Factor completely
y^4+3y^3-5y^2-15y
2007-06-10 14:30:49 · 4 answers · asked by Pimpmeister 1 in Education & Reference ➔ Homework Help
y^4+3y^3-5y^2-15y
Factor
y²(y² - 5) +3y(y² - 5)
(y² + 3y)(y² - 5)
y(y + 3)(y² - 5)
We could do one more step
y(y + 3)(y - √5)(y + √5)
.
2007-06-10 14:36:48 · answer #1 · answered by Robert L 7 · 0⤊ 1⤋
What grade is this?
You could try factoring out a y to get
y(y^3+3y^2-5y-15y)
and then add the -5y and -15y to get -20y
So you now have:
y(y^3+3y^2-20y)
And after that, I'm not sure if you can factor it down anymore.
2007-06-10 21:34:13 · answer #2 · answered by Anonymous · 0⤊ 1⤋
y(y^3+3y^2-5y-15)
y{y^2(y+3)-5(y+3)}
y(y^2-5)(y+3)
2007-06-10 21:41:09 · answer #3 · answered by turnawayandgo 2 · 0⤊ 0⤋
i am not looking forward to that.
2007-06-10 21:33:11 · answer #4 · answered by Sienna 2 · 0⤊ 0⤋