Answer
Solution $x=1$.
The remainder is zero.
$\left \{ -6,-\frac{1}{2},1\right \}$.
Work Step by Step
The given equation is
$2x^3+11x^2-7x-6=0$
By using the table the value of $y_1$ is zero at $x=1$
Thus, the solution is $x=1$.
The value of $c$ is $1$.
Use synthetic division to divide the polynomial $2x^3+11x^2-7x-6$ by $x-c=x-1$.
$\begin{matrix}
1) &2&11&-7&-6 \\
& &2&13&6 \\
& --&--&--& --\\
& 2&13&6&0
\end{matrix}$
The remainder is zero, which means the solution is correct.
The quotient is $2x^2+13x+6$.
We can write the given equation in a factor form shown below.
$\Rightarrow (x-1)(2x^2+13x+6)=0 $
Rewrite the middle term $13x$ as $12x+1x$.
$\Rightarrow (x-1)(2x^2+12x+1x+6)=0 $
Group the terms.
$\Rightarrow (x-1)[(2x^2+12x)+(1x+6)]=0 $
Factor each group.
$\Rightarrow (x-1)[(2x(x+6)+1(x+6)]=0 $
Factor out $(2x-1)$.
$\Rightarrow (x-1)(x+6)(2x+1)=0 $
By using zero product rule set each factor equal to zero.
$\Rightarrow x-1=0$ or $x+6=0$ or $2x+1=0 $
Isolate $x$.
$\Rightarrow x=1$ or $x=-6$ or $x=-\frac{1}{2} $
The solution set is $\left \{ -6,-\frac{1}{2},1\right \}$.
