Question

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

Answer 1

`(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))`