Linear Algebra Examples

$[\begin{array}{c} - 2 & 0 & 0 \\ 0 & - 2 & 0 \\ 0 & 0 & 13 \end{array}]$

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 $[\begin{array}{c} - 2 & 0 & 0 \\ 0 & - 2 & 0 \\ 0 & 0 & 13 \end{array}]$ for $A$ .

$p (λ) = determinant ([\begin{array}{c} - 2 & 0 & 0 \\ 0 & - 2 & 0 \\ 0 & 0 & 13 \end{array}] - λ I_{3})$

Step 3.2

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

$p (λ) = determinant ([\begin{array}{c} - 2 & 0 & 0 \\ 0 & - 2 & 0 \\ 0 & 0 & 13 \end{array}] - λ [\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}])$

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.

$p (λ) = determinant ([\begin{array}{c} - 2 & 0 & 0 \\ 0 & - 2 & 0 \\ 0 & 0 & 13 \end{array}] + [\begin{array}{c} - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2

Simplify each element in the matrix.

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Step 4.1.2.1

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} - 2 & 0 & 0 \\ 0 & - 2 & 0 \\ 0 & 0 & 13 \end{array}] + [\begin{array}{c} - λ & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.2

Multiply $- λ \cdot 0$ .

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Step 4.1.2.2.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} - 2 & 0 & 0 \\ 0 & - 2 & 0 \\ 0 & 0 & 13 \end{array}] + [\begin{array}{c} - λ & 0 λ & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.2.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} - 2 & 0 & 0 \\ 0 & - 2 & 0 \\ 0 & 0 & 13 \end{array}] + [\begin{array}{c} - λ & 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.3

Multiply $- λ \cdot 0$ .

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Step 4.1.2.3.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} - 2 & 0 & 0 \\ 0 & - 2 & 0 \\ 0 & 0 & 13 \end{array}] + [\begin{array}{c} - λ & 0 & 0 λ \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.3.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} - 2 & 0 & 0 \\ 0 & - 2 & 0 \\ 0 & 0 & 13 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.4

Multiply $- λ \cdot 0$ .

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Step 4.1.2.4.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} - 2 & 0 & 0 \\ 0 & - 2 & 0 \\ 0 & 0 & 13 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 λ & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.4.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} - 2 & 0 & 0 \\ 0 & - 2 & 0 \\ 0 & 0 & 13 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.5

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} - 2 & 0 & 0 \\ 0 & - 2 & 0 \\ 0 & 0 & 13 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.6

Multiply $- λ \cdot 0$ .

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Step 4.1.2.6.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} - 2 & 0 & 0 \\ 0 & - 2 & 0 \\ 0 & 0 & 13 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 λ \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.6.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} - 2 & 0 & 0 \\ 0 & - 2 & 0 \\ 0 & 0 & 13 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.7

Multiply $- λ \cdot 0$ .

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Step 4.1.2.7.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} - 2 & 0 & 0 \\ 0 & - 2 & 0 \\ 0 & 0 & 13 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 λ & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.7.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} - 2 & 0 & 0 \\ 0 & - 2 & 0 \\ 0 & 0 & 13 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.8

Multiply $- λ \cdot 0$ .

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Step 4.1.2.8.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} - 2 & 0 & 0 \\ 0 & - 2 & 0 \\ 0 & 0 & 13 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & 0 λ & - λ \cdot 1 \end{array}])$

Step 4.1.2.8.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} - 2 & 0 & 0 \\ 0 & - 2 & 0 \\ 0 & 0 & 13 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.9

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} - 2 & 0 & 0 \\ 0 & - 2 & 0 \\ 0 & 0 & 13 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & 0 & - λ \end{array}])$

Step 4.2

Add the corresponding elements.

$p (λ) = determinant [\begin{array}{c} - 2 - λ & 0 + 0 & 0 + 0 \\ 0 + 0 & - 2 - λ & 0 + 0 \\ 0 + 0 & 0 + 0 & 13 - λ \end{array}]$

Step 4.3

Simplify each element.

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Step 4.3.1

Add $0$ and $0$ .

$p (λ) = determinant [\begin{array}{c} - 2 - λ & 0 & 0 + 0 \\ 0 + 0 & - 2 - λ & 0 + 0 \\ 0 + 0 & 0 + 0 & 13 - λ \end{array}]$

Step 4.3.2

Add $0$ and $0$ .

