Question

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

Answer 1

`f(y) = 2y^3(3y^2+ 16y + 16)`
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).

Answer 2

In this polynomial expression I notice that all the degrees of y are greater than 3, so I decide to "simplify" by `y^3`:
`6y^5+32y^4+32y^3=`
`=y^3(6y^2+32y+32)=`
`=2y^3(3y^2+16y+16)`
Now I can even try to factor the part `3y^2+16y+16` calculating its solutions:
`Delta_y=16^2-4*3*16=256-192=64=8^2.`
`y_(1,2)=(-16+-8)/6=(-8+-4)/3`
So `y_1=-4/3` and `y_2=-4`.
We can state that:
`3y^2+16y+16=(y+4)(y+4/3)`
So the initial polinomial expression can be rewritten as this:
`6y^5+32y^4+32y^3=2y^3(y+4)(y+4/3)`