How do you factor and solve #16x^2-48x= -27#?

1 Answer
May 23, 2015

#y = 16x^2 - 48x + 27# = (x - p)(x - q). I use the new AC Method (Google, Yahoo Search) to factor trinomials.
Converted trinomial: y' = x^2 - 48x + 432 = (x - p')(x - q').
Find p' and q' by composing factor pairs of a.c = 432. Proceed: ...(8, 54)(12, 36). This last sum is 12 + 36 = 48 = -b. Then p' = -12 and q' = -36. Then, #p = (p')/a = -12/16 = -3/4 # and #q = (q')/a = -36/16 = -9/4.#

Finally: #y = (x - 3/4)(x -9/4) = (4x - 3)(4x - 9)#

Check by developing: y = 16x^2 - 36x - 12x + 27. OK

Solving y = 0 --> (4x - 3) = 0 --> #x = 3/4#
(4x - 9) = 0 -># x = 9/4#

The new AC Method is fast, systematic, no guessing, no factoring by grouping and no solving binomials.