How do you factor #y= 6x^2 + 11x + 4# ?

2 Answers
May 30, 2016

Answer:

#y=(2x+1)(3x+4)#

Explanation:

#y=6x^2+11x+4#

#y=(2x+1)(3x+4)#

May 30, 2016

Answer:

y = (2x + 1)(3x + 4)

Explanation:

Use the new AC Method (Socratic Search)
#y = 6x^2 + 11x + 4 =# 6(x + p)(x + q)
Converted trinomial: #y' = x^2 + 11x + 24 =# (x + p')(x + q')
p' and q' have same sign because ac > 0.
Factor pairs of (ac = 24) --> ...(2, 12)(3, 8). This sum is 11 = b. Then, p' 3 and q' = 8.
Back to the original y --> #p = (p')/a = 3/6 = 1/2#, and
#q = (q')/a = 8/6 = 4/3#
Factored form: #y = 6(x + 1/2)(x + 4/3) = (2x + 1)(3x + 4)#