Factor as far as possible using integer coefficients:
4xy^2 -16x^3
2007-04-16 20:59:27 · 6 answers · asked by Vokalskilz 1 in Science & Mathematics ➔ Mathematics
4xy^2 -16x^3
The common factor is 4x. Factor it out.
4xy^2 -16x^3 = 4x(y^2 - 4x^2)
= 4x[(y)^2 - (2x)^2]
= 4x(2x + y)(y - 2x)
2007-04-17 03:34:30 · answer #1 · answered by Akilesh - Internet Undertaker 7 · 0⤊ 0⤋
This cannot be written as the difference of squares, because the degree of x is odd.
Pull out 4x:
4x(y² - 4x²)
Now you have a difference of squares:
4x(y+2x)(y-2x)
2007-04-17 04:06:24 · answer #2 · answered by Anonymous · 0⤊ 0⤋
4xy^2 - 16x^3
4x(y^2 - 4x^2)
4x(y + 2x) (y - 2x)
I can only go this far
2007-04-17 04:13:49 · answer #3 · answered by detektibgapo 5 · 0⤊ 0⤋
4xy^2 -16x^3
first there is 4
xy^2-4x^3
then there is x
y^2-4x^2
now there is
(y-2x) (y+2x)
So the answer is 4x(y-2x)(y+2x)
2007-04-17 04:06:27 · answer #4 · answered by ignoramus 7 · 0⤊ 0⤋
=4x*(y^2-4x^2)
=4x(y+2x)(y-2x)
2007-04-17 05:09:49 · answer #5 · answered by Hanbiao 1 · 0⤊ 0⤋
4x.(y² - 4x²)
4x.(y - 2x).(y + 2x)
2007-04-17 04:48:14 · answer #6 · answered by Como 7 · 0⤊ 0⤋