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

a. $2x(x-5)$. b. $(x+3)(x-2)(x+2)$
We have a. Factoring $2x$ out we get: $$2x^2 - 10x=2x(x-5)$$ b. We factor by grouping, so first we group the first two terms together and the last two terms together: $$=(x^3 + 3x^2)+(- 4x - 12)$$ Then we factor out the GCF in each group: $$=x^2(x+3)-4(x+3)$$. Then we factor out $x+3$ out we have $$=(x+3)(x^2-4)$$ Finally we factor $(x^2-4)$ as a difference of two squares to obtain: $$=(x+3)(x-2)(x+2)$$