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)#