Precalculus Examples

$4 x^{8} - 13 x^{4} + 9$

Step 1

Rewrite $x^{8}$ as ${(x^{4})}^{2}$ .

$4 {(x^{4})}^{2} - 13 x^{4} + 9$

Step 2

Let $u = x^{4}$ . Substitute $u$ for all occurrences of $x^{4}$ .

$4 u^{2} - 13 u + 9$

Step 3

Factor by grouping.

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Step 3.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 4 \cdot 9 = 36$ and whose sum is $b = - 13$ .

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Step 3.1.1

Factor $- 13$ out of $- 13 u$ .

$4 u^{2} - 13 (u) + 9$

Step 3.1.2

Rewrite $- 13$ as $- 4$ plus $- 9$

$4 u^{2} + (- 4 - 9) u + 9$

Step 3.1.3

Apply the distributive property.

$4 u^{2} - 4 u - 9 u + 9$

Step 3.2

Factor out the greatest common factor from each group.

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Step 3.2.1

Group the first two terms and the last two terms.

$(4 u^{2} - 4 u) - 9 u + 9$

Step 3.2.2

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

$4 u (u - 1) - 9 (u - 1)$

Step 3.3

Factor the polynomial by factoring out the greatest common factor, $u - 1$ .

$(u - 1) (4 u - 9)$

Step 4

Replace all occurrences of $u$ with $x^{4}$ .

$(x^{4} - 1) (4 x^{4} - 9)$

Step 5

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

$({(x^{2})}^{2} - 1) (4 x^{4} - 9)$

Step 6

Rewrite $1$ as $1^{2}$ .

$({(x^{2})}^{2} - 1^{2}) (4 x^{4} - 9)$

Step 7

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x^{2}$ and $b = 1$ .

$(x^{2} + 1) (x^{2} - 1) (4 x^{4} - 9)$

Step 8

Simplify.

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Step 8.1

Rewrite $1$ as $1^{2}$ .

$(x^{2} + 1) (x^{2} - 1^{2}) (4 x^{4} - 9)$

Step 8.2

Factor.

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Step 8.2.1

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 $b = 1$ .

$(x^{2} + 1) ((x + 1) (x - 1)) (4 x^{4} - 9)$

Step 8.2.2

Remove unnecessary parentheses.

$(x^{2} + 1) (x + 1) (x - 1) (4 x^{4} - 9)$

Step 9

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

$(x^{2} + 1) (x + 1) (x - 1) ({(2 x^{2})}^{2} - 9)$

Step 10

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

$(x^{2} + 1) (x + 1) (x - 1) ({(2 x^{2})}^{2} - 3^{2})$

Step 11

Factor.

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Step 11.1

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

$(x^{2} + 1) (x + 1) (x - 1) ((2 x^{2} + 3) (2 x^{2} - 3))$

Step 11.2

Remove unnecessary parentheses.

$(x^{2} + 1) (x + 1) (x - 1) (2 x^{2} + 3) (2 x^{2} - 3)$