How do you factor # (15x²+22x-5)?

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How do you factor # (15x²+22x-5)?

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Answer 1 · 2015-05-04T11:55:49

In an attempt to find integer based factors:
We are looking for factors of 15 $(a \times b)$ $(axxb)$
and factors of 5 $(c \times d)$ $(cxxd)$
such that $a c - b d = 22$ $ac-bd = 22$

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

Since the final term of the given expression is negative, one of $(5, 1)$ $(5,1)$ 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:
$15 x^{2} + 22 x - 5 = (5 x - 1) (3 x + 5)$ $15x^2+22x-5=(5x-1)(3x+5)$

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How do you factor # (15x²+22x-5)?

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