How do you factor #4a^2 - 8a - 5#?

1 Answer
May 11, 2016

Answer:

(2a + 1)(2a - 5)

Explanation:

You can use either the AC Method (YouTube), or the new AC Method (Socratic Search). The AC method proceeds by factoring through grouping. The new AC method proceeds by finding the factors through a converted trinomial.
#y = 4a^2 - 8a - 5 =# 4(a + p)(a + q).
Converted trinomial: #y' = a^2 - 8a - 20 =# (a + p')(a + q').
p' and q' have opposite signs because ac < 0.
Factor pairs of (ac = -20) --> ...(-2, 10)(2, -10). This last sum is -8 = b. Then p' = 2 and q' = -10.
Back to original y --> #p = p'/(a) = 2/4 = 1/2# and #q = (q')/a = -10/4 = -5/2#.
Factored form:# y = 4(a + 1/2)(a - 5/2) = (2a + 1)(2a - 5)#