# How do you divide 3x^3-4x^2-17x+6 by 3x-1 and is it a factor of the polynomial?

Aug 21, 2016

$\left(3 x - 1\right)$ is a factor of the given poly., and, when the poly. is divided

by $\left(3 x - 1\right)$, the quotient is $\left({x}^{2} - x - 6\right)$.

#### Explanation:

Let $p \left(x\right) = 3 {x}^{3} - 4 {x}^{2} - 17 x + 6$.

To find whether or not $\left(3 x - 1\right)$ is a factor of $p \left(x\right)$, we have to

check whether $p \left(\frac{1}{3}\right)$ is 0 or not.

$p \left(\frac{1}{3}\right) = 3 \left(\frac{1}{27}\right) - 4 \left(\frac{1}{9}\right) - 17 \left(\frac{1}{3}\right) + 6 = \frac{1}{9} - \frac{4}{9} - \frac{17}{3} + 6 = - \frac{1}{3} - \frac{17}{3} + 6 = - 6 + 6 = 0$

Hence, $\left(3 x - 1\right)$ is a factor of $p \left(x\right)$.

Now, $p \left(x\right) = 3 {x}^{3} - 4 {x}^{2} - 17 x + 6$.

$= \underline{3 {x}^{3} - {x}^{2}} - \underline{3 {x}^{2} + x} - \underline{18 x + 6}$.

$= {x}^{2} \left(3 x - 1\right) - x \left(3 x - 1\right) - 6 \left(3 x - 1\right)$

$= \left(3 x - 1\right) \left({x}^{2} - x - 6\right)$

Thus, $\left(3 x - 1\right)$ is a factor of $p \left(x\right)$, and, when $p \left(x\right)$ is divided by

$\left(3 x - 1\right)$, the quotient is $\left({x}^{2} - x - 6\right)$.

Enjoy Maths.!