# How do you factor 6y^5 + 32y^4 + 32y^3?

Jun 8, 2015

$f \left(y\right) = 2 {y}^{3} \left(3 {y}^{2} + 16 y + 16\right)$
Factor the trinomial in parentheses by the new AC Method (Yahoo, Google Search)
Converted trinomial: y^2 + 16y + 48.
Compose factor pairs of (a.c) = 48: (2, 24)(3, 16)(4, 12). OK
p' = 4 and q' = 12 --> p = 4/3 and q = 12/3 = 4

f(y) = 2y^3(3y + 4)(y + 4).

Jun 8, 2015

In this polynomial expression I notice that all the degrees of y are greater than 3, so I decide to "simplify" by ${y}^{3}$:
$6 {y}^{5} + 32 {y}^{4} + 32 {y}^{3} =$
$= {y}^{3} \left(6 {y}^{2} + 32 y + 32\right) =$
$= 2 {y}^{3} \left(3 {y}^{2} + 16 y + 16\right)$
Now I can even try to factor the part $3 {y}^{2} + 16 y + 16$ calculating its solutions:
${\Delta}_{y} = {16}^{2} - 4 \cdot 3 \cdot 16 = 256 - 192 = 64 = {8}^{2.}$
${y}_{1 , 2} = \frac{- 16 \pm 8}{6} = \frac{- 8 \pm 4}{3}$
So ${y}_{1} = - \frac{4}{3}$ and ${y}_{2} = - 4$.
We can state that:
$3 {y}^{2} + 16 y + 16 = \left(y + 4\right) \left(y + \frac{4}{3}\right)$
So the initial polinomial expression can be rewritten as this:
$6 {y}^{5} + 32 {y}^{4} + 32 {y}^{3} = 2 {y}^{3} \left(y + 4\right) \left(y + \frac{4}{3}\right)$