Chapter 4.4, Problem 68E

To determine

To find: the value of $k$ such that $\frac{f\left(x\right)}{x-k}$ has a remainder of 0.

Answer to Problem 68E

The value is $k=-1$ or $k=4$ .

Explanation of Solution

Given information:

Given function,

  $f\left(x\right)=x^3-3x^2-4x$

Calculation:

Using $f\left(x\right)=x^3-3x^2-4x$ ,

  $\frac{f\left(x\right)}{x-k}=\frac{x^3-3x^2-4x}{x-k}$

Now factor numerator as,

  $\begin{array}{c}\frac{x^3-3x^2-4x}{x-k}=\frac{x\left(x^2-3x-4\right)}{x-k}\\ =\frac{x\left(x-4\right)\left(x+1\right)}{x-k}\end{array}$

Thus, the value is $k=-1$ or $k=4$ .