How do you factor the trinomial #48b^2-74-10#?

1 Answer
Nghi N.
Mar 13, 2016

y = 2(8x - 1)(3x - 5)

Explanation:

I use the new systematic AC Method to factor trinomials (Socratic Search).
#y = 48x^2 - 74x - 10 =# 48(x + p)(x + q)
Converted trinomial: #y' = x^2 - 74x - 480 =# (x + p')(x + q').
p' and q' have opposite signs.
Factor pairs of (ac = - 480) --> ...(- 5, 96)(-6, 80). This sum is 74 = -b.
The opposite sum (6, -80) gives: p' = 8, and q' = - 80.
Then, #p = (p')/a = 6/48 = 1/8#, and #q = (q')/a = -80/48 = -5/3#
Factored form: y = 48(x + 1/8)(x - 5/3) = 2(8x + 1)(3x - 5)

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