How do you factor # (15x²+22x-5)?

1 Answer
Alan P.
May 4, 2015

In an attempt to find integer based factors:
We are looking for factors of 15 #(axxb)#
and factors of 5 #(cxxd)#
such that #ac-bd = 22#

There are only a very few factors available and we should soon come up with #(a,b)=(5,3)# and #(c,d)=(5,1)#

Since the final term of the given expression is negative, one of #(5,1)# needs to be negative;
furthermore, since the coefficient of the middle term is positive
5 must be the positive value and 1 the negative.

Factors:
#15x^2+22x-5=(5x-1)(3x+5)#

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