## Algebra 2 (1st Edition)

$$x^3-10x^2+32x-35$$
Multiplying each term in the first factor by each term in the second factor, we find: $$\left(x-5\right)\left(x^2-5x+7\right) \\ xx^2+x\left(-5x\right)+x\cdot \:7+\left(-5\right)x^2+\left(-5\right)\left(-5x\right)+\left(-5\right)\cdot \:7 \\ x^3-10x^2+32x-35$$