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

2 Answers
Jun 8, 2015

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).

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:
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)