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Adrian.

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Hi this question is really bugging me.

Write each expression as a product of three factors
L) ax<sup>2</sup> - a - 2x<sup>2</sup> + 2

= a(x<sup>2</sup> - 1) - 2(x<sup>2</sup> - 1)

= (a - 2)(x<sup>2</sup> - 1)

= (x - 1)(x + 1)(a - 2)

This answer is correct but when I expand it out I get.

x<sup>2</sup> + x + ax -2x - x - 1 -a + 2 +ax -2x + a - 2

= x<sup>2</sup> + 2ax - 4x - 1

If I expand out the difference of two squares bit first and then expand again, I get the right answer.

(x<sup>2</sup> - 1)(a - 2)

= ax<sup>2</sup> - a -2x<sup>2</sup> + 2

Can anyone explain why I can't just expand it all out in one go? I'm worried that in an exam i might miss the difference of two squares and then bugger up the question.

Thank you.
 

FinalFantasy

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cuz u expanded wrongly, try expand da first 2 brackets first and den times dat to da last bracket
 

Adrian.

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FinalFantasy said:
cuz u expanded wrongly, try expand da first 2 brackets first and den times dat to da last bracket
Please read the whole post. That's what I did in the end but I don't see why it shouldn't work the first way (I can see it doesn't and can see why) but I always lived in the comfort that if you forgot a rule (like diff of two squares) you could just expand the whole thing out and everything was dandy (except you waste time).
 

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(x - 1)(x + 1)(a - 2)


(x^2 + x - x - 1)(a - 2)
= back to what u have

(x - 1)(x + 1)(a - 2)
= (x - 1)(xa - 2x + a - 2)
= x^2(a) - 2x^2 + ax - 2x - xa +2x - a + 2
= x^2(a) -2x^2 -a + 2
 

FinalFantasy

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(1+2)(2+3)(3+4) is NOT=1*2+1*3+1*4+2*2+2*3+2*4+2*3+2*4+3*3+3*4
and u expanded that way
 

Adrian.

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OK Thank you ... and FF. So you always have to expand in groups of two?

Eg.

(a + b)(c + d)(e +f)

= (ac + ad + bc + bd)(e + f)

= (ace + ade + bce + bde + acf + adf + bcf + bdf)

Thank you.
 

FinalFantasy

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not really "always" =P
many things can happen in maths, you just do things accordingly, do what's logical lol
 

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Adrian. said:
OK Thank you ... and FF. So you always have to expand in groups of two?

Eg.

(a + b)(c + d)(e +f)

= (ac + ad + bc + bd)(e + f)

= (ace + ade + bce + bde + acf + adf + bcf + bdf)

Thank you.
its only an extra step anyway
it will only be that extra 30sec where you are "assure" of that 3 or 4 marks instead of getting 0 for stupid mistakes
 

Adrian.

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... said:
its only an extra step anyway
it will only be that extra 30sec where you are "assure" of that 3 or 4 marks instead of getting 0 for stupid mistakes
Yeah true. Thanks for the help guys.
 

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