Answer
The remainder is $0$.
$\left \{-5,\frac{1}{3},\frac{1}{2}\right\} $.
Work Step by Step
The given equation is
$\Rightarrow 6x^3+25x^2-24x+5=0$.
The solution is $c=-5$.
Synthetic division.
Write the coefficients of the dividend as shown below.
$\begin{matrix}
-5) &6&25&-24&5 &
\end{matrix}$
1. Bring down $6$.
$\begin{matrix}
-5) &6&25&-24&5 \\
& & \\
& --& \\
& 6&
\end{matrix}$
2. Multiply $-5(6)=-30$.
$\begin{matrix}
-5) &6&25&-24&5 \\
& &-30&& \\
& --&--& \\
& 6&
\end{matrix}$
3. Add columns
$\begin{matrix}
-5) &6&25&-24&5 \\
& &-30&& \\
& --&--& \\
& 6&-5&
\end{matrix}$
4. Multiply $-5(-5)=25$.
$\begin{matrix}
-5) &6&25&-24&5 \\
& &-30&25& \\
& --&--&--& \\
& 6&-5&&
\end{matrix}$
5. Add columns
$\begin{matrix}
-5) &6&25&-24&5 \\
& &-30&25& \\
& --&--&--& \\
& 6&-5&1&
\end{matrix}$
6. Multiply $-5(1)=-5$.
$\begin{matrix}
-5) &6&25&-24&5 \\
& &-30&25&-5 \\
& --&--&--& --\\
& 6&-5&1&
\end{matrix}$
7. Add columns
$\begin{matrix}
-5) &6&25&-24&5 \\
& &-30&25&-5 \\
& --&--&--& --\\
& 6&-5&1&0
\end{matrix}$
Use $6,-5,1,0$ to write the quotient and 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 $6x^2-5x+1$ and the remainder is $0$.
The factor is $x-(c)=x-(-5)=x+5$.
We can write the given equation as
$\Rightarrow (x+5)(6x^2-5x+1)=0$.
Now factor $6x^2-5x+1$
$\Rightarrow 6x^2-5x+1$
Rewrite the middle term $-5x$ as $-3x-2x$.
$\Rightarrow 6x^2-3x-2x+1$
Group terms.
$\Rightarrow (6x^2-3x)+(-2x+1)$
Factor each group.
$\Rightarrow 3x(2x-1)-1(2x-1)$
Factor out $(2x-1)$.
$\Rightarrow (2x-1)(3x-1)$
Back substitute.
$\Rightarrow (x+5)(2x-1)(3x-1)=0$.
Set each factor equal to zero.
$x+5=0$ or $2x-1=0$ or $3x-1=0$
Isolate.
$x=-5$ or $x=\frac{1}{2}$ or $x=\frac{1}{3}$
The solution set is $\{-5,\frac{1}{3},\frac{1}{2}\}$.
