CPL.Luke Posted January 10, 2007 Posted January 10, 2007 set it equal to zero and solve (if by hand look up and use cardano's method) then plug the roots into the expression (A-a1)(A-a2)(A-a3) where a1,a2,a3 are the roots and should be functions of B ^note in the above your using A as your x.
Cap'n Refsmmat Posted January 10, 2007 Posted January 10, 2007 [math]a^3 + b^3 + a^2 - b^2[/math] Hmm... break it up into a difference of two squares and a sum of two cubes: [math](a + b)(a^2 - ab + b^2) + (a + b)(a - b)[/math] I think it can be simplified further from there, but I won't do the rest of your work for you. edit: I seem to be outclassed.
Asian Posted January 10, 2007 Author Posted January 10, 2007 so then the rest is factor by grouping........ thanks, i just wanted to check my answer because it wasnt in the back of the book and i have a test tomorrow.
CPL.Luke Posted January 10, 2007 Posted January 10, 2007 not outclassed, I can't spot tricks like that for anything so I just figure out ways of guaranteeing an answer no matter what the polynomial is.
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