Linear Algebra Examples

$[\begin{array}{c} - 4 & 0 & 1 \\ 3 & - 6 & 3 \\ 1 & 0 & - 4 \end{array}]$

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 $[\begin{array}{c} - 4 & 0 & 1 \\ 3 & - 6 & 3 \\ 1 & 0 & - 4 \end{array}]$ for $A$ .

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

Step 3.2

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

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

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.

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

Step 4.1.2

Simplify each element in the matrix.

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Step 4.1.2.1

Multiply $- 1$ by $1$ .

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

Step 4.1.2.2

Multiply $- λ \cdot 0$ .

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Step 4.1.2.2.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.2.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.3

Multiply $- λ \cdot 0$ .

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Step 4.1.2.3.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.3.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.4

Multiply $- λ \cdot 0$ .

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Step 4.1.2.4.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.4.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.5

Multiply $- 1$ by $1$ .

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

Step 4.1.2.6

Multiply $- λ \cdot 0$ .

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Step 4.1.2.6.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.6.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.7

Multiply $- λ \cdot 0$ .

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Step 4.1.2.7.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.7.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.8

Multiply $- λ \cdot 0$ .

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Step 4.1.2.8.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.8.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.9

Multiply $- 1$ by $1$ .

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

Step 4.2

Add the corresponding elements.

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

Step 4.3

Simplify each element.

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Step 4.3.1

Add $0$ and $0$ .

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

Step 4.3.2

Add $1$ and $0$ .

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

Step 4.3.3

Add $3$ and $0$ .

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

Step 4.3.4

Add $3$ and $0$ .

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

Step 4.3.5

Add $1$ and $0$ .

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

Step 4.3.6

Add $0$ and $0$ .

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

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

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

Step 5.1.4

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

$0 | \begin{array}{c} 3 & 3 \\ 1 & - 4 - λ \end{array} |$

Step 5.1.5

The minor for $a_{22}$ is the determinant with row $2$ and column $2$ deleted.

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

Step 5.1.6

Multiply element $a_{22}$ by its cofactor.

$(- 6 - λ) | \begin{array}{c} - 4 - λ & 1 \\ 1 & - 4 - λ \end{array} |$

Step 5.1.7

The minor for $a_{32}$ is the determinant with row $3$ and column $2$ deleted.

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

Step 5.1.8

Multiply element $a_{32}$ by its cofactor.

$0 | \begin{array}{c} - 4 - λ & 1 \\ 3 & 3 \end{array} |$

Step 5.1.9

Add the terms together.

$p (λ) = 0 | \begin{array}{c} 3 & 3 \\ 1 & - 4 - λ \end{array} | + (- 6 - λ) | \begin{array}{c} - 4 - λ & 1 \\ 1 & - 4 - λ \end{array} | + 0 | \begin{array}{c} - 4 - λ & 1 \\ 3 & 3 \end{array} |$

Step 5.2

Multiply $0$ by $| \begin{array}{c} 3 & 3 \\ 1 & - 4 - λ \end{array} |$ .

$p (λ) = 0 + (- 6 - λ) | \begin{array}{c} - 4 - λ & 1 \\ 1 & - 4 - λ \end{array} | + 0 | \begin{array}{c} - 4 - λ & 1 \\ 3 & 3 \end{array} |$

Step 5.3

Multiply $0$ by $| \begin{array}{c} - 4 - λ & 1 \\ 3 & 3 \end{array} |$ .

$p (λ) = 0 + (- 6 - λ) | \begin{array}{c} - 4 - λ & 1 \\ 1 & - 4 - λ \end{array} | + 0$

Step 5.4

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

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Step 5.4.1

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

$p (λ) = 0 + (- 6 - λ) ((- 4 - λ) (- 4 - λ) - 1 \cdot 1) + 0$

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 $(- 4 - λ) (- 4 - λ)$ using the FOIL Method.

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Step 5.4.2.1.1.1

Apply the distributive property.

$p (λ) = 0 + (- 6 - λ) (- 4 (- 4 - λ) - λ (- 4 - λ) - 1 \cdot 1) + 0$

Step 5.4.2.1.1.2

Apply the distributive property.

$p (λ) = 0 + (- 6 - λ) (- 4 \cdot - 4 - 4 (- λ) - λ (- 4 - λ) - 1 \cdot 1) + 0$

Step 5.4.2.1.1.3

Apply the distributive property.

$p (λ) = 0 + (- 6 - λ) (- 4 \cdot - 4 - 4 (- λ) - λ \cdot - 4 - λ (- λ) - 1 \cdot 1) + 0$

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 $- 4$ by $- 4$ .

