Chapter 4, Problem 45RE

To determine

To calculate:

The remainder value by taking the ratio $\frac{f\left(x\right)}{g\left(x\right)}$ and tell that $g$ is factor of $f$ or not.

  $\begin{array}{l}f\left(x\right)=8x^3-3x^2+x+4\\ g\left(x\right)=x-1\end{array}$

Answer to Problem 45RE

The remainder value is $10$ . The $g$ is not the factor of $f$ .

Explanation of Solution

Given information:

  $\begin{array}{l}f\left(x\right)=8x^3-3x^2+x+4\\ g\left(x\right)=x-1\end{array}$

Calculation:

The functions are given as:

  $\begin{array}{l}f\left(x\right)=8x^3-3x^2+x+4\\ g\left(x\right)=x-1\end{array}$

Now take the second function:

  $\begin{array}{l}g\left(x\right)=x-1\\ x-1=0\\ x=1\end{array}$

Remainder value is to be estimated by putting the value $x=1$ in $f\left(x\right)$

  $\begin{array}{l}f\left(x\right)=8x^3-3x^2+x+4\\ f\left(1\right)=8{\left(1\right)}^3-3{\left(1\right)}^2+1+4\\ f\left(1\right)=8-3+1+4\\ f\left(1\right)=10\end{array}$

No. the $g$ is not the factor of $f$ because on putting $x=1$ , get $f\left(1\right)=10\ne 0$ .

Hence, remainder value is $10$