Precalculus Examples

$f (x) = x^{4} + 6 x^{3} - 16 x^{2} - 54 x + 63$

Step 1

Set $x^{4} + 6 x^{3} - 16 x^{2} - 54 x + 63$ equal to $0$ .

$x^{4} + 6 x^{3} - 16 x^{2} - 54 x + 63 = 0$

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.

$6 x^{3} - 54 x + x^{4} - 16 x^{2} + 63 = 0$

Step 2.1.2

Factor $6 x$ out of $6 x^{3} - 54 x$ .

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Step 2.1.2.1

Factor $6 x$ out of $6 x^{3}$ .

$6 x (x^{2}) - 54 x + x^{4} - 16 x^{2} + 63 = 0$

Step 2.1.2.2

Factor $6 x$ out of $- 54 x$ .

$6 x (x^{2}) + 6 x (- 9) + x^{4} - 16 x^{2} + 63 = 0$

Step 2.1.2.3

Factor $6 x$ out of $6 x (x^{2}) + 6 x (- 9)$ .

$6 x (x^{2} - 9) + x^{4} - 16 x^{2} + 63 = 0$

Step 2.1.3

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

$6 x (x^{2} - 3^{2}) + x^{4} - 16 x^{2} + 63 = 0$

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 $b = 3$ .

$6 x ((x + 3) (x - 3)) + x^{4} - 16 x^{2} + 63 = 0$

Step 2.1.4.2

Remove unnecessary parentheses.

$6 x (x + 3) (x - 3) + x^{4} - 16 x^{2} + 63 = 0$

Step 2.1.5

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

$6 x (x + 3) (x - 3) + {(x^{2})}^{2} - 16 x^{2} + 63 = 0$

Step 2.1.6

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

$6 x (x + 3) (x - 3) + u^{2} - 16 u + 63 = 0$

Step 2.1.7

Factor $u^{2} - 16 u + 63$ 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 $63$ and whose sum is $- 16$ .

$- 9, - 7$

Step 2.1.7.2

Write the factored form using these integers.

$6 x (x + 3) (x - 3) + (u - 9) (u - 7) = 0$

Step 2.1.8

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

$6 x (x + 3) (x - 3) + (x^{2} - 9) (x^{2} - 7) = 0$

Step 2.1.9

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

$6 x (x + 3) (x - 3) + (x^{2} - 3^{2}) (x^{2} - 7) = 0$

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 $b = 3$ .

$6 x (x + 3) (x - 3) + (x + 3) (x - 3) (x^{2} - 7) = 0$

Step 2.1.11

Factor $(x + 3) (x - 3)$ out of $6 x (x + 3) (x - 3) + (x + 3) (x - 3) (x^{2} - 7)$ .

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Step 2.1.11.1

Factor $(x + 3) (x - 3)$ out of $6 x (x + 3) (x - 3)$ .

$(x + 3) (x - 3) (6 x) + (x + 3) (x - 3) (x^{2} - 7) = 0$

Step 2.1.11.2

Factor $(x + 3) (x - 3)$ out of $(x + 3) (x - 3) (6 x) + (x + 3) (x - 3) (x^{2} - 7)$ .

$(x + 3) (x - 3) (6 x + x^{2} - 7) = 0$

Step 2.1.12

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

$(x + 3) (x - 3) (6 u + u^{2} - 7) = 0$

Step 2.1.13

Factor $6 u + u^{2} - 7$ 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 $- 7$ and whose sum is $6$ .

$- 1, 7$

Step 2.1.13.2

Write the factored form using these integers.

$(x + 3) (x - 3) ((u - 1) (u + 7)) = 0$

Step 2.1.14

Factor.

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Step 2.1.14.1

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

$(x + 3) (x - 3) ((x - 1) (x + 7)) = 0$

Step 2.1.14.2

Remove unnecessary parentheses.

$(x + 3) (x - 3) (x - 1) (x + 7) = 0$

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 $0$ .

$x + 3 = 0$

$x - 3 = 0$

$x - 1 = 0$

$x + 7 = 0$

Step 2.3

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

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Step 2.3.1

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

$x + 3 = 0$

Step 2.3.2

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 $x - 1$ equal to $0$ and solve for $x$ .

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Step 2.5.1

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Step 2.5.2

Add $1$ to both sides of the equation.

$x = 1$

Step 2.6

Set $x + 7$ equal to $0$ and solve for $x$ .

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Step 2.6.1

Set $x + 7$ equal to $0$ .

$x + 7 = 0$

Step 2.6.2

Subtract $7$ from both sides of the equation.

$x = - 7$

Step 2.7

The final solution is all the values that make $(x + 3) (x - 3) (x - 1) (x + 7) = 0$ true.

$x = - 3, 3, 1, - 7$

Step 3