Answer
Solution $x=1$.
The remainder is zero.
$\left \{ \frac{1}{3},\frac{1}{2},1\right \}$.
Work Step by Step
The given equation is
$6x^3-11x^2+6x-1=0$
By using the table the value of $y_1$ is zero at $x=1$
Thus, a solution is $x=1$
The value of $c$ is $1$. We will show that c=1 is a solution of the given equation using synthetic division.
We divide the polynomial $6x^3-11x^2+6x-1$ by $x−c=x−1$.
Use synthetic division
$\begin{matrix}
1) &6&-11&6&-1 \\
& &6&-5&1 \\
& --&--&--& --\\
& 6&-5&1&0
\end{matrix}$
The remainder is zero, which means the solution is correct.
The quotient is $6x^2-5x+1$
We can write the given equation in a factor form shown below.
$\Rightarrow (x-1)(6x^2-5x+1)=0 $
Rewrite the middle term $-5x$ as $-3x+2x$.
$\Rightarrow (x-1)(6x^2-3x-2x+1)=0 $
Group the terms.
$\Rightarrow (x-1)[(6x^2-3x)+(-2x+1)]=0 $
Factor each group.
$\Rightarrow (x-1)[3x(2x-1)-1(2x-1)]=0 $
Factor out $(2x-1)$.
$\Rightarrow (x-1)(2x-1)(3x-1)=0 $
By using zero product rule set each factor equal to zero.
$\Rightarrow x-1=0$ or $2x-1=0$ or $3x-1=0 $
Isolate $x$.
$\Rightarrow x=1$ or $x=\frac{1}{2}$ or $x=\frac{1}{3} $
The solution set is $\left \{ \frac{1}{3},\frac{1}{2},1\right \}$.
