College Algebra & Trigonometry, 2018a

1.2. FACTORING 17
x 2 +3x − 10 = (x − 2)(x +5)
The multiplication of the (−2) and the (+5) produce the (−10) and the fact that
the 2 and 5 have opposite signs creates the difference that gives us (+3). A companion
problem to this one is x 2 − 3x − 10. In this case, the sign of the constant
term is still negative, which means that we still need factors of 10 that have a
difference of 3. This means we still need to use 2 and 5. However in this case,
instead of the (+3) as the coefficient of the middle term, we’ll need a (−3). Todo
this we simply reverse the signs of the 2 and 5 from the previous problem:
x 2 − 3x − 10 = (x +2)(x − 5)
Now the (+2)(−5) gives us (−10), but the +2x − 5x gives us (−3x) instead of
(+3x).
Example
Factor x 2 +11x − 42
In this problem, the sign of the constant term is negative. That means that we
need factors of 42 that have a difference of 11. A systematic exploration of all the
factor pairs of 42 can help us to find the correct pair:
1 42
2 21
3 14
6 7