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

1 Answer

Rewrite #6b^2 - 13b + 3 = -3#
as
#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#, #q#, #r#, and #s# (all #>= 0#)
such that
#(pb-q)(rb-s) = 6b^2-13b+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).

#(pb-q)(rb-s) = (pbr)b^2 - (ps+qr)b + qs#
#rarr pb = 6#
#rarr ps+qr = 13#
#rarr qs = 6#

Since the only positive integer factors of #6# are #6xx1# and #3xx2#

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

That is
#(3b-2)^2 = 6b^2-12b+6 = 0#

Which implies there is only one solution (#3b-2=0#)
#b=2/3#