Question

fully simplify the expression below and write your answer as a single fraction.\\(\dfrac{6x^2 - 12x - 210}{x^5 - 13x^4 + 42x^3} \cdot \dfrac{x - 6}{9x^2 + 9x - 180}\\)

Explanation:

Step1: Factor all numerators/denominators

First expression numerator: $6x^2 -12x -210 = 6(x^2 -2x -35) = 6(x-7)(x+5)$
First expression denominator: $x^5 -13x^4 +42x^3 = x^3(x^2 -13x +42) = x^3(x-6)(x-7)$
Second expression numerator: $x-6$
Second expression denominator: $9x^2 +9x -180 = 9(x^2 +x -20) = 9(x+5)(x-4)$

Substitute factored forms:
$$\frac{6(x-7)(x+5)}{x^3(x-6)(x-7)} \cdot \frac{x-6}{9(x+5)(x-4)}$$

Step2: Cancel common factors

Cancel $(x-7)$, $(x+5)$, $(x-6)$ from numerator and denominator:
$$\frac{6}{x^3} \cdot \frac{1}{9(x-4)}$$

Step3: Multiply remaining terms

Multiply numerators and denominators, simplify constants:
$$\frac{6}{9x^3(x-4)} = \frac{2}{3x^3(x-4)}$$
Expand the denominator (optional, but single fraction):
$$\frac{2}{3x^4 -12x^3}$$

Answer:

$\frac{2}{3x^3(x-4)}$ (or equivalently $\frac{2}{3x^4 - 12x^3}$)