Question

1. find the product. if possible, write your answer in factored form.

a \\(\frac{x^2 - 4x}{x - 1} \cdot \frac{x^2 + 3x - 4}{2x} = \square\\) , b \\(x \
eq 1\\) , c \\(x \
eq 0\\)

Explanation:

Step1: Factor numerators

Factor \(x^2 - 4x\) as \(x(x - 4)\), and factor \(x^2 + 3x - 4\) using quadratic factoring: find two numbers that multiply to \(-4\) and add to \(3\), which are \(4\) and \(-1\), so \(x^2 + 3x - 4=(x + 4)(x - 1)\).
Now the expression becomes \(\frac{x(x - 4)}{x - 1}\cdot\frac{(x + 4)(x - 1)}{2x}\).

Step2: Cancel common factors

Cancel out the common factors \(x\) (note \(x
eq0\)) and \(x - 1\) (note \(x
eq1\)) from the numerators and denominators.
After canceling, we have \(\frac{(x - 4)(x + 4)}{2}\).

Step3: Simplify the numerator

Simplify \((x - 4)(x + 4)\) using the difference of squares formula \(a^2 - b^2=(a - b)(a + b)\), so \((x - 4)(x + 4)=x^2 - 16\).
So the simplified form is \(\frac{x^2 - 16}{2}\) (or we can leave it as \(\frac{(x - 4)(x + 4)}{2}\) which is factored form).

Answer:

\(\frac{(x - 4)(x + 4)}{2}\) (or \(\frac{x^2 - 16}{2}\))