Question

1. multiply or divide each expression.

a) \\(\frac{x^2 - 2x - 35}{3x^2 - 20x - 7} \cdot \frac{18x + 6}{x^2 - 25}\\)

Explanation:

Step1: Factor each numerator and denominator

Factor \(x^2 - 2x - 35\): Find two numbers that multiply to \(-35\) and add to \(-2\). They are \(-7\) and \(5\), so \(x^2 - 2x - 35=(x - 7)(x + 5)\).
Factor \(3x^2 - 20x - 7\): Use AC method. \(3\times(-7)=-21\), find two numbers that multiply to \(-21\) and add to \(-20\) (which are \(-21\) and \(1\)). Rewrite middle term: \(3x^2 - 21x + x - 7 = 3x(x - 7)+1(x - 7)=(3x + 1)(x - 7)\).
Factor \(18x + 6\): Factor out \(6\), so \(18x + 6 = 6(3x + 1)\).
Factor \(x^2 - 25\): This is a difference of squares, so \(x^2 - 25=(x - 5)(x + 5)\).

Now the expression becomes:
\(\frac{(x - 7)(x + 5)}{(3x + 1)(x - 7)} \cdot \frac{6(3x + 1)}{(x - 5)(x + 5)}\)

Step2: Cancel common factors

Cancel \((x - 7)\) from numerator and denominator, \((x + 5)\) from numerator and denominator, and \((3x + 1)\) from numerator and denominator:
After canceling, we have \(\frac{6}{x - 5}\) (note that \(x
eq 7, -\frac{1}{3}, \pm5\) to avoid division by zero)

Answer:

\(\frac{6}{x - 5}\) (with restrictions \(x
eq 7, -\frac{1}{3}, 5, -5\))