Question

fully simplify the expression below and write your answer as a single fraction.
$\frac{x^{3}-16x}{x^{5}+14x^{4}+40x^{3}} cdot \frac{9x^{2}+18x-720}{6x-24}$

Explanation:

Step1: Factor all numerators/denominators

First fraction numerator:

$x^3 - 16x = x(x^2 - 16) = x(x-4)(x+4)$

First fraction denominator:

$x^5 + 14x^4 + 40x^3 = x^3(x^2 +14x +40) = x^3(x+10)(x+4)$

Second fraction numerator:

$9x^2 +18x -720 = 9(x^2 +2x -80) = 9(x+10)(x-8)$

Second fraction denominator:

$6x -24 = 6(x-4)$

Substitute back into the expression:
$$\frac{x(x-4)(x+4)}{x^3(x+10)(x+4)} \cdot \frac{9(x+10)(x-8)}{6(x-4)}$$

Step2: Cancel common factors

Cancel $x$, $(x-4)$, $(x+4)$, $(x+10)$ from numerator and denominator:
$$\frac{1}{x^2} \cdot \frac{9(x-8)}{6}$$

Step3: Simplify the remaining terms

Simplify $\frac{9}{6} = \frac{3}{2}$:
$$\frac{3(x-8)}{2x^2}$$

Answer:

$\frac{3(x-8)}{2x^2}$