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Factorising

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Original post by Illidan2
Simplify them? (2x-3) (2x-3) (x-1) (x-1)


Almost - check your signs.
Reply 21
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Illidan2
OP
14
(2x+3)(2x−3)(x−1)(x+1) (2x+3) (2x-3) (x-1) (x+1)

Awesome! Thank you so much! :smile: About the a4 a^4 b b thing though, it was a question in my textbook. I don't know if that means it'll come up in the exam, but it's in there. 2x4+14x+24 2x^4+14x+24 that was my fault though, I didn't mean b b , I meant bx bx

Edit: Ignore this. I missed the x2 x^2 in the question.
(edited 6 years ago)
Original post by Illidan2
(2x+3)(2x−3)(x−1)(x+1) (2x+3) (2x-3) (x-1) (x+1)

Awesome! Thank you so much! :smile: About the a4 a^4 b b thing though, it was a question in my textbook. I don't know if that means it'll come up in the exam, but it's in there. 2x4+14x+24 2x^4+14x+24 that was my fault though, I didn't mean b b , I meant bx bx

Edit: Ignore this. I missed the x2 x^2 in the question.


You've got it :smile:

You might already know this, but any expression of the form a2x2−b2a^2x^2 - b^2 can be factorised as (ax+b)(ax−b)(ax + b)(ax - b); that's called "difference of two squares" and it's useful to look out for square numbers in questions involving factorising so that you can see when you need to do that.
Reply 23
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Illidan2
OP
14
I didn't know about the difference of two squares thing, so it's very helpful that I now have knowledge of the technique. :smile: Apart from substitution and difference of two squares, are there other techniques I will need in order to factorise any given problem? (To A-Level standard, at least.)
Reply 24
Avatar for Notnek
Notnek
21
Original post by Illidan2
I didn't know about the difference of two squares thing, so it's very helpful that I now have knowledge of the technique. :smile: Apart from substitution and difference of two squares, are there other techniques I will need in order to factorise any given problem? (To A-Level standard, at least.)

First get your quadratic into the form ax2+bx+cax^2+bx+c where bb or cc could be 0.

Next always look for a common factor that is a number e.g.

6x2+24x+186x^2+24x+18 becomes much easier to factorise once you take out a factor of 6.

Then if you have something of the form ax2+bx+cax^2+bx+c where b and c are non-zero, you can factorise into two brackets using the method you've learnt.

If you have something of the form ax2+bxax^2+bx then take out a factor of xx (or something like kxkx).

If you have ax2−cax^2-c then consider difference of two squares.
(edited 6 years ago)

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