How do you divide 4v^3+6v^2-8v-12 by 2v-3 and is it a factor of the polynomial?

1 Answer
Feb 11, 2017

$\text{ not a factor of the polynomial}$

Explanation:

Using some $\textcolor{b l u e}{\text{algebraic manipulation}}$ by substituting the divisor into the numerator as a factor.

$\frac{\textcolor{red}{2 {v}^{2}} \left(2 v - 3\right) + \left(\textcolor{b l u e}{6 {v}^{2}} + 6 {v}^{2}\right) - 8 v - 12}{2 v - 3}$

$= \frac{\textcolor{red}{2 {v}^{2}} \left(2 v - 3\right) \textcolor{red}{+ 6 v} \left(2 v - 3\right) + \left(\textcolor{b l u e}{+ 18 v} - 8 v\right) - 12}{2 v - 3}$

$= \frac{\textcolor{red}{2 {v}^{2}} \left(2 v - 3\right) \textcolor{red}{+ 6 v} \left(2 v - 3\right) \textcolor{red}{+ 5} \left(2 v - 3\right) + \left(\textcolor{b l u e}{+ 15} - 12\right)}{2 v - 3}$

$\Rightarrow 4 {v}^{3} + 6 {v}^{2} - 8 v - 12 = 2 {v}^{2} + 6 v + 5 + \frac{3}{2 v - 3}$

Since the remainder of the division is not zero.

$\text{Then "2v-3" is not a factor of the polynomial}$