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# Basic Integration And Factorising??

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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.

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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.

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$\int^{2}_{0}\frac{(x^3-8)}{(x-2)}$

like that ;)

$x^3-8 = x^3 - 2^3$

$(x-2)(x^2+2x+4)$

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 ;)

thanks

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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.

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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.

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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.

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