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This is what I got: \[x^2y^2(15y-21y^3+30x^2)\] Is that the correct answer?
You can also factor a 3.
DOH! You are right!!!
Don't forget to factor the coefficients!
I am so used to them not being able to be factored, I didn't even stop to think if they could :P
I hear ya - I do the same thing when I use the quadratic formula to factor something like x^2+6x+5. Duh.
My next one seems alot more complicated... \[8x^3+12x^2-10x-15\]
I 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?
Yech...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.
I think it is (2x)^3 though, and the rest I just factor as normal and include in the equation. I hope so, anyway.
You can "factor by grouping". Group the 8x^3 together with the -10x. You can factor a 2X out of each getting 2x(4x^2-5). You can then factor a 3 out of the remaining two terms getting 3(4x^2-5). Resulting in (2x+3)(4x^2-5)