Answer
The remainder is $0$.
$\{-4,2,-\frac{1}{3}\}$.
Work Step by Step
We will show that $c=-4$ is a solution of the given equation using synthetic division and the Remainder Theorem.
We divide the polynomial $3x^3+7x^2-22x-8$ by $x-c=x-(-4)$.
Synthetic division.
Write the coefficients of the dividend as shown below.
$\begin{matrix}
-4) &3&7&-22&-8 &
\end{matrix}$
1. Bring down $3$.
$\begin{matrix}
-4) &3&7&-22&-8 & \\
& & \\
& --& \\
& 3&
\end{matrix}$
2. Multiply $-4(3)=-12$.
$\begin{matrix}
-4) &3&7&-22&-8 & \\
& &-12&& \\
& --&--& \\
& 3&
\end{matrix}$
3. Add column wise.
$\begin{matrix}
-4) &3&7&-22&-8 & \\
& &-12&& \\
& --&--& \\
& 3&-5&
\end{matrix}$
4. Multiply $-4(-5)=20$.
$\begin{matrix}
-4) &3&7&-22&-8 & \\
& &-12&20& \\
& --&--&--& \\
& 3&-5&&
\end{matrix}$
5. Add column wise.
$\begin{matrix}
-4) &3&7&-22&-8 & \\
& &-12&20& \\
& --&--&--& \\
& 3&-5&-2&
\end{matrix}$
6. Multiply $-4(-2)=8$.
$\begin{matrix}
-4) &3&7&-22&-8 & \\
& &-12&20&8 \\
& --&--&--& --\\
& 3&-5&-2&
\end{matrix}$
7. Add column wise.
$\begin{matrix}
-4) &3&7&-22&-8 & \\
& &-12&20&8 \\
& --&--&--& --\\
& 3&-5&-2&0
\end{matrix}$
Use $3,-5,-2,0$ to write the quotient and the remainder.
The degree of the first term of the quotient is one less than the degree of the first term of the dividend.
The quotient is $3x^2-5x-2$ and the remainder is $0$.
The factor is $x-(c)=x-(-4)=x+4$.
Because the remainder is $0$, the polynomial has a value of $0$ when x = -4, therefore $-4$ is a solution of the given equation.
We can write the given equation as
$\Rightarrow (x+4)(3x^2-5x-2)=0$.
Solve the equation.
First factor $3x^2-5x-2$
Rewrite the middle term $-5x$ as $-6x+1x$.
$\Rightarrow 3x^2-6x+1x-2$
Group terms.
$\Rightarrow (3x^2-6x)+(1x-2)$
Factor each group.
$\Rightarrow 3x(x-2)+1(x-2)$
Factor out $(x-2)$.
$\Rightarrow (x-2)(3x+1)$
Back substitute.
$\Rightarrow (x+4)(x-2)(3x+1)=0$.
Set each factor equal to zero.
$x+4=0$ or $x-2=0$ or $3x+1=0$
Isolate $x$.
$x=-4$ or $x=2$ or $x=-\frac{1}{3}$
The solution set is $\left\{-4,2,-\frac{1}{3}\right\}$.
