Question

simplify each as much as possible.

1. (dfrac{x^2 + 2x - 15}{4x^3 + 20x^2} cdot dfrac{2x}{x^3 - 27})

Explanation:

Step1: Factor numerator 1

$x^2 + 2x - 15 = (x+5)(x-3)$

Step2: Factor denominator 1

$4x^3 + 20x^2 = 4x^2(x+5)$

Step3: Factor denominator 2

$x^3 - 27 = (x-3)(x^2+3x+9)$

Step4: Rewrite & cancel terms

$$\frac{(x+5)(x-3)}{4x^2(x+5)} \cdot \frac{2x}{(x-3)(x^2+3x+9)} = \frac{\cancel{(x+5)}\cancel{(x-3)}}{4x^2\cancel{(x+5)}} \cdot \frac{2x}{\cancel{(x-3)}(x^2+3x+9)}$$

Step5: Simplify remaining terms

$$\frac{2x}{4x^2(x^2+3x+9)} = \frac{1}{2x(x^2+3x+9)}$$

Answer:

$\frac{1}{2x(x^2+3x+9)}$ (where $x
eq -5, 0, 3$)