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# Factoring Quartics (FP3)

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1. I was doing this question
There are 2 ways of doing it (according to MS). Either you manipulate it using hyperbolic identities, or you use the exponential definition.

If you use the latter method, you end up with a quartic that looks like this.
Now, how do you factorise/ solve this?

2. (Original post by P____P)
I was doing this question
There are 2 ways of doing it (according to MS). Either you manipulate it using hyperbolic identities, or you use the exponential definition.

If you use the latter method, you end up with a quartic that looks like this.
Now, how do you factorise/ solve this?

Convert everything to exponentials (and multiply by 2 to save me from fractions):

from which an obvious factor should make itself present and then factorise out.
3. Zacken

How did you get 14e^x?
Posted from TSR Mobile
4. (Original post by Zacken)
Well, that's obviously wrong.

Convert everything to exponentials (and multiply by 2 to save me from fractions):

from which an obvious factor should make itself present and then factorise out.
You'll want to double check those 14s.
5. (Original post by Farhan.Hanif93)
You'll want to double check those 14s.
Was just about to delete my post, darn it.
6. (Original post by Zacken)
Was just about to delete my post, darn it.
lol
7. (Original post by P____P)
Zacken

How did you get 14e^x?
Posted from TSR Mobile
Yep, sorry. Blatant fail on my part. I can't see how you'd be able to solve that quartic without spotting that or the other two factors are factors of it where unless you had a GDC.
8. (Original post by Zacken)
Yep, sorry. Blatant fail on my part. I can't see how you'd be able to solve that quartic without spotting that or the other two factors are factors of it where unless you had a GDC.
hmm...
TeeEm
ghostwalker
Anyone else??
9. This is why one should just use the identities method..
10. (Original post by P____P)
hmm...
TeeEm
ghostwalker
Anyone else??
post the question
11. (Original post by P____P)
hmm...
TeeEm
ghostwalker
Anyone else??
you use the identity for cosh2x to make sinhes
12. (Original post by 13 1 20 8 42)
This is why one should just use the identities method..
(Original post by TeeEm)
you use the identity for cosh2x to make sinhes
Agreed, but the 1st attempt I had at this I flopped and got myself into this mess
(Original post by TeeEm)
post the question
http://puu.sh/p50Fs/0432fcd543.png
13. (Original post by P____P)
Either you manipulate it using hyperbolic identities, or you use the exponential definition.
By the latter option, I presume they didn't expect you to solve the quartic via direct factorisation. Instead, I suspect you're supposed to note:

And let , so that the equation reduces to the quadratic:

Which you can solve for and then solve further quadratics from the definition of for .

However, it's worth pointing out that this is 'secretly' rebuilding the hyperbolic identity for in terms of so it pays to know those instead.
14. (Original post by P____P)
Agreed, but the 1st attempt I had at this I flopped and got myself into this mess

http://puu.sh/p50Fs/0432fcd543.png
Usually if some polynomial of degree more than 2 comes up and its factorisation isn't immediately obvious you probably made a wrong turn..
15. (Original post by Zacken)
Convert everything to exponentials (and multiply by 2 to save me from fractions):

from which an obvious factor should make itself present and then factorise out.
you want to double check those 14
16. Use identities which is very easy

or using exponentials you get what is known as a symmetric polynomial
17. (Original post by The gains kinggg)
you want to double check those 14
18. Either use identities or quartic formula. I know which one i'd do.
19. (Original post by Zacken)
You'll want to check the math before you try to 'help' people
20. (Original post by The gains kinggg)
You'll want to check the math before you try to 'help' people
You'll want to spell maths properly.

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