Chapter 11.3, Problem 30PPS

To determine

To Find: The zeroes of the function.

Answer to Problem 30PPS

  $x=-3,4$

Explanation of Solution

Given: $\frac{x^2-x-12}{x^2+2x-35}$

Given= $f(x)=\frac{x^2-x-12}{x^2+2x-35}$

Set $f(x)=0$

  $\begin{array}{l}0=\frac{x^2-x-12}{x^2+2x-35}\\ 0=\frac{x(x-4)+3(x-4)}{x(x+7)-5(x+7)}\\ 0=\frac{(x-4)(x+3)}{(x+7)(x-5)}\\ (x-4)(x+3)=0\end{array}$

  $x=-3,4$

Hence the zeroes of the function is at $x=-3,4$