# How do you completely factor 6x^3+7x^2-1 given 2x+1 is one of the factors?

Oct 22, 2016

The complete factorization is $\left(2 x + 1\right) \left(3 x - 1\right) \left(x + 1\right)$

#### Explanation:

We have to make long division

$6 {x}^{3} + 7 {x}^{2}$ $\textcolor{w h i t e}{a a a a}$ $- 1$ $\textcolor{w h i t e}{a a a a a a}$ ∣2x+1
$6 {x}^{3} + 3 {x}^{2}$ $\textcolor{w h i t e}{a a a a a a a a a a a a a a}$ $3 {x}^{2} + 2 x - 1$
$0 + 4 {x}^{2} + 2 x$
$\textcolor{w h i t e}{a a a a a a a}$ $- 2 x - 1$
$\textcolor{w h i t e}{a a a a a a a a}$ $+ 0 - 0$

So the result of the long division is $3 {x}^{2} + 2 x - 1$

which upon factorization gives $\left(3 x - 1\right) \left(x + 1\right)$