How do you factor #42c^2-29c-5#?

1 Answer
Nghi N.
Jul 7, 2015

Factor #y = 42x^2 - 29x - 5#

Explanation:

#y = 42x^2 - 29x - 5 =# a(x - p)(x - q).
I use the new AC Method to factor trinomials (Google, Yahoo Search).
Converted #y' = x^2 - 29x - 210 = #(x - p')(x - q').
a and c have different signs.
Factor pairs of ac = -210 ->...(-5, 42)(-6, 35). This sum is 29 = -b.
Then p' = 6 and q' = -35.
Then, #p = (p')/a = 6/42 = 1/7# and #q = (q')/a = -35/42 = -5/6#

Factored form: #y = 42(x + 1/7)(x - 5/6) = (7x + 1)(6x - 5)#

Note. This new method is fast, systematic, no factoring by grouping and no solving binomials.

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