Question

what is the product?
\\(\frac{x^2 - 16}{2x + 8} cdot \frac{x^3 - 2x^2 + x}{x^2 + 3x - 4}\\)
\\(\bigcirc \frac{x(x - 4)(x - 1)}{2(x + 4)}\\)
\\(\bigcirc \frac{x(x - 1)}{2}\\)
\\(\bigcirc \frac{(x + 4)(x - 4)}{2x(x - 1)}\\)
\\(\bigcirc \frac{(x - 4)(x - 1)}{2x(x + 4)}\\)

Explanation:

Step1: Factor each expression

• Factor \(x^2 - 16\) (difference of squares): \(x^2 - 16=(x + 4)(x - 4)\)
• Factor \(2x + 8\): \(2x+8 = 2(x + 4)\)
• Factor \(x^3-2x^2 + x\) (factor out \(x\) first, then factor the quadratic): \(x^3-2x^2 + x=x(x^2-2x + 1)=x(x - 1)^2\)
• Factor \(x^2+3x - 4\) (find two numbers that multiply to \(-4\) and add to \(3\)): \(x^2+3x - 4=(x + 4)(x - 1)\)

Step2: Substitute factored forms into the product

The original product \(\frac{x^2 - 16}{2x + 8}\cdot\frac{x^3-2x^2 + x}{x^2+3x - 4}\) becomes:

$$ \frac{(x + 4)(x - 4)}{2(x + 4)}\cdot\frac{x(x - 1)^2}{(x + 4)(x - 1)} $$

Step3: Cancel common factors

• Cancel \((x + 4)\) from the numerator and denominator of the first fraction.
• Cancel one \((x - 1)\) from the numerator and denominator of the second fraction.

After canceling, we have:

$$ \frac{(x - 4)}{2}\cdot\frac{x(x - 1)}{(x + 4)}=\frac{x(x - 4)(x - 1)}{2(x + 4)} $$

Answer:

\(\frac{x(x - 4)(x - 1)}{2(x + 4)}\) (corresponding to the first option: \(\boldsymbol{\frac{x(x - 4)(x - 1)}{2(x + 4)}}\))