LJP07 Posted November 30, 2006 Report Share Posted November 30, 2006 Factorise: x cubed - 8 Hence using the above, evaluate: Integration sign ( like an s ) then 2 on the top and 0 on the bottom, this is called the Definite Integral. x cubed - 8 / x - 2 dx. Quote Link to comment Share on other sites More sharing options...

sebbysteiny Posted December 1, 2006 Report Share Posted December 1, 2006 You first need to find one factor. Then, the rest should follow without much difficulty. So how do you find the first factor? By using the theorum, (I think it is completeness theorum. Nope, not called that [just looked it up]). Basically, find an answer to solve the equation in which the thing you want to fractorise = 0. Eg, factorise x^4 -2x + 1. So find an answer to x^4 -2x +1 = 0. Try x = 2. LHS = 15, so (x-2) is not a factor. Try x = 1. LHS = 0. So (x-1) is a factor. This leaves (x-1)(x^3 + x^2 +x -1). And you now need to do the same for x^3 + x^2 + x -1. But I don't think that expression has any real roots. Your question will. Hope this is enough to do the factorisation. UPDATE: Just remembered it. It's called factor theorum. Quote Link to comment Share on other sites More sharing options...

Jay-qu Posted December 1, 2006 Report Share Posted December 1, 2006 [math]\int^{2}_{0}\frac{(x^3-8)}{(x-2)}[/math] like that ;) [math]x^3-8 = x^3 - 2^3 [/math][math](x-2)(x^2+2x+4)[/math] if you sub that into the integral you will find the bottom term cancels and you should be able to carry out the intergration no probs ;) Quote Link to comment Share on other sites More sharing options...

LJP07 Posted December 3, 2006 Author Report Share Posted December 3, 2006 thanks Quote Link to comment Share on other sites More sharing options...

sebbysteiny Posted December 3, 2006 Report Share Posted December 3, 2006 if you sub that into the integral you will find the bottom term cancels and you should be able to carry out the intergration no probs Unless x = 2. Quote Link to comment Share on other sites More sharing options...

Jay-qu Posted December 4, 2006 Report Share Posted December 4, 2006 I think you need to read that again sebby.. when x = 2 there is no problem, that is the reason for my method, it eliminates the divide by zero. Quote Link to comment Share on other sites More sharing options...

sebbysteiny Posted December 4, 2006 Report Share Posted December 4, 2006 if you sub that into the integral you will find the bottom term cancels Unless x = 2. think you need to read that again sebby.. when x = 2 there is no problem, that is the reason for my method, it eliminates the divide by zero. I might be wrong, but didn't you 'eliminate the divide by zero' by cancelling the (x-2) on the top with the (x-2) on the bottom to make 1? (x-2)/(x-2) = 1. However, that step not valid at x = 2. Quote Link to comment Share on other sites More sharing options...

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