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

Mar 30, 2016

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

Explanation:

Call $f \left(x\right) = \frac{g \left(x\right)}{h \left(x\right)} .$
Factor g(x) by the new AC Method to factor trinomials (Socratic)
$g \left(x\right) = 4 {x}^{2} + 16 x + 15 =$ 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) -->$p = \frac{p '}{a} = \frac{6}{4} = \frac{3}{2}$ and$q = \frac{q '}{a} = \frac{10}{4} = \frac{5}{2.}$
$g \left(x\right) = 4 \left(x + \frac{3}{2}\right) \left(x + \frac{5}{2}\right) = \left(2 x + 3\right) \left(2 x + 5\right)$
Factor h(x) = 2x^2 + x - 3
Since a + b + c = 0, use shortcut. One factor is (x - 1) and the other is $\left(- \frac{c}{a} = \frac{3}{2}\right) .$
$h \left(x\right) = \left(x - 1\right) \left(x + \frac{3}{2}\right) = 2 \left(x - 1\right) \left(2 x + 3\right)$
Finally, $f \left(x\right) = \frac{g \left(x\right)}{h \left(x\right)} = \frac{2 x + 5}{2 \left(x - 1\right)}$