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.

Related questions
See all questions in Factorization of Quadratic Expressions
Impact of this question
1307 views around the world
You can reuse this answer
Creative Commons License