Chapter 11.3, Problem 35PPS

To determine

To Find: The zeroes of the function.

Answer to Problem 35PPS

  $x=0,-2,6$

Explanation of Solution

Given: $f(x)=\frac{x^3-4x^2-12x}{x+2}$

Given= $f(x)=\frac{x^3-4x^2-12x}{x+2}$

  $\begin{array}{l}f(x)=\frac{x^3-4x^2-12x}{x+2}\\ f(x)=\frac{x(x^2-4x-12)}{x+2}\\ f(x)=\frac{x(x^2-6x+2x-12)}{x+2}\\ f(x)=\frac{x(x+2)(x-6)}{x+2}\end{array}$

Set $f(x)=0$

  $\begin{array}{l}f(x)=\frac{x(x+2)(x-6)}{x+2}\\ x(x+2)(x-6)=0\\ x=0,-2,6\end{array}$

Hence the zeroes of the function is at $x=0,-2,6$