$p (λ) = 0 + (- 6 - λ) (16 - 4 (- λ) - λ \cdot - 4 - λ (- λ) - 1 \cdot 1) + 0$

Step 5.4.2.1.2.1.2

Multiply $- 1$ by $- 4$ .

$p (λ) = 0 + (- 6 - λ) (16 + 4 λ - λ \cdot - 4 - λ (- λ) - 1 \cdot 1) + 0$

Step 5.4.2.1.2.1.3

Multiply $- 4$ by $- 1$ .

$p (λ) = 0 + (- 6 - λ) (16 + 4 λ + 4 λ - λ (- λ) - 1 \cdot 1) + 0$

Step 5.4.2.1.2.1.4

Rewrite using the commutative property of multiplication.

$p (λ) = 0 + (- 6 - λ) (16 + 4 λ + 4 λ - 1 \cdot - 1 λ \cdot λ - 1 \cdot 1) + 0$

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 $λ$ .

$p (λ) = 0 + (- 6 - λ) (16 + 4 λ + 4 λ - 1 \cdot - 1 (λ \cdot λ) - 1 \cdot 1) + 0$

Step 5.4.2.1.2.1.5.2

Multiply $λ$ by $λ$ .

$p (λ) = 0 + (- 6 - λ) (16 + 4 λ + 4 λ - 1 \cdot - 1 λ^{2} - 1 \cdot 1) + 0$

Step 5.4.2.1.2.1.6

Multiply $- 1$ by $- 1$ .

$p (λ) = 0 + (- 6 - λ) (16 + 4 λ + 4 λ + 1 λ^{2} - 1 \cdot 1) + 0$

Step 5.4.2.1.2.1.7

Multiply $λ^{2}$ by $1$ .

$p (λ) = 0 + (- 6 - λ) (16 + 4 λ + 4 λ + λ^{2} - 1 \cdot 1) + 0$

Step 5.4.2.1.2.2

Add $4 λ$ and $4 λ$ .

$p (λ) = 0 + (- 6 - λ) (16 + 8 λ + λ^{2} - 1 \cdot 1) + 0$

Step 5.4.2.1.3

Multiply $- 1$ by $1$ .

$p (λ) = 0 + (- 6 - λ) (16 + 8 λ + λ^{2} - 1) + 0$

Step 5.4.2.2

Subtract $1$ from $16$ .

$p (λ) = 0 + (- 6 - λ) (8 λ + λ^{2} + 15) + 0$

Step 5.4.2.3

Reorder $8 λ$ and $λ^{2}$ .

$p (λ) = 0 + (- 6 - λ) (λ^{2} + 8 λ + 15) + 0$

Step 5.5

Simplify the determinant.

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Step 5.5.1

Combine the opposite terms in $0 + (- 6 - λ) (λ^{2} + 8 λ + 15) + 0$ .

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Step 5.5.1.1

Add $0$ and $(- 6 - λ) (λ^{2} + 8 λ + 15)$ .

$p (λ) = (- 6 - λ) (λ^{2} + 8 λ + 15) + 0$

Step 5.5.1.2

Add $(- 6 - λ) (λ^{2} + 8 λ + 15)$ and $0$ .

$p (λ) = (- 6 - λ) (λ^{2} + 8 λ + 15)$

Step 5.5.2

Expand $(- 6 - λ) (λ^{2} + 8 λ + 15)$ by multiplying each term in the first expression by each term in the second expression.

$p (λ) = - 6 λ^{2} - 6 (8 λ) - 6 \cdot 15 - λ \cdot λ^{2} - λ (8 λ) - λ \cdot 15$

Step 5.5.3

Simplify each term.

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Step 5.5.3.1

Multiply $8$ by $- 6$ .

$p (λ) = - 6 λ^{2} - 48 λ - 6 \cdot 15 - λ \cdot λ^{2} - λ (8 λ) - λ \cdot 15$

Step 5.5.3.2

Multiply $- 6$ by $15$ .

$p (λ) = - 6 λ^{2} - 48 λ - 90 - λ \cdot λ^{2} - λ (8 λ) - λ \cdot 15$

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 $λ^{2}$ .

$p (λ) = - 6 λ^{2} - 48 λ - 90 - (λ^{2} λ) - λ (8 λ) - λ \cdot 15$

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 $1$ .

$p (λ) = - 6 λ^{2} - 48 λ - 90 - (λ^{2} λ^{1}) - λ (8 λ) - λ \cdot 15$

Step 5.5.3.3.2.2

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

$p (λ) = - 6 λ^{2} - 48 λ - 90 - λ^{2 + 1} - λ (8 λ) - λ \cdot 15$

Step 5.5.3.3.3

Add $2$ and $1$ .

