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 $f\left(x\right)$.

$f\left(x\right)=4x^6-64x^4+x^2-15;x+4$

Short answer

The remainder is $1$and $x+4$is not a factor.

Step-by-step solution

Step 1. Given information.

The given function is,

$f\left(x\right)=4x^6-64x^4+x^2-15;x+4$

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 $x+4$, then the remainder is $f(-4)$that is

$$\begin{gathered} f(-4)=4{\left(-4\right)}^6-64{\left(-4\right)}^4+{\left(-4\right)}^2-15 \\ =4\left(4096\right)-64\left(256\right)+16-15 \\ =16384-16384+1 \\ =1 \end{gathered}$$

Therefore, the remainder is $1$.

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 $c=-4$, then

$$\begin{gathered} f\left(c\right)=f(-4) \\ =1 \end{gathered}$$

We observe that $f(-4)\ne 0$,then $x+4$is not a factor.