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 $f\left(x\right)$.

$f\left(x\right)=2x^6-18x^4+x^2-9;x+3$

Short answer

The remainder is$0$and$x+3$is a factor.

Step-by-step solution

Step 1. Given information.

The given function is,

$f\left(x\right)=2x^6-18x^4+x^2-9;x+3$.

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 $x+3$, then the remainder is $f(-3)$that is

$$\begin{gathered} f(-3)=2{\left(-3\right)}^6-18{\left(-3\right)}^4+{\left(-3\right)}^2-9 \\ =2\left(729\right)-18\left(81\right)+9-9 \\ =1458-1458 \\ =0 \end{gathered}$$

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 $c=-3$, then

$$\begin{gathered} f\left(c\right)=f(-3) \\ =0 \end{gathered}$$

We observe that $f(-3)=0$,then $x+3$is a factor.