Factor x^4-8x^2-9

$x^{4} - 8 x^{2} - 9$

Step 1

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

${(x^{2})}^{2} - 8 x^{2} - 9$

Step 2

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

$u^{2} - 8 u - 9$

Step 3

Factor $u^{2} - 8 u - 9$ using the AC method.

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 $- 9$ and whose sum is $- 8$ .

$- 9, 1$

Write the factored form using these integers.

$(u - 9) (u + 1)$

Step 4

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

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

Step 5

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

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

Step 6

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 $b = 3$ .

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

Factor x^4-8x^2-9

Related Questions

Support LCMGCF.com by disabling your adblocker.