$p (λ) = determinant [\begin{array}{c} - 2 - λ & 0 & 0 \\ 0 + 0 & - 2 - λ & 0 + 0 \\ 0 + 0 & 0 + 0 & 13 - λ \end{array}]$

Step 4.3.3

Add $0$ and $0$ .

$p (λ) = determinant [\begin{array}{c} - 2 - λ & 0 & 0 \\ 0 & - 2 - λ & 0 + 0 \\ 0 + 0 & 0 + 0 & 13 - λ \end{array}]$

Step 4.3.4

Add $0$ and $0$ .

$p (λ) = determinant [\begin{array}{c} - 2 - λ & 0 & 0 \\ 0 & - 2 - λ & 0 \\ 0 + 0 & 0 + 0 & 13 - λ \end{array}]$

Step 4.3.5

Add $0$ and $0$ .

$p (λ) = determinant [\begin{array}{c} - 2 - λ & 0 & 0 \\ 0 & - 2 - λ & 0 \\ 0 & 0 + 0 & 13 - λ \end{array}]$

Step 4.3.6

Add $0$ and $0$ .

$p (λ) = determinant [\begin{array}{c} - 2 - λ & 0 & 0 \\ 0 & - 2 - λ & 0 \\ 0 & 0 & 13 - λ \end{array}]$

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.

$| \begin{array}{c} - 2 - λ & 0 \\ 0 & 13 - λ \end{array} |$

Step 5.1.4

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

$(- 2 - λ) | \begin{array}{c} - 2 - λ & 0 \\ 0 & 13 - λ \end{array} |$

Step 5.1.5

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

$| \begin{array}{c} 0 & 0 \\ 0 & 13 - λ \end{array} |$

Step 5.1.6

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

$0 | \begin{array}{c} 0 & 0 \\ 0 & 13 - λ \end{array} |$

Step 5.1.7

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

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

Step 5.1.8

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

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

Step 5.1.9

Add the terms together.

$p (λ) = (- 2 - λ) | \begin{array}{c} - 2 - λ & 0 \\ 0 & 13 - λ \end{array} | + 0 | \begin{array}{c} 0 & 0 \\ 0 & 13 - λ \end{array} | + 0 | \begin{array}{c} 0 & - 2 - λ \\ 0 & 0 \end{array} |$

Step 5.2

Multiply $0$ by $| \begin{array}{c} 0 & 0 \\ 0 & 13 - λ \end{array} |$ .

$p (λ) = (- 2 - λ) | \begin{array}{c} - 2 - λ & 0 \\ 0 & 13 - λ \end{array} | + 0 + 0 | \begin{array}{c} 0 & - 2 - λ \\ 0 & 0 \end{array} |$

Step 5.3

Multiply $0$ by $| \begin{array}{c} 0 & - 2 - λ \\ 0 & 0 \end{array} |$ .

$p (λ) = (- 2 - λ) | \begin{array}{c} - 2 - λ & 0 \\ 0 & 13 - λ \end{array} | + 0 + 0$

Step 5.4

Evaluate $| \begin{array}{c} - 2 - λ & 0 \\ 0 & 13 - λ \end{array} |$ .

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Step 5.4.1

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

$p (λ) = (- 2 - λ) ((- 2 - λ) (13 - λ) + 0 \cdot 0) + 0 + 0$

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

Expand $(- 2 - λ) (13 - λ)$ using the FOIL Method.

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Step 5.4.2.1.1.1

Apply the distributive property.

$p (λ) = (- 2 - λ) (- 2 (13 - λ) - λ (13 - λ) + 0 \cdot 0) + 0 + 0$

Step 5.4.2.1.1.2

Apply the distributive property.

$p (λ) = (- 2 - λ) (- 2 \cdot 13 - 2 (- λ) - λ (13 - λ) + 0 \cdot 0) + 0 + 0$

Step 5.4.2.1.1.3

Apply the distributive property.

$p (λ) = (- 2 - λ) (- 2 \cdot 13 - 2 (- λ) - λ \cdot 13 - λ (- λ) + 0 \cdot 0) + 0 + 0$

Step 5.4.2.1.2

Simplify and combine like terms.

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Step 5.4.2.1.2.1

Simplify each term.

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Step 5.4.2.1.2.1.1

Multiply $- 2$ by $13$ .

$p (λ) = (- 2 - λ) (- 26 - 2 (- λ) - λ \cdot 13 - λ (- λ) + 0 \cdot 0) + 0 + 0$

Step 5.4.2.1.2.1.2

Multiply $- 1$ by $- 2$ .

