How do i factor this math problem
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How do i factor this math problem

[From: ] [author: ] [Date: 11-05-14] [Hit: ]
y^2,The lowest power in z^2, 1,Im trying to make the thing in brackets look like a quadratic, but its not cooperating much.It does look like a quadratic in y but I dont think it will separate from z in a simple way.......
72y^3 z^2+12y^2 minus 24y^4 z^2 thanks

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72 y^3 z^2 + 12y^2 - 24 y^4 z^2

The greatest common factor of the three terms is 12 y^2, so factor that out:

72 y^3 z^2 + 12y^2 - 24 y^4 z^2 =
12 y^2 (6 y z^2 + 1 - 2 y^2 z^2)

If you want, you can also reorder the second factor to place the +1 last, and factor 2 y z^2 out of its other two terms:

12 y^2 (6 y z^2 + 1 - 2 y^2 z^2) =
12 y^2 (6 y z^2 - 2 y^2 z^2 + 1) =
12 y^2 (2 y z^2 (3 - y) + 1)

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72 y^3 z^2 + 12 y^2 - 24 y^4 z^2

Common factors:
72 = 6 x 12, 12 = 1 x 12, -24 = -2 x 12 so factor out 12

The lowest power in y^3, y^2, y^4 is y^2 so factor out y^2

The lowest power in z^2, 1, z^2 is 1 so you can't factor out z or z^2

So far you get this
12 y^2 (6 y z^2 + 1 - 2 y^2 z^2)
-12 y^2 (y^2 z^2 - 6 y z^2 - 1)

I'm trying to make the thing in brackets look like a quadratic, but it's not cooperating much.
It does look like a quadratic in y but I don't think it will separate from z in a simple way.

-
72y^3z^2 + 12y^2 - 24y^4z^2

12y^2(6yz^2 + 1 - 2y^2z^2)
12y^2[2yz^2(3 - y) + 1]
1
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