$p (λ) = - 6 λ^{2} - 48 λ - 90 - λ^{3} - λ (8 λ) - λ \cdot 15$

Step 5.5.3.4

Rewrite using the commutative property of multiplication.

$p (λ) = - 6 λ^{2} - 48 λ - 90 - λ^{3} - 1 \cdot 8 λ \cdot λ - λ \cdot 15$

Step 5.5.3.5

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

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Step 5.5.3.5.1

Move $λ$ .

$p (λ) = - 6 λ^{2} - 48 λ - 90 - λ^{3} - 1 \cdot 8 (λ \cdot λ) - λ \cdot 15$

Step 5.5.3.5.2

Multiply $λ$ by $λ$ .

$p (λ) = - 6 λ^{2} - 48 λ - 90 - λ^{3} - 1 \cdot 8 λ^{2} - λ \cdot 15$

Step 5.5.3.6

Multiply $- 1$ by $8$ .

$p (λ) = - 6 λ^{2} - 48 λ - 90 - λ^{3} - 8 λ^{2} - λ \cdot 15$

Step 5.5.3.7

Multiply $15$ by $- 1$ .

$p (λ) = - 6 λ^{2} - 48 λ - 90 - λ^{3} - 8 λ^{2} - 15 λ$

Step 5.5.4

Subtract $8 λ^{2}$ from $- 6 λ^{2}$ .

$p (λ) = - 14 λ^{2} - 48 λ - 90 - λ^{3} - 15 λ$

Step 5.5.5

Subtract $15 λ$ from $- 48 λ$ .

$p (λ) = - 14 λ^{2} - 63 λ - 90 - λ^{3}$

Step 5.5.6

Move $- 90$ .

$p (λ) = - 14 λ^{2} - 63 λ - λ^{3} - 90$

Step 5.5.7

Move $- 63 λ$ .

$p (λ) = - 14 λ^{2} - λ^{3} - 63 λ - 90$

Step 5.5.8

Reorder $- 14 λ^{2}$ and $- λ^{3}$ .

$p (λ) = - λ^{3} - 14 λ^{2} - 63 λ - 90$

Step 6

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

$- λ^{3} - 14 λ^{2} - 63 λ - 90 = 0$

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 $- 1$ out of $- λ^{3} - 14 λ^{2} - 63 λ - 90$ .

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Step 7.1.1.1

Factor $- 1$ out of $- λ^{3}$ .

$- (λ^{3}) - 14 λ^{2} - 63 λ - 90 = 0$

Step 7.1.1.2

Factor $- 1$ out of $- 14 λ^{2}$ .

$- (λ^{3}) - (14 λ^{2}) - 63 λ - 90 = 0$

Step 7.1.1.3

Factor $- 1$ out of $- 63 λ$ .

$- (λ^{3}) - (14 λ^{2}) - (63 λ) - 90 = 0$

Step 7.1.1.4

Rewrite $- 90$ as $- 1 (90)$ .

$- (λ^{3}) - (14 λ^{2}) - (63 λ) - 1 \cdot 90 = 0$

Step 7.1.1.5

Factor $- 1$ out of $- (λ^{3}) - (14 λ^{2})$ .

$- (λ^{3} + 14 λ^{2}) - (63 λ) - 1 \cdot 90 = 0$

Step 7.1.1.6

Factor $- 1$ out of $- (λ^{3} + 14 λ^{2}) - (63 λ)$ .

$- (λ^{3} + 14 λ^{2} + 63 λ) - 1 \cdot 90 = 0$

Step 7.1.1.7

Factor $- 1$ out of $- (λ^{3} + 14 λ^{2} + 63 λ) - 1 (90)$ .

$- (λ^{3} + 14 λ^{2} + 63 λ + 90) = 0$

Step 7.1.2

Factor $λ^{3} + 14 λ^{2} + 63 λ + 90$ using the rational roots test.

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Step 7.1.2.1

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.

$p = \pm 1, \pm 90, \pm 2, \pm 45, \pm 3, \pm 30, \pm 5, \pm 18, \pm 6, \pm 15, \pm 9, \pm 10$

$q = \pm 1$

Step 7.1.2.2

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

$\pm 1, \pm 90, \pm 2, \pm 45, \pm 3, \pm 30, \pm 5, \pm 18, \pm 6, \pm 15, \pm 9, \pm 10$

Step 7.1.2.3

Substitute $- 3$ and simplify the expression. In this case, the expression is equal to $0$ so $- 3$ is a root of the polynomial.

