# How do you factor the expression 2x^2 +11x-9?

Sep 23, 2016

$2 {x}^{2} + 11 x - 9 = \frac{1}{8} \left(4 x + 11 - \sqrt{193}\right) \left(4 x + 11 + \sqrt{193}\right)$

#### Explanation:

Note that if the $-$ sign was a $+$, then this would factor with integer coefficients:

$2 {x}^{2} + 11 x + 9 = \left(x + 1\right) \left(2 x + 9\right)$

To factor the given quadratic with a minus sign, we can complete the square, but to cut down on fractions, I would like to multiply through by $8$ first, then divide through by $8$ at the end...

$8 \left(2 {x}^{2} + 11 x - 9\right) = 16 {x}^{2} + 88 x - 72$

$\textcolor{w h i t e}{8 \left(2 {x}^{2} + 11 x - 9\right)} = {\left(4 x\right)}^{2} + 2 \left(11\right) \left(4 x\right) + 121 - 193$

$\textcolor{w h i t e}{8 \left(2 {x}^{2} + 11 x - 9\right)} = {\left(4 x + 11\right)}^{2} - {\left(\sqrt{193}\right)}^{2}$

$\textcolor{w h i t e}{8 \left(2 {x}^{2} + 11 x - 9\right)} = \left(\left(4 x + 11\right) - \sqrt{193}\right) \left(\left(4 x + 11\right) + \sqrt{193}\right)$

$\textcolor{w h i t e}{8 \left(2 {x}^{2} + 11 x - 9\right)} = \left(4 x + 11 - \sqrt{193}\right) \left(4 x + 11 + \sqrt{193}\right)$

So:

$2 {x}^{2} + 11 x - 9 = \frac{1}{8} \left(4 x + 11 - \sqrt{193}\right) \left(4 x + 11 + \sqrt{193}\right)$