$p (λ) = (- 2 - λ) (- 26 + 2 λ - λ \cdot 13 - λ (- λ) + 0 \cdot 0) + 0 + 0$

Step 5.4.2.1.2.1.3

Multiply $13$ by $- 1$ .

$p (λ) = (- 2 - λ) (- 26 + 2 λ - 13 λ - λ (- λ) + 0 \cdot 0) + 0 + 0$

Step 5.4.2.1.2.1.4

Rewrite using the commutative property of multiplication.

$p (λ) = (- 2 - λ) (- 26 + 2 λ - 13 λ - 1 \cdot - 1 λ \cdot λ + 0 \cdot 0) + 0 + 0$

Step 5.4.2.1.2.1.5

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

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Step 5.4.2.1.2.1.5.1

Move $λ$ .

$p (λ) = (- 2 - λ) (- 26 + 2 λ - 13 λ - 1 \cdot - 1 (λ \cdot λ) + 0 \cdot 0) + 0 + 0$

Step 5.4.2.1.2.1.5.2

Multiply $λ$ by $λ$ .

$p (λ) = (- 2 - λ) (- 26 + 2 λ - 13 λ - 1 \cdot - 1 λ^{2} + 0 \cdot 0) + 0 + 0$

Step 5.4.2.1.2.1.6

Multiply $- 1$ by $- 1$ .

$p (λ) = (- 2 - λ) (- 26 + 2 λ - 13 λ + 1 λ^{2} + 0 \cdot 0) + 0 + 0$

Step 5.4.2.1.2.1.7

Multiply $λ^{2}$ by $1$ .

$p (λ) = (- 2 - λ) (- 26 + 2 λ - 13 λ + λ^{2} + 0 \cdot 0) + 0 + 0$

Step 5.4.2.1.2.2

Subtract $13 λ$ from $2 λ$ .

$p (λ) = (- 2 - λ) (- 26 - 11 λ + λ^{2} + 0 \cdot 0) + 0 + 0$

Step 5.4.2.1.3

Multiply $0$ by $0$ .

$p (λ) = (- 2 - λ) (- 26 - 11 λ + λ^{2} + 0) + 0 + 0$

Step 5.4.2.2

Add $- 26 - 11 λ + λ^{2}$ and $0$ .

$p (λ) = (- 2 - λ) (- 26 - 11 λ + λ^{2}) + 0 + 0$

Step 5.4.2.3

Move $- 26$ .

$p (λ) = (- 2 - λ) (- 11 λ + λ^{2} - 26) + 0 + 0$

Step 5.4.2.4

Reorder $- 11 λ$ and $λ^{2}$ .

$p (λ) = (- 2 - λ) (λ^{2} - 11 λ - 26) + 0 + 0$

Step 5.5

Simplify the determinant.

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Step 5.5.1

Combine the opposite terms in $(- 2 - λ) (λ^{2} - 11 λ - 26) + 0 + 0$ .

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Step 5.5.1.1

Add $(- 2 - λ) (λ^{2} - 11 λ - 26)$ and $0$ .

$p (λ) = (- 2 - λ) (λ^{2} - 11 λ - 26) + 0$

Step 5.5.1.2

Add $(- 2 - λ) (λ^{2} - 11 λ - 26)$ and $0$ .

$p (λ) = (- 2 - λ) (λ^{2} - 11 λ - 26)$

Step 5.5.2

Expand $(- 2 - λ) (λ^{2} - 11 λ - 26)$ by multiplying each term in the first expression by each term in the second expression.

$p (λ) = - 2 λ^{2} - 2 (- 11 λ) - 2 \cdot - 26 - λ \cdot λ^{2} - λ (- 11 λ) - λ \cdot - 26$

Step 5.5.3

Simplify each term.

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Step 5.5.3.1

Multiply $- 11$ by $- 2$ .

$p (λ) = - 2 λ^{2} + 22 λ - 2 \cdot - 26 - λ \cdot λ^{2} - λ (- 11 λ) - λ \cdot - 26$

Step 5.5.3.2

Multiply $- 2$ by $- 26$ .

$p (λ) = - 2 λ^{2} + 22 λ + 52 - λ \cdot λ^{2} - λ (- 11 λ) - λ \cdot - 26$

Step 5.5.3.3

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

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Step 5.5.3.3.1

Move $λ^{2}$ .

