Precalculus Examples

$x^{4} + 7 x^{3} + 3 x^{2} - 63 x - 108$

Step 1

Set $x^{4} + 7 x^{3} + 3 x^{2} - 63 x - 108$ equal to $0$ .

$x^{4} + 7 x^{3} + 3 x^{2} - 63 x - 108 = 0$

Step 2

Solve for $x$ .

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Step 2.1

Factor the left side of the equation.

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Step 2.1.1

Regroup terms.

$7 x^{3} - 63 x + x^{4} + 3 x^{2} - 108 = 0$

Step 2.1.2

Factor $7 x$ out of $7 x^{3} - 63 x$ .

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Step 2.1.2.1

Factor $7 x$ out of $7 x^{3}$ .

$7 x (x^{2}) - 63 x + x^{4} + 3 x^{2} - 108 = 0$

Step 2.1.2.2

Factor $7 x$ out of $- 63 x$ .

$7 x (x^{2}) + 7 x (- 9) + x^{4} + 3 x^{2} - 108 = 0$

Step 2.1.2.3

Factor $7 x$ out of $7 x (x^{2}) + 7 x (- 9)$ .

$7 x (x^{2} - 9) + x^{4} + 3 x^{2} - 108 = 0$

Step 2.1.3

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

$7 x (x^{2} - 3^{2}) + x^{4} + 3 x^{2} - 108 = 0$

Step 2.1.4

Factor.

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Step 2.1.4.1

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 $b = 3$ .

$7 x ((x + 3) (x - 3)) + x^{4} + 3 x^{2} - 108 = 0$

Step 2.1.4.2

Remove unnecessary parentheses.

$7 x (x + 3) (x - 3) + x^{4} + 3 x^{2} - 108 = 0$

Step 2.1.5

Rewrite $x^{4}$ as ${(x^{2})}^{2}$ .

$7 x (x + 3) (x - 3) + {(x^{2})}^{2} + 3 x^{2} - 108 = 0$

Step 2.1.6

Let $u = x^{2}$ . Substitute $u$ for all occurrences of $x^{2}$ .

$7 x (x + 3) (x - 3) + u^{2} + 3 u - 108 = 0$

Step 2.1.7

Factor $u^{2} + 3 u - 108$ using the AC method.

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Step 2.1.7.1

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 $- 108$ and whose sum is $3$ .

$- 9, 12$

Step 2.1.7.2

Write the factored form using these integers.

$7 x (x + 3) (x - 3) + (u - 9) (u + 12) = 0$

Step 2.1.8

Replace all occurrences of $u$ with $x^{2}$ .

$7 x (x + 3) (x - 3) + (x^{2} - 9) (x^{2} + 12) = 0$

Step 2.1.9

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

$7 x (x + 3) (x - 3) + (x^{2} - 3^{2}) (x^{2} + 12) = 0$

Step 2.1.10

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 $b = 3$ .

$7 x (x + 3) (x - 3) + (x + 3) (x - 3) (x^{2} + 12) = 0$

Step 2.1.11

Factor $(x + 3) (x - 3)$ out of $7 x (x + 3) (x - 3) + (x + 3) (x - 3) (x^{2} + 12)$ .

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Step 2.1.11.1

Factor $(x + 3) (x - 3)$ out of $7 x (x + 3) (x - 3)$ .

$(x + 3) (x - 3) (7 x) + (x + 3) (x - 3) (x^{2} + 12) = 0$

Step 2.1.11.2

Factor $(x + 3) (x - 3)$ out of $(x + 3) (x - 3) (7 x) + (x + 3) (x - 3) (x^{2} + 12)$ .

$(x + 3) (x - 3) (7 x + x^{2} + 12) = 0$

Step 2.1.12

Let $u = x$ . Substitute $u$ for all occurrences of $x$ .

$(x + 3) (x - 3) (7 u + u^{2} + 12) = 0$

Step 2.1.13

Factor $7 u + u^{2} + 12$ using the AC method.

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Step 2.1.13.1

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 $12$ and whose sum is $7$ .

$3, 4$

Step 2.1.13.2

Write the factored form using these integers.

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

Step 2.1.14

Factor.

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Step 2.1.14.1

Replace all occurrences of $u$ with $x$ .

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

Step 2.1.14.2

Remove unnecessary parentheses.

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

Step 2.1.15

Combine exponents.

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Step 2.1.15.1

Raise $x + 3$ to the power of $1$ .

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

Step 2.1.15.2

Raise $x + 3$ to the power of $1$ .

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

Step 2.1.15.3

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

${(x + 3)}^{1 + 1} (x - 3) (x + 4) = 0$

Step 2.1.15.4

Add $1$ and $1$ .

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

Step 2.2

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

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

$x - 3 = 0$

$x + 4 = 0$

Step 2.3

Set ${(x + 3)}^{2}$ equal to $0$ and solve for $x$ .

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Step 2.3.1

Set ${(x + 3)}^{2}$ equal to $0$ .

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

Step 2.3.2

Solve ${(x + 3)}^{2} = 0$ for $x$ .

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Step 2.3.2.1

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

$x + 3 = 0$

Step 2.3.2.2

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 2.4

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

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Step 2.4.1

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

$x - 3 = 0$

Step 2.4.2

Add $3$ to both sides of the equation.

$x = 3$

Step 2.5

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

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Step 2.5.1

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

$x + 4 = 0$

Step 2.5.2

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 2.6

The final solution is all the values that make ${(x + 3)}^{2} (x - 3) (x + 4) = 0$ true.

$x = - 3, 3, - 4$

Step 3