How do you factor completely #10m^2-21m-10#?

1 Answer
Dec 15, 2015

Factor #y = 10m^2 - 21m - 10#

Ans: y = (5m - 2)(2m + 5)

Explanation:

I use the new AC Method to factor trinomials (Socratic Search)
#y = 10m^2 - 21m - 10 =# 10(m + p)(m + q)
Converted trinomial #y' = m^2 - 21m - 100 #= (m + p')(m + q'). The numbers p' and q' have opposite signs. Factor pairs of (-100) -->
(-2, 50)(-4, 25). This sum is 21 = -b. The opposite sum (b) gives p' = 4 and q' = -25.
Back to original trinomial, #p = (p')/10 = -4/10 = -2/5# and
#q = (q')/10 = 25/10 = 5/2#
Factored form:
#y = 10(m - 2/5)(m + 5/2) = (5m - 2)(2m + 5)#