Factor y^4+8y^2+16

$y^{4} + 8 y^{2} + 16$

Step 1

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

${(y^{2})}^{2} + 8 y^{2} + 16$

Step 2

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

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

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 $y^{2}$ .

${(y^{2} + 4)}^{2}$