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anonymous
 5 years ago
Factor Completely:
15x^2y^321x^2y^5+30x^4y^2
anonymous
 5 years ago
Factor Completely: 15x^2y^321x^2y^5+30x^4y^2

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This is what I got: \[x^2y^2(15y21y^3+30x^2)\] Is that the correct answer?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You can also factor a 3.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0DOH! You are right!!!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Don't forget to factor the coefficients!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I am so used to them not being able to be factored, I didn't even stop to think if they could :P

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I hear ya  I do the same thing when I use the quadratic formula to factor something like x^2+6x+5. Duh.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0My next one seems alot more complicated... \[8x^3+12x^210x15\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I know 8 has a cube root of 2 and x has a cube root of x. But there is no a and b term to use with that, nothing else can be cubed. Or would it just be (2x)^3?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yech...hopefully someone else can help, but I will guess and check. Don't forget other factors of 8x^3 though, such as 8*1*1, 4*2*1.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think it is (2x)^3 though, and the rest I just factor as normal and include in the equation. I hope so, anyway.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You can "factor by grouping". Group the 8x^3 together with the 10x. You can factor a 2X out of each getting 2x(4x^25). You can then factor a 3 out of the remaining two terms getting 3(4x^25). Resulting in (2x+3)(4x^25)
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