Question

Use the Remainder Theorem to find the remainder when $f\left(x\right)$is divided by $x-c$. Then use the Factor Theorem to determine whether $x-c$is a factor of $f\left(x\right)$.

$f\left(x\right)=3x^4+x^3-3x+1;x+\frac{1}{3}$

Short answer

The remainder is$2$and$x+\frac{1}{3}$is not a factor.

Step-by-step solution

Step 1. Given information.

The given function is,

$f\left(x\right)=3x^4+x^3-3x+1;x+\frac{1}{3}$

Step 2. Using remainder theorem.

We know that if $f\left(x\right)$is divided by $x-c$then the remainder will be $f\left(c\right)$.

Here $f\left(x\right)$is divided by $x+\frac{1}{3}$, then the remainder is $f\left(-\frac{1}{3}\right)$that is

$$\begin{gathered} f\left(-\frac{1}{3}\right)=3{\left(-\frac{1}{3}\right)}^4+{\left(-\frac{1}{3}\right)}^3-3\left(-\frac{1}{3}\right)+1 \\ =\frac{1}{27}-\frac{1}{27}+1+1 \\ =2 \end{gathered}$$

Therefore, the remainder is $2$.

Step 3. Using factor theorem.

We know that if $x-c$is the factor of the polynomial $f\left(x\right)$then

$f\left(c\right)=0$,

Here $c=-\frac{1}{3}$, then

$$\begin{gathered} f\left(c\right)=f\left(-\frac{1}{3}\right) \\ =2 \end{gathered}$$

We observe that $f\left(-\frac{1}{3}\right)\ne 0$,then $x+\frac{1}{3}$is not a factor.