How do you factor 6x^2+x-1 ?

Jan 29, 2017

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

Explanation:

Here are a couple of methods (in no particular order):

Method 1

Note that:

$\left(a x + 1\right) \left(b x - 1\right) = a b {x}^{2} + \left(b - a\right) x - 1$

Comparing with:

$6 {x}^{2} + x - 1$

we want to find $a , b$ such that $a b = 6$ and $b - a = 1$

The values $a = 2$, $b = 3$ work, so we find:

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

$\textcolor{w h i t e}{}$
Method 2 - Completing the square

To avoid much arithmetic with fractions, multiply first by $24 = 6 \cdot {2}^{2}$ then divide by it at the end:

$24 \left(6 {x}^{2} + x - 1\right) = 144 {x}^{2} + 24 x - 24$

$\textcolor{w h i t e}{24 \left(6 {x}^{2} + x - 1\right)} = {\left(12 x\right)}^{2} + 2 \left(12 x\right) + 1 - 25$

$\textcolor{w h i t e}{24 \left(6 {x}^{2} + x - 1\right)} = {\left(12 x + 1\right)}^{2} - {5}^{2}$

$\textcolor{w h i t e}{24 \left(6 {x}^{2} + x - 1\right)} = \left(\left(12 x + 1\right) - 5\right) \left(\left(12 x + 1\right) + 5\right)$

$\textcolor{w h i t e}{24 \left(6 {x}^{2} + x - 1\right)} = \left(12 x - 4\right) \left(12 x + 6\right)$

$\textcolor{w h i t e}{24 \left(6 {x}^{2} + x - 1\right)} = \left(4 \left(3 x - 1\right)\right) \left(6 \left(2 x + 1\right)\right)$

$\textcolor{w h i t e}{24 \left(6 {x}^{2} + x - 1\right)} = 24 \left(3 x - 1\right) \left(2 x + 1\right)$

Divide both ends by $24$ to get:

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