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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