## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$4(x^2-2x+8)$
To factor the second-degree polynomial $4x^2 - 8x + 32=4(x^2-2x+8)$, we have to find integers whose product is $8$ and whose sum is $-2$. But there are no such factors. Thus, the polynomial $x^2-2x+8$ is prime and hence we have $$4x^2 - 8x + 32=4(x^2-2x+8) .$$