How do you factor #(4x^2+16x+15)/(2x^2+x-3)#?

1 Answer
Mar 30, 2016

Answer:

#(2x + 5)/(2(x - 1)#

Explanation:

Call #f(x) = (g(x))/(h(x)).#
Factor g(x) by the new AC Method to factor trinomials (Socratic)
#g(x) = 4x^2 + 16x + 15 =# 4(x + p)(x + q)
Converted trinomial g'(x) = x^2 + 16 x + 60 = (x + p')(x + q')
Factor pairs od (6)) -->... (5, 12)(6, 10). This sum is 16 = b. Then p' = 6 and q' = 10.
Back to g(x) --># p = (p')/a = 6/4 = 3/2# and# q = (q')/a = 10/4 = 5/2.#
#g(x) = 4(x + 3/2)(x + 5/2) = (2x + 3)(2x + 5)#
Factor h(x) = 2x^2 + x - 3
Since a + b + c = 0, use shortcut. One factor is (x - 1) and the other is #(-c/a = 3/2).#
#h(x) = (x - 1)(x + 3/2) = 2(x - 1)(2x + 3)#
Finally, #f(x) = (g(x))/(h(x)) = (2x + 5)/(2(x - 1))#