factorise please

mufc802 · 13 Apr 2011 at 00:34

2x^4+14x^2+24.

I'm not sure what to do when the power is greater than 2. Please explain.

trigger-happy · 13 Apr 2011 at 00:35

c1 mathematics?

Tikkabhuna · 13 Apr 2011 at 00:37

(2x^2 + #) ( x^2 + #)

Like that but obviously you need to work out the rest. Gotta leave something for you to do.

hellarda · 13 Apr 2011 at 00:38

take the 2 out first i would assume so you get 2(x^4 + 7x^2 + 12) will get back to you if i can work out the rest in a sec

mufc802 · 13 Apr 2011 at 00:39

trigger-happy said:
c1 mathematics?

Yeah I'm trying to teach myself.

killari · 13 Apr 2011 at 00:39

You could try synthetic division.

hellarda · 13 Apr 2011 at 00:40

then if you factorise that you get 2 (x^2 -3) (x^2 -4) i think?

Nitefly · 13 Apr 2011 at 00:43

(2X^2 + 6)(X^2 + 4)

Well that's what I got in my head

EDIT: taking the 2 out the front is going to be...

2(X^2 + 3)(x^2 + 4)

Tikkabhuna · 13 Apr 2011 at 00:47

Ah I thought it was homework. I have to say I do miss a level maths.

Of course multiplying x^2 by x^3 is X^5, for example, so you can work it back for your problem. As we only have x^2's and X^4's we can safely split it into ( 2x^2 + A) ( x^2 + B). Then you have to work out out what A * B = 24 and (A * x^2) + (B * 2x^2) = 14x^2, as normal. So in this case A = 6 and B = 4.

Another edit: A simple way to do (A * x^2) + (B * 2x^2) = 14x^2 is to take out the x^2 so you just have A + 2B = 14 and then check that it works to add to 24.

TheCenturion · 13 Apr 2011 at 00:48

mufc802 said:
I'm not sure what to do when the power is greater than 2. Please explain.

It just adds more variables mate. EG: when you have x^2, both factors must be x. But when you have x^4, they could be X^2, or x^3 and x^1

Creelers · 13 Apr 2011 at 00:52

(2x^2 + 12)(x^2 + 2)

hellarda · 13 Apr 2011 at 00:53

Nitefly said:
2(X^2 + 3)(x^2 + 4)

i was close... considering im only doing gcse maths atm....

Nitefly · 13 Apr 2011 at 00:53

Creelers said:
(2x^2 + 12)(x^2 + 2)

That would give you 16X^2

Creelers · 13 Apr 2011 at 00:57

Nitefly said:
That would give you 16X^2

Didn't notice that, I'm a bit tired

(2x^2 + 6)(x^2 + 4)

That should work.

mufc802 · 13 Apr 2011 at 01:08

OK I worked it out using my method.

So 2x^4 + 14x^2 + 24

A = 2 | B = 14 | C = 24

AC = 48
B = 14

8*6 = 48 & 8+6 = 14

(2x^4 + 8x^2) + (6x^2 + 24)
2x^2(x^2 + 4) + 6(x^2 + 4)
(2x^2 + 6) (x^2 + 4) or 2(x^2 + 3) (x^2 + 4)

So is this the correct method or is there a better way?

Tikkabhuna · 13 Apr 2011 at 01:17

That is the correct one. Eventually you'll be able to look at equations and factorise it in your head.

GreatAuk · 13 Apr 2011 at 01:35

mufc802 said:
2x^4+14x^2+24.

I'm not sure what to do when the power is greater than 2. Please explain.

Just factorise but with your x's being squared - exactly the same as factorising 2x^2 + 14x + 24 , but just use x^2 instead of x. It doesn't make any difference. And if you really get stuck on working out what numbers to use, you could use the quadratic formula to find solutions. But I don't think they'd want that on this kind of question.

Nitefly · 13 Apr 2011 at 01:40

GreatAuk said:
Just factorise but with your x's being squared - exactly the same as factorising 2x^2 + 14x + 24 , but just use x^2 instead of x. It doesn't make any difference. And if you really get stuck on working out what numbers to use, you could use the quadratic formula to find solutions. But I don't think they'd want that on this kind of question.

Well, not quite.

The rule is that X^Y multiplied by X^Z is X^(Y+Z)

This means X^7 could be X^3 multiplied by X^4, or X^2 multiplied by X^5.

So it can make a difference depending on what he asked to factorise. In the question in the OP, you know it can't be x and x^3 instead of x^2 and X^2 because there are no 'x^3' in the question. However, this might not always be the case.