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

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

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Answer 1 · 2015-03-28T14:34:16

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

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

$(p b - q) (r b - s) = (p b r) b^{2} - (p s + q r) b + q s$ $(pb-q)(rb-s) = (pbr)b^2 - (ps+qr)b + qs$
$\to p b = 6$ $rarr pb = 6$
$\to p s + q r = 13$ $rarr ps+qr = 13$
$\to q s = 6$ $rarr qs = 6$

Since the only positive integer factors of $6$ $6$ are $6 \times 1$ $6xx1$ and $3 \times 2$ $3xx2$

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

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

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

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

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