Precalculus Examples

$x^{4} - 81$

Step 1

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

${(x^{2})}^{2} - 81$

Step 2

Rewrite $81$ as $9^{2}$ .

${(x^{2})}^{2} - 9^{2}$

Step 3

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - i^{2} = (a + i) (a - i)$ where $a = x^{2}$ and $i = 9$ .

$(x^{2} + 9) (x^{2} - 9)$

Step 4

Simplify.

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Step 4.1

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

$(x^{2} + 9) (x^{2} - 3^{2})$

Step 4.2

Factor.

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Step 4.2.1

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

$(x^{2} + 9) ((x + 3) (x - 3))$

Step 4.2.2

Remove unnecessary parentheses.

$(x^{2} + 9) (x + 3) (x - 3)$

Step 5

Multiply the constant in the polynomial $x^{2} + 9$ by $- i^{2}$ where $i^{2}$ is equal to $- 1$ .

$(x^{2} - 9 i^{2}) (x + 3) (x - 3)$

Step 6

Rewrite $9 i^{2}$ as ${(3 i)}^{2}$ .

$(x^{2} - {(3 i)}^{2}) (x + 3) (x - 3)$

Step 7

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

$(x + 3 i) (x - (3 i)) (x + 3) (x - 3)$

Step 8

Multiply $3$ by $- 1$ .

$(x + 3 i) (x - 3 i) (x + 3) (x - 3)$