Factor x^4-8x^2+16

$x^{4} - 8 x^{2} + 16$

Step 1

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

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

Step 2

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

$u^{2} - 8 u + 16$

Step 3

Factor using the perfect square rule.

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

$u^{2} - 8 u + 4^{2}$

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.

$u^{2} - 2 \cdot u \cdot 4 + 4^{2}$

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

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

Step 4

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

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

Step 5

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

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

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 $b = 2$ .

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

Step 7

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

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