2006-11-16 14:12:32 · 9 answers · asked by odie_109 1 in Science & Mathematics ➔ Mathematics
xy^2(x^2+2y^2-3xy^2)
2006-11-16 14:15:01 · answer #1 · answered by raj 7 · 0⤊ 0⤋
I guess what you need are steps, right?
We have:
x^3y^2+2xy^3-3x^2y^4
What does it mean to factor?
Factor means REVERSE MULTIPLICATION. They give you a product and you must find the factors.
Notice that in x^3y^2+2xy^3-3x^2y^4, there is an x and y in every term. See it?
We take out x and y^2 because we can divide EVERY TERM inside the parentheses by x and y^2.
Ready?
I will do x first and then y^2. Okay?
x into x^3 = x^2
x into x = 1
x into x^2 = x
Next: y^2
y^2 into y^2 = 1
y^2 into y^3 = y
y^2 into y^4 = y^2
We now put everything together.
x (y^2) (x^2 + 2y^2 - 3xy^2)
Notice:
(x), (y^2) and (x^2 + 2y^2 - 3xy^2) are the THREE factors of your original question.
Guido
2006-11-16 22:36:12 · answer #2 · answered by Anonymous · 0⤊ 0⤋
x^3y^2+2xy^3-3x^2y^4
xy^2(x^2+2y-3xy^2) That's it for this 1,
2006-11-16 22:16:12 · answer #3 · answered by yupchagee 7 · 0⤊ 0⤋
x^3y^2 + 2xy^3 - 3x^2y^4
(xy^2)(x^2 + 2y - 3y^2)
2006-11-16 23:09:51 · answer #4 · answered by Sherman81 6 · 0⤊ 0⤋
I can't get any farther than to factor out
XY^2(X^2 + 2Y - 3XY^2)
Are you sure you have the problem typed correctly?
2006-11-16 22:29:37 · answer #5 · answered by Scottie 1 · 0⤊ 0⤋
x^3y^2+2xy^3-3x^2y^4
=xy^2(x^2+2y-3xy^2)
2006-11-16 22:22:21 · answer #6 · answered by Dupinder jeet kaur k 2 · 0⤊ 0⤋
x^3y^2+2xy^3-3x^2y^4
x=elvis.
2006-11-16 22:25:39 · answer #7 · answered by Anonymous · 1⤊ 0⤋
just take out the xs and ys
xy^2(x^2-3xy^2+2y)
2006-11-16 22:16:17 · answer #8 · answered by RichUnclePennybags 4 · 0⤊ 0⤋
xy^2(x^2+2y-3xy^2)
2006-11-16 23:08:52 · answer #9 · answered by noble357 1 · 0⤊ 0⤋