How do you factor #10x^2+23x+6 #?

1 Answer
Apr 12, 2016

Answer:

(10x + 3)(x + 2)

Explanation:

Use the new systematic, non-guessing AC Method (Socratic Search)
#y = 10x^2 + 23x + 6 =# 10(x + p)(x + q)
Converted trinomial: #y' = x^2 + 23x + 60 =#(x + p')(x + q').
p' and q' have same sign because ac > 0.
Factor pairs of (ac = 60) --> (2, 30)(3, 20). This last sum is 23 = b.
Then, #p = (p')/a = 3/10# and #q = (q')/a = 20/10 = 2#

Factored form: #y = 10(x + 3/10)(x + 2) = (10x + 3)(x + 2)#