^{2} + 7 λ - λ^{3} + 35 - 35$ .

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Step 5.5.5.1

Subtract $35$ from $35$ .

$p (λ) = 13 λ^{2} + 7 λ - λ^{3} + 0$

Step 5.5.5.2

Add $13 λ^{2} + 7 λ - λ^{3}$ and $0$ .

$p (λ) = 13 λ^{2} + 7 λ - λ^{3}$

Step 5.5.6

Move $7 λ$ .

$p (λ) = 13 λ^{2} - λ^{3} + 7 λ$

Step 5.5.7

Reorder $13 λ^{2}$ and $- λ^{3}$ .

$p (λ) = - λ^{3} + 13 λ^{2} + 7 λ$

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