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 Post subject: Cubic factorisation
PostPosted: Tue, 20 Dec 2011 01:17:34 UTC 
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Hello all,

Can someone please tell m,e what I am doing wrong here?? I cannot seem to get it right...

The question is: (x+1)^3-8

a^3-b^3=(a-b)(a^2+xy+b^2)

(x+1)^3-8=((x+1)-2)((x+1)^2-2(x-1)+4)

= (x-1)(x^2+2x+1-2X-1+4)

= (x-1)(x^2+4)...........I dont really know what happens here as the answer apparently should be (x-1)(x^2+4x+7)

Any help would be greatly appreciated

Thanks


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 Post subject: Re: Cubic factorisation
PostPosted: Tue, 20 Dec 2011 01:20:18 UTC 
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Where did you get the xy term in your initial factorization?

In any case, your mistake is in your first step, you have a -2(x-1) where you should have a +2(x+1).

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 Post subject: Re: Cubic factorisation
PostPosted: Tue, 20 Dec 2011 01:45:39 UTC 
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Thank you Shadow.

I'm sorry, I don't think I understand you.

Firstly, the xy term came from the ((x+1)-2). Also, how do you get +2(x-1)??? Doesn't a '+' and a '-' make a '-'?

Thanks


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 Post subject: Re: Cubic factorisation
PostPosted: Tue, 20 Dec 2011 03:43:46 UTC 
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gareth wrote:
The question is: (x+1)^3-8
a^3-b^3=(a-b)(a^2+xy+b^2)

a^3 - b^3 = (a - b)(a^2 + xy + b^2)
The xy you have in there should be ab; that the standard way to
show the difference of cubes; use it as a GUIDE ONLY :shock:

Start this way: let k = x + 1; then your expression becomes k^3 - 2^3;
now factor that, using a^3 - b^3 = (a - b)(a^2 + ab + b^2) as a GUIDE:
k^3 - 2^3 = (k - 2)(k^2 + 2k + 2^2)
Now substitute x-1 back in and finish off...

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 Post subject: Re: Cubic factorisation
PostPosted: Tue, 20 Dec 2011 06:47:28 UTC 
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gareth wrote:
Thank you Shadow.

I'm sorry, I don't think I understand you.

Firstly, the xy term came from the ((x+1)-2). Also, how do you get +2(x-1)??? Doesn't a '+' and a '-' make a '-'?

Thanks


No, I get a +2(x+1), you have a=x+1 and b=2, multiply them and you get 2(x+1)=ab and the formula we have (the correct version) a^3-b^3=(a-b)(a^2+ab+b^2) tells us we need to add that, so naturally I did.

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 Post subject: Re: Cubic factorisation
PostPosted: Wed, 21 Dec 2011 20:02:36 UTC 
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Or, if you wanted to be really long-winded about it (and didn't have a set method that you had to use), you could:

(1) Expand your cubic:
$ f(x) = (x+1)^{3} - 8 = x^{3} + 3x^{2} + 3x - 7

(2) Use the remainder theorem to find roots by substituting in divisors of 7 (which, as a prime, means \pm1, \pm7):
$ f(1) = 1 + 3 + 3 - 7 = 0, so x=1 is a root \Rightarrow (x-1) is a factor

(3) Factorise your cubic using your new-found factor
$ f(x) = (x-1)(x^{2} + ax + 7)

(4) Re-expand and then compare to get the coefficient of x in the quadratic term (i.e. to find the a)
x^{2} : a - 1 = 3 \Rightarrow a = 4

(5) Check the discriminant of your quadratic to see if you could factorise it further
b^{2} - 4ac = (4)^{2} - 4(1)(7) = 12 < 0, so it can't be factorised over the reals any further.

Hence: (x+1)^{3} - 8 = (x-1)(x^{2} + 4x + 7)

It's not quick and it's not pretty, but it definitely works :D

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 Post subject: Re: Cubic factorisation
PostPosted: Fri, 13 Jan 2012 12:53:15 UTC 
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Hello everybody,

just getting back to my math again...

A dumb question: Going by the cubic factorisation formula, why does b=2???? Isn't the \sqrt[3]{-8}=-2??. Obviously b=2 gives the correct answer, but I thought that you subbed b=-2 directly onto the formula. Why is this not so????

Any help is appreciated.

Regards


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 Post subject: Re: Cubic factorisation
PostPosted: Fri, 13 Jan 2012 20:44:27 UTC 
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gareth wrote:
Hello everybody,

just getting back to my math again...

A dumb question: Going by the cubic factorisation formula, why does b=2???? Isn't the \sqrt[3]{-8}=-2??. Obviously b=2 gives the correct answer, but I thought that you subbed b=-2 directly onto the formula. Why is this not so????

Any help is appreciated.

Regards


DIFFERENCE of cubes, not SUM.

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