$p (λ) = - 2 λ^{2} + 22 λ + 52 - (λ^{2} λ) - λ (- 11 λ) - λ \cdot - 26$

Step 5.5.3.3.2

Multiply $λ^{2}$ by $λ$ .

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Step 5.5.3.3.2.1

Raise $λ$ to the power of $1$ .

$p (λ) = - 2 λ^{2} + 22 λ + 52 - (λ^{2} λ^{1}) - λ (- 11 λ) - λ \cdot - 26$

Step 5.5.3.3.2.2

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

$p (λ) = - 2 λ^{2} + 22 λ + 52 - λ^{2 + 1} - λ (- 11 λ) - λ \cdot - 26$

Step 5.5.3.3.3

Add $2$ and $1$ .

$p (λ) = - 2 λ^{2} + 22 λ + 52 - λ^{3} - λ (- 11 λ) - λ \cdot - 26$

Step 5.5.3.4

Rewrite using the commutative property of multiplication.

$p (λ) = - 2 λ^{2} + 22 λ + 52 - λ^{3} - 1 \cdot - 11 λ \cdot λ - λ \cdot - 26$

Step 5.5.3.5

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

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Step 5.5.3.5.1

Move $λ$ .

$p (λ) = - 2 λ^{2} + 22 λ + 52 - λ^{3} - 1 \cdot - 11 (λ \cdot λ) - λ \cdot - 26$

Step 5.5.3.5.2

Multiply $λ$ by $λ$ .

$p (λ) = - 2 λ^{2} + 22 λ + 52 - λ^{3} - 1 \cdot - 11 λ^{2} - λ \cdot - 26$

Step 5.5.3.6

Multiply $- 1$ by $- 11$ .

$p (λ) = - 2 λ^{2} + 22 λ + 52 - λ^{3} + 11 λ^{2} - λ \cdot - 26$

Step 5.5.3.7

Multiply $- 26$ by $- 1$ .

$p (λ) = - 2 λ^{2} + 22 λ + 52 - λ^{3} + 11 λ^{2} + 26 λ$

Step 5.5.4

Add $- 2 λ^{2}$ and $11 λ^{2}$ .

$p (λ) = 9 λ^{2} + 22 λ + 52 - λ^{3} + 26 λ$

Step 5.5.5

Add $22 λ$ and $26 λ$ .

$p (λ) = 9 λ^{2} + 48 λ + 52 - λ^{3}$

Step 5.5.6

Move $52$ .

$p (λ) = 9 λ^{2} + 48 λ - λ^{3} + 52$

Step 5.5.7

Move $48 λ$ .

$p (λ) = 9 λ^{2} - λ^{3} + 48 λ + 52$

Step 5.5.8

Reorder $9 λ^{2}$ and $- λ^{3}$ .

$p (λ) = - λ^{3} + 9 λ^{2} + 48 λ + 52$

Step 6

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

$- λ^{3} + 9 λ^{2} + 48 λ + 52 = 0$

Step 7

Solve for $λ$ .

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Step 7.1

Factor the left side of the equation.

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Step 7.1.1

Factor $- λ^{3} + 9 λ^{2} + 48 λ + 52$ using the rational roots test.

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Step 7.1.1.1

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.

$p = \pm 1, \pm 52, \pm 2, \pm 26, \pm 4, \pm 13$

$q = \pm 1$

Step 7.1.1.2

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

$\pm 1, \pm 52, \pm 2, \pm 26, \pm 4, \pm 13$

Step 7.1.1.3

Substitute $- 2$ and simplify the expression. In this case, the expression is equal to $0$ so $- 2$ is a root of the polynomial.

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Step 7.1.1.3.1

Substitute $- 2$ into the polynomial.

$- {(- 2)}^{3} + 9 {(- 2)}^{2} + 48 \cdot - 2 + 52$

Step 7.1.1.3.2

Raise $- 2$ to the power of $3$ .

$- - 8 + 9 {(- 2)}^{2} + 48 \cdot - 2 + 52$

Step 7.1.1.3.3

Multiply $- 1$ by $- 8$ .

$8 + 9 {(- 2)}^{2} + 48 \cdot - 2 + 52$

Step 7.1.1.3.4

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

$8 + 9 \cdot 4 + 48 \cdot - 2 + 52$

Step 7.1.1.3.5

Multiply $9$ by $4$ .

$8 + 36 + 48 \cdot - 2 + 52$

Step 7.1.1.3.6

Add $8$ and $36$ .

$44 + 48 \cdot - 2 + 52$

Step 7.1.1.3.7

Multiply $48$ by $- 2$ .

