What are the factors for #10x^2 - 7x - 12#?

1 Answer
Nghi N.
May 9, 2015

I use the new AC Method (Google Search) to factor

#f(x) = 10x^2 - 7x - 12#= (x - p)( - q)

Converted trinomial: f'(x) = x^2 - 7x - 120. (ac = -12(10) = -120).
Find 2 numbers p' and q' knowing their sum (-7) and their product (-120).
a and c have different sign. Compose factor pairs of a*c = -120. Proceed: (-1, 120)(-2, 60)...(-8, 15), This sum is 15 - 8 = 7 = -b. Then, p' = 8 and q' = -15.
Next, find p = p'/a = 8/10 = 4/5; and q = q'/a = -15/10 = -3/2.

Factored form of f(x):
f(x) = (x - p)(x - q) = (x + 4/5)(x - 3/2) = (5x + 4)(2x - 3)

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