How do you factor #24y^2 + 41y + 12#?

1 Answer
Aug 6, 2015

Answer:

Factor f(y) = 24y^2 + 41y + 12

Ans: (8x + 3)(3x + 4)

Explanation:

#y = 24y^2 + 41y + 12 = #24(x - p)(x - q)
I use the new AC Method (Google, Yahoo Search)
Converted trinomial #y' = x^2 + 41 x + 288 =# (x - p')(x - q')
p' and q' have same sign (Rule of signs)
Factor pairs of 288 --> ...(6, 48)(8, 36)(9, 32). This sum is 41 = -b.
Then p' = 9 and q' = 32
Therefor, #p = (p')/a = 9/24 = 3/8#, and #q = (q')/a = 32/24 = 4/3.#

Factored form: #y = 24(x + 3/8)(x + 4/3) = (8x + 3)(3x + 4)#

NOTE. Solving by the new AC Method is fast, systematic, no guessing, no lengthy solving by grouping.