# How do you factor completely: 9x^2 + 25x -6?

Jul 15, 2015

I found: $\left(x + 3\right) \left(9 x - 2\right)$

#### Explanation:

What I did was to solve using the Quadratic Formula the second degree equation:
$9 {x}^{2} + 25 x - 6 = 0$
so: ${x}_{1 , 2} = \frac{- 25 \pm \sqrt{841}}{18}$ that gives you:
${x}_{1} = - 3$
${x}_{2} = \frac{2}{9}$
So basically my equation can be written as:
$\left(x + 3\right) \left(x - \frac{2}{9}\right) = 0$ or
$\left(x + 3\right) \left(\frac{9 x - 2}{9}\right) = 0$
$\left(x + 3\right) \left(9 x - 2\right) = 0 \cdot 9 = 0$

Jul 16, 2015

Factor: $y = 9 {x}^{2} + 25 x - 6$

y = (9x + 2)( x - 3)

#### Explanation:

$y = 9 {x}^{2} - 25 x - 6 =$ a(x - p)(x - q)
I use the new AC Method to factor trinomials. (Google , Yahoo Search)
Converted $y ' = {x}^{2} - 25 x - 54 =$ (x - p')(x - q'). p' and q' have opposite signs. Factor pairs of (-54) --> (-1, 54)(-2, 27). This sum is 25 = -b. Then p' = 2 and q' = -27
$p = \frac{2}{9}$ and $q = - \frac{27}{9} = - 3$

Factor form: $y = 9 \left(x + \frac{2}{9}\right) \left(x - 3\right) = \left(9 x + 2\right) \left(x - 3\right)$