How do you factor 14x^2 - 41x + 15?

1 Answer
Nghi N.
Aug 5, 2015

Factor y = 14x^2 - 41x + 15.

Ans: (7x - 3)(2x - 5)

Explanation:

y = 14x^2 - 41x + 15 = 14(x - p)(x - q)
I use the new AC Method. (Google, Yahoo, Bing Search)
Converted trinomial: y' = x^2 - 41x + 210 = (x - p')(x - q')
p' and q' have same sign (Rule of signs)
Factor pairs of 210 --> ...(5, 42)(6, 35). This sum is 41 = -b.
Then, p' = -6 and q' = -35.
Therefor, p = (p')/a = -6/14 = -3/7, and q = -35/14 = -5/2
Factored form: y = `14(x - 3/7)(x - 5/2) = (7x - 3)(2x - 5)

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