# How do you factor 9y^2+18y+8?

May 9, 2015

$9 {y}^{2} + 18 y + 8$

$= 9 \left({y}^{2} + 2 y + \frac{8}{9}\right)$

Looking for $a \times b = \frac{8}{9}$ with $a + b = 2$
reveals the solution:

$= 9 \left(y + \frac{4}{3}\right) \left(y + \frac{2}{3}\right)$

May 9, 2015

I use the new AC Method (Google, Yahoo Search) to factor:
f(x) = 9y^2 + 18y + 8. = (y - p)(y - q)
Converted trinomial: f'(y) = y^2 + 18y + 72 = (y -p') (y - q') with
(a.c = 9.(8) = 72.
Compose factor pairs of ac = 72. Proceed: (1, 72)(2, 36)...(6, 12).OK
$p ' = 6 \to p = \frac{p '}{a}$ = $\frac{6}{9} = \frac{2}{3}$
$\mathmr{and} q ' = 12 \to q = \frac{q '}{a} = \frac{12}{9} = \frac{4}{3}$

f(y) = (y + 2/3)(y + 4/3) = (3y + 2)(3y + 4)