Answer
The remainder is $0$.
$\{-2,-1,2,5\}$.
Work Step by Step
Use synthetic division to divide the polynomial $x^4-4x^3-9x^2+16x+20$ by $x-c$, where $c=5$.
$\begin{matrix}
5) &1&-4&-9&16&20 \\
& &5&5&-20 &-20\\
& --&--&--& --&--\\
& 1&1&-4&-4&0
\end{matrix}$
Use $1,1,-4,-4,0$ 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 $x^3+x^2-4x-4$ and the remainder is $0$.
The factor is $x-(5)=x-5$.
We can write the given equation as
$\Rightarrow (x-5)(x^3+x^2-4x-4)=0$.
Now factor $x^3+x^2-4x-4$
Group terms.
$\Rightarrow (x^3+x^2)+(-4x-4)$
Factor each group.
$\Rightarrow x^2(x+1)-4(x+1)$
Factor out $(x+1)$.
$\Rightarrow (x+1)(x^2-4)$
$\Rightarrow (x+1)(x^2-2^2)$
Use the special formula $A^2-B^2=(A+B)(A-B)$.
$\Rightarrow (x+1)(x+2)(x-2)$
Back substitute in tthe given equation.
$\Rightarrow (x-5)(x+1)(x+2)(x-2)=0$.
Set each factor equal to zero.
$x-5=0$ or $x+1=0$ or $x+2=0$ or $x-2=0$
Isolate $x$.
$x=5$ or $x=-1$ or $x=-2$ or $x=2$
The solution set is $\{-2,-1,2,5\}$.
