Answer
The remainder is zero.
$\{-1,\frac{1}{3},\frac{1}{2}\}$.
Work Step by Step
Use synthetic division to divide the polynomial $6x^3+x^2-4x+1$ by $x-c$, where $c=\frac{1}{2}$.
$\begin{matrix}
\frac{1}{2}) &6&1&-4&1 \\
& &3&2&-1 \\
& --&--&--& --\\
& 6&4&-2&0
\end{matrix}$
The quotient is $6x^2+4x-2$ and the remainder is zero.
Since the remainder is $0$, it means $\frac{1}{2}$ is a solution of the given equation.
Factor the given expression using the result from the synthetic division.
$\Rightarrow (x-\frac{1}{2})(6x^2+4x-2)=0$
Factor out $2$.
$\Rightarrow 2(x-\frac{1}{2})(3x^2+2x-1)=0$
Factor $3x^2+2x-1$.
Rewrite the middle term $+2x$ as $3x-1x$.
$\Rightarrow 3x^2+3x-1x-1$
Group the terms.
$\Rightarrow (3x^2+3x)+(-1x-1)$
Factor each term.
$\Rightarrow 3x(x+1)-1(x+1)$
Factor out $(x+1)$.
$\Rightarrow (x+1)(3x-1)$
Back substitute into the equation.
$\Rightarrow 2(x-\frac{1}{2})(x+1)(3x-1)=0$
By using zero product rule set each factor equal to zero.
$\Rightarrow x-\frac{1}{2}=0$ or $x+1=0$ or $3x-1=0$
Isolate $x$.
$\Rightarrow x=\frac{1}{2}$ or $x=-1$ or $x=\frac{1}{3}$
The solution set is $\{-1,\frac{1}{3},\frac{1}{2}\}$.