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Step 7.1.2.3.1

Substitute $- 3$ into the polynomial.

${(- 3)}^{3} + 14 {(- 3)}^{2} + 63 \cdot - 3 + 90$

Step 7.1.2.3.2

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

$- 27 + 14 {(- 3)}^{2} + 63 \cdot - 3 + 90$

Step 7.1.2.3.3

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

$- 27 + 14 \cdot 9 + 63 \cdot - 3 + 90$

Step 7.1.2.3.4

Multiply $14$ by $9$ .

$- 27 + 126 + 63 \cdot - 3 + 90$

Step 7.1.2.3.5

Add $- 27$ and $126$ .

$99 + 63 \cdot - 3 + 90$

Step 7.1.2.3.6

Multiply $63$ by $- 3$ .

$99 - 189 + 90$

Step 7.1.2.3.7

Subtract $189$ from $99$ .

$- 90 + 90$

Step 7.1.2.3.8

Add $- 90$ and $90$ .

$0$

Step 7.1.2.4

Since $- 3$ is a known root, divide the polynomial by $λ + 3$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{λ^{3} + 14 λ^{2} + 63 λ + 90}{λ + 3}$

Step 7.1.2.5

Divide $λ^{3} + 14 λ^{2} + 63 λ + 90$ by $λ + 3$ .

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Step 7.1.2.5.1

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

$λ$

$3$

$λ^{3}$

$14 λ^{2}$

$63 λ$

$90$

Step 7.1.2.5.2

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

					$λ^{2}$
	$λ$	+	$3$		$λ^{3}$	+	$14 λ^{2}$	+	$63 λ$	+	$90$

Step 7.1.2.5.3

Multiply the new quotient term by the divisor.

				$λ^{2}$
$λ$	+	$3$		$λ^{3}$	+	$14 λ^{2}$	+	$63 λ$	+	$90$
			+	$λ^{3}$	+	$3 λ^{2}$

Step 7.1.2.5.4

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

				$λ^{2}$
$λ$	+	$3$		$λ^{3}$	+	$14 λ^{2}$	+	$63 λ$	+	$90$
			-	$λ^{3}$	-	$3 λ^{2}$

Step 7.1.2.5.5

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

				$λ^{2}$
$λ$	+	$3$		$λ^{3}$	+	$14 λ^{2}$	+	$63 λ$	+	$90$
			-	$λ^{3}$	-	$3 λ^{2}$
					+	$11 λ^{2}$

Step 7.1.2.5.6

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

				$λ^{2}$
$λ$	+	$3$		$λ^{3}$	+	$14 λ^{2}$	+	$63 λ$	+	$90$
			-	$λ^{3}$	-	$3 λ^{2}$
					+	$11 λ^{2}$	+	$63 λ$

Step 7.1.2.5.7

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

				$λ^{2}$	+	$11 λ$
$λ$	+	$3$		$λ^{3}$	+	$14 λ^{2}$	+	$63 λ$	+	$90$
			-	$λ^{3}$	-	$3 λ^{2}$
					+	$11 λ^{2}$	+	$63 λ$

Step 7.1.2.5.8

Multiply the new quotient term by the divisor.

				$λ^{2}$	+	$11 λ$
$λ$	+	$3$		$λ^{3}$	+	$14 λ^{2}$	+	$63 λ$	+	$90$
			-	$λ^{3}$	-	$3 λ^{2}$
					+	$11 λ^{2}$	+	$63 λ$
					+	$11 λ^{2}$	+	$33 λ$

Step 7.1.2.5.9

The expression needs to be subtracted from the dividend, so change all the signs in $11 λ^{2} + 33 λ$

				$λ^{2}$	+	$11 λ$
$λ$	+	$3$		$λ^{3}$	+	$14 λ^{2}$	+	$63 λ$	+	$90$
			-	$λ^{3}$	-	$3 λ^{2}$
					+	$11 λ^{2}$	+	$63 λ$
					-	$11 λ^{2}$	-	$33 λ$

Step 7.1.2.5.10

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

				$λ^{2}$	+	$11 λ$
$λ$	+	$3$		$λ^{3}$	+	$14 λ^{2}$	+	$63 λ$	+	$90$
			-	$λ^{3}$	-	$3 λ^{2}$
					+	$11 λ^{2}$	+	$63 λ$
					-	$11 λ^{2}$	-	$33 λ$
							+	$30 λ$

Step 7.1.2.5.11

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

				$λ^{2}$	+	$11 λ$
$λ$	+	$3$		$λ^{3}$	+	$14 λ^{2}$	+	$63 λ$	+	$90$
			-	$λ^{3}$	-	$3 λ^{2}$
					+	$11 λ^{2}$	+	$63 λ$
					-	$11 λ^{2}$	-	$33 λ$
							+	$30 λ$	+	$90$

Step 7.1.2.5.12

Divide the highest order term in the dividend $30 λ$ by the highest order term in divisor $λ$ .

