Algebra Examples

$x^{3} + 8 x^{2} + 19 x + 12 = 0$ , $x = - 4$

Step 1

Replace all occurrences of $x$ with $- 4$ in each equation.

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Step 1.1

Replace all occurrences of $x$ in $x^{3} + 8 x^{2} + 19 x + 12 = 0$ with $- 4$ .

${(- 4)}^{3} + 8 {(- 4)}^{2} + 19 (- 4) + 12 = 0$

$x = - 4$

Step 1.2

Simplify the left side.

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Step 1.2.1

Simplify ${(- 4)}^{3} + 8 {(- 4)}^{2} + 19 (- 4) + 12$ .

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Step 1.2.1.1

Simplify each term.

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Step 1.2.1.1.1

Raise $- 4$ to the power of $3$ .

$- 64 + 8 {(- 4)}^{2} + 19 (- 4) + 12 = 0$

$x = - 4$

Step 1.2.1.1.2

Raise $- 4$ to the power of $2$ .

$- 64 + 8 \cdot 16 + 19 (- 4) + 12 = 0$

$x = - 4$

Step 1.2.1.1.3

Multiply $8$ by $16$ .

$- 64 + 128 + 19 (- 4) + 12 = 0$

$x = - 4$

Step 1.2.1.1.4

Multiply $19$ by $- 4$ .

$- 64 + 128 - 76 + 12 = 0$

$x = - 4$

$- 64 + 128 - 76 + 12 = 0$

$x = - 4$

Step 1.2.1.2

Simplify by adding and subtracting.

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Step 1.2.1.2.1

Add $- 64$ and $128$ .

$64 - 76 + 12 = 0$

$x = - 4$

Step 1.2.1.2.2

Subtract $76$ from $64$ .

$- 12 + 12 = 0$

$x = - 4$

Step 1.2.1.2.3

Add $- 12$ and $12$ .

$0 = 0$

$x = - 4$

$0 = 0$

$x = - 4$

$0 = 0$

$x = - 4$

$0 = 0$

$x = - 4$

$0 = 0$

$x = - 4$

Step 2

Remove any equations from the system that are always true.

$x = - 4$

Step 3