- #1
- 2
- 0
the original equation is 4x^2-6x^2-12x+9=0
I got to this point: (2x^2)(2x-3)+(-3)(4x-3)=0
I got to this point: (2x^2)(2x-3)+(-3)(4x-3)=0
Ah, this looks like the "factoring by grouping" method, which, unfortunately, doesn't work for ALL cubic polynomials. The rational root test, as micromass mentioned, is your best bet.the original equation is 4x^2-6x^2-12x+9=0
I got to this point: (2x^2)(2x-3)+(-3)(4x-3)=0
the original equation is 4x^2-6x^2-12x+9=0
I got to this point: (2x^2)(2x-3)+(-3)(4x-3)=0
You mean 4x^{3}-6x^{2}-12x+9=0?
The roots of a polynomial must be a factor of the product of the first term and the last term. Therefore, we have to try all the factors for 4*9=36. There's this really convenient method for factoring polynomials like this. It's kind of like long division, but not really.
(Snip)