Question

fully simplify the expression below and write your answer as a single fraction.
$\frac{x^{3}-16x}{6x + 24}cdot\frac{x^{2}-100}{x^{4}-14x^{3}+40x^{2}}$

Explanation:

Step1: Factor the expressions

Factor each part:

• For $x^{3}-16x$, we have $x(x^{2}-16)=x(x + 4)(x - 4)$ (using $a^{2}-b^{2}=(a + b)(a - b)$ with $a=x$ and $b = 4$).
• For $6x + 24$, we factor out 6 to get $6(x + 4)$.
• For $x^{2}-100$, we have $(x + 10)(x - 10)$ (using $a^{2}-b^{2}=(a + b)(a - b)$ with $a=x$ and $b = 10$).
• For $x^{4}-14x^{3}+40x^{2}$, we factor out $x^{2}$ first to get $x^{2}(x^{2}-14x + 40)$, and then factor the quadratic $x^{2}-14x + 40=(x - 4)(x - 10)$. So $x^{4}-14x^{3}+40x^{2}=x^{2}(x - 4)(x - 10)$.

The original expression $\frac{x^{3}-16x}{6x + 24}\cdot\frac{x^{2}-100}{x^{4}-14x^{3}+40x^{2}}$ becomes $\frac{x(x + 4)(x - 4)}{6(x + 4)}\cdot\frac{(x + 10)(x - 10)}{x^{2}(x - 4)(x - 10)}$.

Step2: Cancel out the common factors

Cancel out the common factors:

• Cancel out $(x + 4)$ in the first - fraction, cancel out $(x - 4)$ and $(x - 10)$ and one factor of $x$ (since we have $x$ in the numerator and $x^{2}$ in the denominator).

We get $\frac{1}{6x}$.

Answer:

$\frac{1}{6x}$