Factor 9x^8-72x^4+144

$9 x^{8} - 72 x^{4} + 144$

Step 1

Factor $9$ out of $9 x^{8} - 72 x^{4} + 144$ .

Factor $9$ out of $9 x^{8}$ .

$9 (x^{8}) - 72 x^{4} + 144$

Factor $9$ out of $- 72 x^{4}$ .

$9 (x^{8}) + 9 (- 8 x^{4}) + 144$

Factor $9$ out of $144$ .

$9 x^{8} + 9 (- 8 x^{4}) + 9 \cdot 16$

Factor $9$ out of $9 x^{8} + 9 (- 8 x^{4})$ .

$9 (x^{8} - 8 x^{4}) + 9 \cdot 16$

Factor $9$ out of $9 (x^{8} - 8 x^{4}) + 9 \cdot 16$ .

$9 (x^{8} - 8 x^{4} + 16)$

Step 2

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

$9 ({(x^{4})}^{2} - 8 x^{4} + 16)$

Step 3

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

$9 (u^{2} - 8 u + 16)$

Step 4

Factor using the perfect square rule.

Rewrite $16$ as $4^{2}$ .

$9 (u^{2} - 8 u + 4^{2})$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$8 u = 2 \cdot u \cdot 4$

Rewrite the polynomial.

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

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = u$ and $b = 4$ .

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

Step 5

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

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

Step 6

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

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

Step 7

Rewrite $4$ as $2^{2}$ .

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

Step 8

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 $b = 2$ .

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

Step 9

Factor.

Apply the product rule to $(x^{2} + 2) (x^{2} - 2)$ .

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

Remove unnecessary parentheses.

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

Factor 9x^8-72x^4+144

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