How do you factor #10x^(2)+21x-10#?

1 Answer
Jun 29, 2016

(5x - 2)(2x + 5)

Explanation:

Use the new AC Method to factor trinomials (Socratic Search)
#y = 10x^2 + 21x - 10 =# 10(x + p)(x + q)
Converted trinomial: #y' = x^2 + 21x - 100 =# (x + p') (x + q').
Find p' and q', that have opposite signs (ac < 0), and that have sum (b = 21) and product (ac = -100). They are: p' = -4 and q' = 25.
Back to original y: #p = (p')/a = -4/10 = -2/5#, and #q = (q')/a = 25/10 = 5/2#
#y = 10(x - 2/5)(x + 5/2) = (5x - 2)(2x + 5)#.