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32w^3 - 2w^3 y^4

2007-03-10 14:34:58 · 5 answers · asked by Anonymous in Science & Mathematics Mathematics

5 answers

32w^3 - 2w^3 y^4

First, factor our the greatest common monomial. In this case, it's 2w^3.

2w^3 (16 - y^4)

Now, in the brackets we have a difference of squares. Factor it as such.

2w^3 (4 - y^2)(4 + y^2)

We have yet another difference of squares.

2w^3 (2 - y)(2 + y)(4 + y^2)

2007-03-10 14:39:09 · answer #1 · answered by Puggy 7 · 0 0

32w³ - 2w³y^4 =
2w³(16 - y^4) =
2w³(2² - y²)(2² + y²)
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2007-03-10 22:49:56 · answer #2 · answered by aeiou 7 · 0 0

32w^3 - 2w^3 y^4
2w^3(16-y^4)
2w^3(4+y^2)(4-y^2)
2w^3(4+y^2)(2-y)(2+y)

2007-03-10 22:42:01 · answer #3 · answered by ironduke8159 7 · 0 0

take out 2w^3

2w^3 ( 16 - y^4)

2w^3 ( 4 - y^2)(4+y^2)

2w^3 (2+y)(2-y)(2i + y)(2i-y)

2007-03-10 22:38:59 · answer #4 · answered by      7 · 0 1

2w^3(16 - y^4)
(2w^3)(4 + y^2)(4 - y^2)

(2w^3)(4 + y^2)(2 + y)(2 - y)

2007-03-10 22:38:43 · answer #5 · answered by Newbody 4 · 0 0

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