How do you factor #-63x^2 - 19x + 30#?

1 Answer
Nghi N.
May 28, 2015

#y = -(63x^2 + 19x - 30# = (x - p)(x - q).
I use the new AC Method to factor trinomials (Google, Yahoo Search).
Convert y to #y' = x^2 + 19x - 1890 #= (x - p')(x - q').
p' and q' have different signs (Rule of Signs). Compose factor pairs of (a.c = -1890) and find the pair whose sum is b.
Use a calculator and start from the middle:...(-30, 63)(-35, 54). This sum is (54 - 35 = 19 = b). Then, p' = -35 and q' = 54.

Back to y,#p = (p')/a = -35/63 = -5/9#, and #q = (o')/a = 54/63 = 6/7#

Factored form: y = -[(x - 5/9)(x + 6/7)] = -[(9x - 5)(7x + 6)]

Check by developing:# y = -(63x^2 + 54x - 35x -30)# OK

NOTE . The new AC Method is fast, systematic, no guessing, no factoring by grouping, and no solving binomials.

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