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Factoring Quartics (FP3)

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?

Thanks for your help.

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Reply 1
Avatar for Zacken
Zacken
22
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?

Thanks for your help.


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

e2x+e−2x−14ex+14e−x=10  ⟺  e4x−14e3x+10e2x+14ex+1=0e^{2x} + e^{-2x} - 14e^x + 14e^{-x} = 10 \iff e^{4x} - 14e^{3x} + 10e^{2x} + 14e^x + 1 = 0 from which an obvious factor should make itself present and then factorise out.
(edited 7 years ago)
Reply 2
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P____P
OP
2
@Zacken
1464209190446.jpg
How did you get 14e^x?
Posted from TSR Mobile
(edited 7 years ago)
Original post by Zacken
Well, that's obviously wrong.

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

e2x+e−2x−14ex+14e−x=10  ⟺  e4x−14e3x+10e2x+14ex+1=0e^{2x} + e^{-2x} - 14e^x + 14e^{-x} = 10 \iff e^{4x} - 14e^{3x} + 10e^{2x} + 14e^x + 1 = 0 from which an obvious factor should make itself present and then factorise out.

You'll want to double check those 14s.
Reply 4
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Zacken
22
Original post by Farhan.Hanif93
You'll want to double check those 14s.


Was just about to delete my post, darn it. :tongue:
Reply 5
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P____P
OP
2
Original post by Zacken
Was just about to delete my post, darn it. :tongue:


lol
Reply 6
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Zacken
22
Original post by P____P
@Zacken
1464209190446.jpg
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 u=4±17u = 4 \pm \sqrt{17} or the other two factors are factors of it where u=exu = e^x unless you had a GDC.
Reply 7
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P____P
OP
2
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 u=4±17u = 4 \pm \sqrt{17} or the other two factors are factors of it where u=exu = e^x unless you had a GDC.


hmm...
@TeeEm
@ghostwalker
Anyone else??:redface:
Reply 8
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math42
20
This is why one should just use the identities method..
Reply 9
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TeeEm
19
Original post by P____P
hmm...
@TeeEm
@ghostwalker
Anyone else??:redface:


post the question
Reply 10
Avatar for TeeEm
TeeEm
19
Original post by P____P
hmm...
@TeeEm
@ghostwalker
Anyone else??:redface:


you use the identity for cosh2x to make sinhes
Reply 11
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P____P
OP
2
Original post by 1 8 13 20 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 :frown:
Original post by TeeEm
post the question

http://puu.sh/p50Fs/0432fcd543.png
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:

(ex−e−x)2=e2x+e−2x−2\left(e^x-e^{-x}\right)^2 = e^{2x}+e^{-2x}-2

And let w=ex−e−xw=e^x - e^{-x}, so that the equation reduces to the quadratic:

w2−7w−8=0w^2 -7w -8=0

Which you can solve for ww and then solve further quadratics from the definition of ww for exe^x.

However, it's worth pointing out that this is 'secretly' rebuilding the hyperbolic identity for cosh⁡2x\cosh 2x in terms of sinh⁡x\sinh x so it pays to know those instead.
(edited 7 years ago)
Reply 13
Avatar for math42
math42
20
Original post by P____P
Agreed, but the 1st attempt I had at this I flopped and got myself into this mess :frown:

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..
Original post by Zacken
Convert everything to exponentials (and multiply by 2 to save me from fractions):

e2x+e−2x−14ex+14e−x=10  ⟺  e4x−14e3x+10e2x+14ex+1=0e^{2x} + e^{-2x} - 14e^x + 14e^{-x} = 10 \iff e^{4x} - 14e^{3x} + 10e^{2x} + 14e^x + 1 = 0 from which an obvious factor should make itself present and then factorise out.


you want to double check those 14
Reply 15
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TeeEm
19
Use identities which is very easy

or using exponentials you get what is known as a symmetric polynomial
Reply 16
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Zacken
22
Original post by The gains kinggg
you want to double check those 14


You'll want to read the thread.
Reply 17
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B_9710
18
Either use identities or quartic formula. I know which one i'd do.
Original post by Zacken
You'll want to read the thread.


You'll want to check the math before you try to 'help' people
Reply 19
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Zacken
22
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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