Question

fully simplify the expression below and write your answer as a single fraction.
$\frac{x^{3}-3x^{2}-28x}{15x^{4}-240x^{3}+945x^{2}}cdot\frac{x - 9}{x + 4}$

Explanation:

Step1: Factor the numerator of the first - fraction

$x^{3}-3x^{2}-28x=x(x^{2}-3x - 28)=x(x - 7)(x+4)$

Step2: Factor the denominator of the first - fraction

$15x^{4}-240x^{3}+945x^{2}=15x^{2}(x^{2}-16x + 63)=15x^{2}(x - 7)(x - 9)$

Step3: Multiply the two fractions

$\frac{x(x - 7)(x + 4)}{15x^{2}(x - 7)(x - 9)}\cdot\frac{x - 9}{x + 4}$

Step4: Cancel out the common factors

Cancel out $(x - 7)$, $(x + 4)$ and $(x - 9)$ in the numerator and denominator, and also cancel out one factor of $x$ from the numerator and denominator. We get $\frac{1}{15x}$

Answer:

$\frac{1}{15x}$