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 $f\left(x\right)$.

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

Short answer

The remainder is $0$and $x-\frac{1}{2}$is a factor.

Step-by-step solution

Step 1. Given information.

The given function is,

$f\left(x\right)=2x^4-x^3+2x-1;x-\frac{1}{2}$.

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 $x-\frac{1}{2}$, then the remainder is $f\left(\frac{1}{2}\right)$that is

$$\begin{gathered} f\left(\frac{1}{2}\right)=2{\left(\frac{1}{2}\right)}^4-{\left(\frac{1}{2}\right)}^3+2\left(\frac{1}{2}\right)-1 \\ =2\left(\frac{1}{16}\right)-\frac{1}{8}+1-1 \\ =\frac{1}{8}-\frac{1}{8} \\ =0 \end{gathered}$$

Therefore, the remainder is $0$.

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 $c=\frac{1}{2}$, then

$$\begin{gathered} f\left(c\right)=f\left(\frac{1}{2}\right) \\ =0 \end{gathered}$$

We observe that $f\left(\frac{1}{2}\right)=0$,then $x-\frac{1}{2}$is a factor.