Chapter 6.7, Problem 36WE

To determine

To calculate: the value of $k$.

Answer to Problem 36WE

The value of $\text{k}=3$.

Explanation of Solution

Given information:

  $(2x-1)\text{ is a factor of }4x^3-6x^2-4x+k$

Formula used:

Factor Theorem : If $(x-a)$ is a factor $f(x)=(a_n{x}^n+a_{n-1}{x}^{n-1}+\mathrm{....}+ax+a_0)$ then $f(a)=0$ .

Calculation:

Consider ,

  $(2x-1)\text{ is a factor of }4x^3-6x^2-4x+k$

Now,

  $\begin{array}{l}=2x-1=0\\ =2x=1\\ =x=\frac{1}{2}\end{array}$

Putting value of x we get,

  $=4x^3-6x^2-4x+k=4{\left(\frac{1}{2}\right)}^3-6{\left(\frac{1}{2}\right)}^2-4\left(\frac{1}{2}\right)+k$

  $\begin{array}{l}=4\left(\frac{1}{8}\right)-6\left(\frac{1}{4}\right)-4\left(\frac{1}{2}\right)+k=0\\ =\frac{1}{2}-\frac{3}{2}-2+k=0\\ =\frac{1-3}{2}-2+k=0\\ =-\frac{2}{2}-2+k=0\\ =\frac{-2-4+2k}{2}=0\end{array}$

  $\begin{array}{l}=-6+2k=0\\ =2k=6\\ =k=\frac{6}{2}\\ =k=3\end{array}$

Thus, the value of $\text{k}=3$ .