factor $z^{4}-14z^{2}-32$ completely.

Answer:

$(z^{2}+2)(z^{2}-16)=(z^{2}+2)(z + 4)(z - 4)$

Explanation:

Step1: Let $u = z^{2}$

The expression becomes $u^{2}-14u - 32$.

Step2: Factor quadratic in $u$

We need two numbers that multiply to $-32$ and add to $-14$. The numbers are $-16$ and $2$. So $u^{2}-14u - 32=(u + 2)(u - 16)$.

Step3: Substitute back $u = z^{2}$

We get $(z^{2}+2)(z^{2}-16)$.

Step4: Factor difference - of - squares

Since $z^{2}-16$ is a difference of squares ($a^{2}-b^{2}=(a + b)(a - b)$ with $a = z$ and $b = 4$), $z^{2}-16=(z + 4)(z - 4)$. So the complete factorization is $(z^{2}+2)(z + 4)(z - 4)$.