# Is it possible to factor y=3x^2 + 9x-6? If so, what are the factors?

Feb 22, 2017

$3 {x}^{2} + 9 x - 6 = 3 \left(x + \frac{3}{2} - \frac{\sqrt{17}}{2}\right) \left(x + \frac{3}{2} + \frac{\sqrt{17}}{2}\right)$

#### Explanation:

Complete the square then use the difference of squares identity:

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

with $a = \left(2 x + 3\right)$ and $b = \sqrt{17}$ as follows:

$3 {x}^{2} + 9 x - 6 = \frac{3}{4} \left(4 {x}^{2} + 12 x - 8\right)$

$\textcolor{w h i t e}{3 {x}^{2} + 9 x - 6} = \frac{3}{4} \left({\left(2 x\right)}^{2} + 2 \left(2 x\right) \left(3\right) + {3}^{2} - 17\right)$

$\textcolor{w h i t e}{3 {x}^{2} + 9 x - 6} = \frac{3}{4} \left({\left(2 x + 3\right)}^{2} - {\left(\sqrt{17}\right)}^{2}\right)$

$\textcolor{w h i t e}{3 {x}^{2} + 9 x - 6} = \frac{3}{4} \left(2 x + 3 - \sqrt{17}\right) \left(2 x + 3 + \sqrt{17}\right)$

$\textcolor{w h i t e}{3 {x}^{2} + 9 x - 6} = 3 \left(x + \frac{3}{2} - \frac{\sqrt{17}}{2}\right) \left(x + \frac{3}{2} + \frac{\sqrt{17}}{2}\right)$