$44 - 96 + 52$

Step 7.1.1.3.8

Subtract $96$ from $44$ .

$- 52 + 52$

Step 7.1.1.3.9

Add $- 52$ and $52$ .

$0$

Step 7.1.1.4

Since $- 2$ is a known root, divide the polynomial by $λ + 2$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{- λ^{3} + 9 λ^{2} + 48 λ + 52}{λ + 2}$

Step 7.1.1.5

Divide $- λ^{3} + 9 λ^{2} + 48 λ + 52$ by $λ + 2$ .

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Step 7.1.1.5.1

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

$λ$

$2$

$λ^{3}$

$9 λ^{2}$

$48 λ$

$52$

Step 7.1.1.5.2

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

				-	$λ^{2}$
	$λ$	+	$2$	-	$λ^{3}$	+	$9 λ^{2}$	+	$48 λ$	+	$52$

Step 7.1.1.5.3

Multiply the new quotient term by the divisor.

			-	$λ^{2}$
$λ$	+	$2$	-	$λ^{3}$	+	$9 λ^{2}$	+	$48 λ$	+	$52$
			-	$λ^{3}$	-	$2 λ^{2}$

Step 7.1.1.5.4

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

			-	$λ^{2}$
$λ$	+	$2$	-	$λ^{3}$	+	$9 λ^{2}$	+	$48 λ$	+	$52$
			+	$λ^{3}$	+	$2 λ^{2}$

Step 7.1.1.5.5

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

			-	$λ^{2}$
$λ$	+	$2$	-	$λ^{3}$	+	$9 λ^{2}$	+	$48 λ$	+	$52$
			+	$λ^{3}$	+	$2 λ^{2}$
					+	$11 λ^{2}$

Step 7.1.1.5.6

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

			-	$λ^{2}$
$λ$	+	$2$	-	$λ^{3}$	+	$9 λ^{2}$	+	$48 λ$	+	$52$
			+	$λ^{3}$	+	$2 λ^{2}$
					+	$11 λ^{2}$	+	$48 λ$

Step 7.1.1.5.7

Divide the highest order term in the dividend $11 λ^{2}$ by the highest order term in divisor $λ$ .

			-	$λ^{2}$	+	$11 λ$
$λ$	+	$2$	-	$λ^{3}$	+	$9 λ^{2}$	+	$48 λ$	+	$52$
			+	$λ^{3}$	+	$2 λ^{2}$
					+	$11 λ^{2}$	+	$48 λ$

Step 7.1.1.5.8

Multiply the new quotient term by the divisor.

			-	$λ^{2}$	+	$11 λ$
$λ$	+	$2$	-	$λ^{3}$	+	$9 λ^{2}$	+	$48 λ$	+	$52$
			+	$λ^{3}$	+	$2 λ^{2}$
					+	$11 λ^{2}$	+	$48 λ$
					+	$11 λ^{2}$	+	$22 λ$

Step 7.1.1.5.9

The expression needs to be subtracted from the dividend, so change all the signs in $11 λ^{2} + 22 λ$

			-	$λ^{2}$	+	$11 λ$
$λ$	+	$2$	-	$λ^{3}$	+	$9 λ^{2}$	+	$48 λ$	+	$52$
			+	$λ^{3}$	+	$2 λ^{2}$
					+	$11 λ^{2}$	+	$48 λ$
					-	$11 λ^{2}$	-	$22 λ$

Step 7.1.1.5.10

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

			-	$λ^{2}$	+	$11 λ$
$λ$	+	$2$	-	$λ^{3}$	+	$9 λ^{2}$	+	$48 λ$	+	$52$
			+	$λ^{3}$	+	$2 λ^{2}$
					+	$11 λ^{2}$	+	$48 λ$
					-	$11 λ^{2}$	-	$22 λ$
							+	$26 λ$

Step 7.1.1.5.11

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

			-	$λ^{2}$	+	$11 λ$
$λ$	+	$2$	-	$λ^{3}$	+	$9 λ^{2}$	+	$48 λ$	+	$52$
			+	$λ^{3}$	+	$2 λ^{2}$
					+	$11 λ^{2}$	+	$48 λ$
					-	$11 λ^{2}$	-	$22 λ$
							+	$26 λ$	+	$52$

Step 7.1.1.5.12

Divide the highest order term in the dividend $26 λ$ by the highest order term in divisor $λ$ .

