Chapter 4, Problem 12RE

To determine

To calculate: The remainder R when $f(x)=8x^3-3x^2+x+4$ is divided by $g(x)=x-1$ .

Also, to find out if g a factor of f .

Answer to Problem 12RE

The remainder is $10$ and g is not a factor of f .

Explanation of Solution

Given information:

  $f(x)=8x^3-3x^2+x+4$ and $g(x)=x-1$ .

Formula used:

When $f(x)$ is divided by a simple polynomial say $(x-c)$ we get,

  $f(x)=(x-c).q(x)+r(x)$

Where $r(x)$ is the remainder with either the zero polynomial or a polynomial function of degree 0.

Remainder Theorem:

The theorem states when we divide a polynomial $f(x)$ by $(x-c)$ the remainder is $f(c)$ .

Factor Theorem:

The theorem states $(x-c)$ is a factor of $f(x)$ if and only if $f(c)=0$ .

Calculation:

Consider ,

  $f(x)=8x^3-3x^2+x+4$

And $g(x)=x-1$

Since $g(x)$ is of the form $(x-c)$

  $=c=1$ .

Now ,

  $\begin{array}{l}f(1)=8{(}^1-3{(}^1+1+4\\ =8-3+1+4\\ =5+5\\ =10\end{array}$

The remainder is 10.

Also,

  $\begin{array}{l}=f(c)\ne 0\\ =10\ne 0\end{array}$ .

Hence , the remainder is 10 and g is not a factor of f .