				$λ^{2}$	+	$11 λ$	+	$30$
$λ$	+	$3$		$λ^{3}$	+	$14 λ^{2}$	+	$63 λ$	+	$90$
			-	$λ^{3}$	-	$3 λ^{2}$
					+	$11 λ^{2}$	+	$63 λ$
					-	$11 λ^{2}$	-	$33 λ$
							+	$30 λ$	+	$90$

Step 7.1.2.5.13

Multiply the new quotient term by the divisor.

				$λ^{2}$	+	$11 λ$	+	$30$
$λ$	+	$3$		$λ^{3}$	+	$14 λ^{2}$	+	$63 λ$	+	$90$
			-	$λ^{3}$	-	$3 λ^{2}$
					+	$11 λ^{2}$	+	$63 λ$
					-	$11 λ^{2}$	-	$33 λ$
							+	$30 λ$	+	$90$
							+	$30 λ$	+	$90$

Step 7.1.2.5.14

The expression needs to be subtracted from the dividend, so change all the signs in $30 λ + 90$

				$λ^{2}$	+	$11 λ$	+	$30$
$λ$	+	$3$		$λ^{3}$	+	$14 λ^{2}$	+	$63 λ$	+	$90$
			-	$λ^{3}$	-	$3 λ^{2}$
					+	$11 λ^{2}$	+	$63 λ$
					-	$11 λ^{2}$	-	$33 λ$
							+	$30 λ$	+	$90$
							-	$30 λ$	-	$90$

Step 7.1.2.5.15

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

				$λ^{2}$	+	$11 λ$	+	$30$
$λ$	+	$3$		$λ^{3}$	+	$14 λ^{2}$	+	$63 λ$	+	$90$
			-	$λ^{3}$	-	$3 λ^{2}$
					+	$11 λ^{2}$	+	$63 λ$
					-	$11 λ^{2}$	-	$33 λ$
							+	$30 λ$	+	$90$
							-	$30 λ$	-	$90$
										$0$

Step 7.1.2.5.16

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

$λ^{2} + 11 λ + 30$

Step 7.1.2.6

Write $λ^{3} + 14 λ^{2} + 63 λ + 90$ as a set of factors.

$- ((λ + 3) (λ^{2} + 11 λ + 30)) = 0$

Step 7.1.3

Factor.

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Step 7.1.3.1

Factor $λ^{2} + 11 λ + 30$ using the AC method.

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Step 7.1.3.1.1

Factor $λ^{2} + 11 λ + 30$ using the AC method.

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Step 7.1.3.1.1.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $30$ and whose sum is $11$ .

$5, 6$

Step 7.1.3.1.1.2

Write the factored form using these integers.

$- ((λ + 3) ((λ + 5) (λ + 6))) = 0$

Step 7.1.3.1.2

Remove unnecessary parentheses.

$- ((λ + 3) (λ + 5) (λ + 6)) = 0$

Step 7.1.3.2

Remove unnecessary parentheses.

$- (λ + 3) (λ + 5) (λ + 6) = 0$

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 $0$ .

$λ + 3 = 0$

$λ + 5 = 0$

$λ + 6 = 0$

Step 7.3

Set $λ + 3$ equal to $0$ and solve for $λ$ .

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Step 7.3.1

Set $λ + 3$ equal to $0$ .

$λ + 3 = 0$

Step 7.3.2

Subtract $3$ from both sides of the equation.

$λ = - 3$

Step 7.4

Set $λ + 5$ equal to $0$ and solve for $λ$ .

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Step 7.4.1

Set $λ + 5$ equal to $0$ .

$λ + 5 = 0$

Step 7.4.2

Subtract $5$ from both sides of the equation.

$λ = - 5$

Step 7.5

Set $λ + 6$ equal to $0$ and solve for $λ$ .

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Step 7.5.1

Set $λ + 6$ equal to $0$ .

$λ + 6 = 0$

Step 7.5.2

Subtract $6$ from both sides of the equation.

$λ = - 6$

Step 7.6

The final solution is all the values that make $- (λ + 3) (λ + 5) (λ + 6) = 0$ true.

$λ = - 3, - 5, - 6$