Question

fully simplify the expression below and write your answer as a single fraction.
$\frac{x^{2}-7x - 18}{x^{2}-3x - 54}cdot\frac{x^{2}+10x + 24}{x^{2}+6x + 8}$

Explanation:

Step1: Factor the quadratic expressions

$x^{2}-7x - 18=(x - 9)(x+2)$; $x^{2}-3x - 54=(x - 9)(x + 6)$; $x^{2}+10x + 24=(x + 4)(x+6)$; $x^{2}+6x + 8=(x + 2)(x + 4)$
So the expression becomes $\frac{(x - 9)(x+2)}{(x - 9)(x + 6)}\cdot\frac{(x + 4)(x+6)}{(x + 2)(x + 4)}$

Step2: Cancel out the common factors

Cancel out $(x - 9)$ in the first - fraction, $(x + 2)$ in the first and second fractions, $(x + 6)$ in the first and second fractions, and $(x + 4)$ in the first and second fractions.
$\frac{(x - 9)(x+2)}{(x - 9)(x + 6)}\cdot\frac{(x + 4)(x+6)}{(x + 2)(x + 4)} = 1$

Answer:

$1$