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Newbie
Bandwidth Hog
Joined: 23 Jan 2004
Posts: 2247
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 Posted: 22 Jan 2009 06:51:24 pm Post subject: |
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Need help on forming a polynomial whose zeros are -4, -1, 2, 3; degree 4. I know how to solve a degree 3 by foiling and then distributing the third term over the foiled one, but I am confused on how to do 4. I've set the problem up so far:
I don't want an answer, just need help on what the next step is.
(X+4) (X+1) (X-2) (X-3) |
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simplethinker
snjwffl
Active Member
Joined: 25 Jul 2006
Posts: 700
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| Posted: 22 Jan 2009 07:21:58 pm Post subject: |
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You just have to remember the distributive law :)
We have p(x) = (X+4)(X+1)(X-2)(X-3). Now, let A=X+4, B=X+1, C=X-2, D=X-3.
This gives p(x) = ABCD = (AB)CD = ((X+4)B)CD = (BX+4B)CD = ( X(X+1) + 4(X+1) )CD = ((X^2 + X + 4X + 4)C)D = (CX^2 + CX + C4X + C4)D = ...
Just continue that pattern of distributing one term at a time and simplify and you can factor (or un-factor, I always forget which goes which way) any polynomial :biggrin: |
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Newbie
Bandwidth Hog
Joined: 23 Jan 2004
Posts: 2247
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| Posted: 23 Jan 2009 09:32:23 am Post subject: |
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Simplethinker thank you so much for your response. That helped me greatly and I was able to get the correct answer. Kudos to you. Better than my math book. lol
Anyway the correct answer is: x^4-15x^2+10x+24 |
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simplethinker
snjwffl
Active Member
Joined: 25 Jul 2006
Posts: 700
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| Posted: 23 Jan 2009 01:44:20 pm Post subject: |
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Newbie wrote: Simplethinker thank you so much for your response. That helped me greatly and I was able to get the correct answer. Kudos to you. Better than my math book. lol
Anyway the correct answer is: x^4-15x^2+10x+24
No problem  I've only had one teacher that explained it that way, even though it's actually the simplest. |
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DarkerLine
ceci n'est pas une |
Super Elite (Last Title)
Joined: 04 Nov 2003
Posts: 8328
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| Posted: 23 Jan 2009 01:58:36 pm Post subject: |
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Here's a more advanced method that might be faster:
Start with (X-A)(X-B)(X-C)(X-D).
First we want to compute the coefficient of X^4. The only way we can get an X^4 term is if we choose the X term from each factor. The X terms all have a coefficient of 1, so the coefficient of X^4 will be 1*1*1*1=1.
Then the coefficient of X^3. To get an X^4 term, we choose the X term for three of the factors, and the constant term from one of the factors. There are four ways to do it: there constant term could be -A (giving -AX^3), -B (giving -BX^3), -C, or -D -- so together, they give (-A-B-C-D)X^3.
Then the coefficient of X^2. This is the hardest one. We choose the X term from two factors, and the constant term from the other two. There are 6 ways to do it, which give (-A)(-B)X^2, (-A)(-C)X^2, (-A)(-D)X^2, (-B)(-C)X^2, (-B)(-D)X^2, and (-C)(-D)X^2. So together we get (AB+AC+AD+BC+BD+CD)X^2.
Then the coefficient of X. Here, we only choose the X term from one of the factors, so we get all but one of the constant terms each time. This will be (-A)(-B)(-C)X, (-A)(-B)(-D)X, (-A)(-C)(-D)X, and (-B)(-C)(-D)X, for a total of (-ABC-ABD-ACD-BCD)X.
Then the constant coefficient. Here, we always choose the constant term, so we'll get (-A)(-B)(-C)(-D) or ABCD.
Of course, this method can be used for any number of factors.
Last edited by Guest on 23 Jan 2009 01:58:50 pm; edited 1 time in total |
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thornahawk
μολών λαβέ
Active Member
Joined: 27 Mar 2005
Posts: 569
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| Posted: 26 Jan 2009 12:37:03 pm Post subject: |
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Short version of DarkerLine's post: you can use Vieta's formulae to form the coefficients of the corresponding monic (that is, the polynomial with 1 as the coefficient of the highest-power term) polynomial given the roots.
That makes for a quick check of the coefficient of the next highest power and the constant term: the product of all the roots is the constant term, and the negative of the sum of the roots is the coefficient of the next highest power. For other coefficients, the formulae become more complicated as the polynomial degree goes up.
thornahawk |
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mr. sir
Newbie
Joined: 21 Feb 2008
Posts: 41
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| Posted: 04 Feb 2009 04:36:26 pm Post subject: |
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Incidentally, I was tired of doing these by hand, and so I decided to write a calc. program for it.
If you wish, try to figure out why it works.
Code:
SetUpEditor LCOEF, LROOT
ClrList LCOEF, LROOT
Prompt N
{1->LCOEF
For(D,1,N
Input "ROOT:",X
X->LROOT(D
augment(LCOEF,{0})-LROOT(D)augment({0},LCOEF)->LCOEF
End
For(D,1,N
Pause LCOEF(D
End
Last edited by Guest on 04 Feb 2009 04:59:00 pm; edited 1 time in total |
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