			-	$λ^{2}$	+	$11 λ$	+	$26$
$λ$	+	$2$	-	$λ^{3}$	+	$9 λ^{2}$	+	$48 λ$	+	$52$
			+	$λ^{3}$	+	$2 λ^{2}$
					+	$11 λ^{2}$	+	$48 λ$
					-	$11 λ^{2}$	-	$22 λ$
							+	$26 λ$	+	$52$

Step 7.1.1.5.13

Multiply the new quotient term by the divisor.

			-	$λ^{2}$	+	$11 λ$	+	$26$
$λ$	+	$2$	-	$λ^{3}$	+	$9 λ^{2}$	+	$48 λ$	+	$52$
			+	$λ^{3}$	+	$2 λ^{2}$
					+	$11 λ^{2}$	+	$48 λ$
					-	$11 λ^{2}$	-	$22 λ$
							+	$26 λ$	+	$52$
							+	$26 λ$	+	$52$

Step 7.1.1.5.14

The expression needs to be subtracted from the dividend, so change all the signs in $26 λ + 52$

			-	$λ^{2}$	+	$11 λ$	+	$26$
$λ$	+	$2$	-	$λ^{3}$	+	$9 λ^{2}$	+	$48 λ$	+	$52$
			+	$λ^{3}$	+	$2 λ^{2}$
					+	$11 λ^{2}$	+	$48 λ$
					-	$11 λ^{2}$	-	$22 λ$
							+	$26 λ$	+	$52$
							-	$26 λ$	-	$52$

Step 7.1.1.5.15

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

			-	$λ^{2}$	+	$11 λ$	+	$26$
$λ$	+	$2$	-	$λ^{3}$	+	$9 λ^{2}$	+	$48 λ$	+	$52$
			+	$λ^{3}$	+	$2 λ^{2}$
					+	$11 λ^{2}$	+	$48 λ$
					-	$11 λ^{2}$	-	$22 λ$
							+	$26 λ$	+	$52$
							-	$26 λ$	-	$52$
										$0$

Step 7.1.1.5.16

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

$- λ^{2} + 11 λ + 26$

Step 7.1.1.6

Write $- λ^{3} + 9 λ^{2} + 48 λ + 52$ as a set of factors.

$(λ + 2) (- λ^{2} + 11 λ + 26) = 0$

Step 7.1.2

Factor by grouping.

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Step 7.1.2.1

Factor by grouping.

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Step 7.1.2.1.1

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 $a \cdot c = - 1 \cdot 26 = - 26$ and whose sum is $b = 11$ .

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Step 7.1.2.1.1.1

Factor $11$ out of $11 λ$ .

$(λ + 2) (- λ^{2} + 11 (λ) + 26) = 0$

Step 7.1.2.1.1.2

Rewrite $11$ as $- 2$ plus $13$

$(λ + 2) (- λ^{2} + (- 2 + 13) λ + 26) = 0$

Step 7.1.2.1.1.3

Apply the distributive property.

$(λ + 2) (- λ^{2} - 2 λ + 13 λ + 26) = 0$

Step 7.1.2.1.2

Factor out the greatest common factor from each group.

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Step 7.1.2.1.2.1

Group the first two terms and the last two terms.

$(λ + 2) ((- λ^{2} - 2 λ) + 13 λ + 26) = 0$

Step 7.1.2.1.2.2

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

$(λ + 2) (λ (- λ - 2) - 13 (- λ - 2)) = 0$

Step 7.1.2.1.3

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

$(λ + 2) ((- λ - 2) (λ - 13)) = 0$

Step 7.1.2.2

Remove unnecessary parentheses.

$(λ + 2) (- λ - 2) (λ - 13) = 0$

Step 7.2

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

$λ + 2 = 0$

$- λ - 2 = 0$

$λ - 13 = 0$

Step 7.3

Set $λ + 2$ equal to $0$ and solve for $λ$ .

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Step 7.3.1

Set $λ + 2$ equal to $0$ .

$λ + 2 = 0$

Step 7.3.2

Subtract $2$ from both sides of the equation.

$λ = - 2$

Step 7.4

Set $λ - 13$ equal to $0$ and solve for $λ$ .

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Step 7.4.1

Set $λ - 13$ equal to $0$ .

$λ - 13 = 0$

Step 7.4.2

Add $13$ to both sides of the equation.

$λ = 13$

Step 7.5

The final solution is all the values that make $(λ + 2) (- λ - 2) (λ - 13) = 0$ true.

$λ = - 2, 13$