# How do you factor and solve 6b^2 - 13b + 3 = -3?

Mar 28, 2015

Rewrite $6 {b}^{2} - 13 b + 3 = - 3$
as
$6 {b}^{2} - 13 b + 6 = 0$

Assuming only integer values appear in the factoring (not necessarily true, but easier if it is)
we are looking for integer values for $p$, $q$, $r$, and $s$ (all $\ge 0$)
such that
$\left(p b - q\right) \left(r b - s\right) = 6 {b}^{2} - 13 b + 6$
(the negative sign on the coefficient $- 13$ tells us that at least one of the internal signs in the factors is negative, assuming positive values for $p , q , r , s$
and the positive sign on $+ 6$ tells us that the internal signs are the same i.e. they are both negative).

$\left(p b - q\right) \left(r b - s\right) = \left(p b r\right) {b}^{2} - \left(p s + q r\right) b + q s$
$\rightarrow p b = 6$
$\rightarrow p s + q r = 13$
$\rightarrow q s = 6$

Since the only positive integer factors of $6$ are $6 \times 1$ and $3 \times 2$

It is a fairly simply matter to discover
$p = 3$
$q = 2$
$r = 3$
$s = 2$

That is
${\left(3 b - 2\right)}^{2} = 6 {b}^{2} - 12 b + 6 = 0$

Which implies there is only one solution ($3 b - 2 = 0$)
$b = \frac{2}{3}$