Precalculus Examples

$x^{3} + 4 x^{2} - 9 x - 36 = 0$ , $x = 3$

Step 1

Factor the left side of the equation.

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Step 1.1

Factor out the greatest common factor from each group.

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Step 1.1.1

Group the first two terms and the last two terms.

$(x^{3} + 4 x^{2}) - 9 x - 36 = 0$

$x = 3$

Step 1.1.2

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

$x^{2} (x + 4) - 9 (x + 4) = 0$

$x = 3$

$x^{2} (x + 4) - 9 (x + 4) = 0$

$x = 3$

Step 1.2

Factor the polynomial by factoring out the greatest common factor, $x + 4$ .

$(x + 4) (x^{2} - 9) = 0$

$x = 3$

Step 1.3

Rewrite $9$ as $3^{2}$ .

$(x + 4) (x^{2} - 3^{2}) = 0$

$x = 3$

Step 1.4

Factor.

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Step 1.4.1

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 3$ .

$(x + 4) ((x + 3) (x - 3)) = 0$

$x = 3$

Step 1.4.2

Remove unnecessary parentheses.

$(x + 4) (x + 3) (x - 3) = 0$

$x = 3$

$(x + 4) (x + 3) (x - 3) = 0$

$x = 3$

$(x + 4) (x + 3) (x - 3) = 0$

$x = 3$

Step 2

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

$x + 4 = 0$

$x + 3 = 0$

$x - 3 = 0$

$x = 3$

Step 3

Set $x + 4$ equal to $0$ and solve for $x$ .

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Step 3.1

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

$x = 3$

Step 3.2

Subtract $4$ from both sides of the equation.

$x = - 4$

$x = 3$

$x = - 4$

$x = 3$

Step 4

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

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Step 4.1

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

$x = 3$

Step 4.2

Subtract $3$ from both sides of the equation.

$x = - 3$

$x = 3$

$x = - 3$

$x = 3$

Step 5

Set $x - 3$ equal to $0$ and solve for $x$ .

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Step 5.1

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

$x = 3$

Step 5.2

Add $3$ to both sides of the equation.

$x = 3$

Step 6

The final solution is all the values that make $(x + 4) (x + 3) (x - 3) = 0$ true.

$x = - 4, - 3, 3$

$x = 3$

Step 7

Since the roots of an equation are the points where the solution is $0$ , set each root as a factor of the equation that equals $0$ .

$(x - (- 4, - 3, 3)) (x - 3) = 0$

Step 8

Simplify.

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Step 8.1

Simplify by multiplying through.

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Step 8.1.1

Apply the distributive property.

$(x + 4, - 3, 3) x + (x + 4, - 3, 3) \cdot - 3 = 0$

Step 8.1.2

Move $- 3$ to the left of $(x + 4, - 3, 3)$ .

$(x + 4, - 3, 3) x - 3 \cdot (x + 4, - 3, 3) = 0$

Step 8.2

Simplify each term.

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Step 8.2.1

Multiply $- 3$ by each element of the matrix.

$(x + 4, - 3, 3) x + (- 3 (x + 4), - 3 \cdot - 3, - 3 \cdot 3) = 0$

Step 8.2.2

Simplify each element in the matrix.

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Step 8.2.2.1

Apply the distributive property.

$(x + 4, - 3, 3) x + (- 3 x - 3 \cdot 4, - 3 \cdot - 3, - 3 \cdot 3) = 0$

Step 8.2.2.2

Multiply $- 3$ by $4$ .

$(x + 4, - 3, 3) x + (- 3 x - 12, - 3 \cdot - 3, - 3 \cdot 3) = 0$

Step 8.2.2.3

Multiply $- 3$ by $- 3$ .

$(x + 4, - 3, 3) x + (- 3 x - 12, 9, - 3 \cdot 3) = 0$

Step 8.2.2.4

Multiply $- 3$ by $3$ .

$(x + 4, - 3, 3) x + (- 3 x - 12, 9, - 9) = 0$

Step 8.3

Reorder factors in $(x + 4, - 3, 3) x + (- 3 x - 12, 9, - 9)$ .

$x (x + 4, - 3, 3) + (- 3 x - 12, 9, - 9) = 0$