[SOLVED] polynomial with two unknowns


by AFG34
Tags: polynomial, solved, unknowns
AFG34
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#1
Jan26-08, 05:44 PM
P: 128
1. The problem statement, all variables and given/known data
The graph of f(x)= 3x^4 + 14x^3 + px^2 + qx + 24 has x-intercepts -4 and 2. Determine the function.

3. The attempt at a solution

I could solve it if there were only one unknown but i don't know how to do it if there are two unknowns.

What i did so far is plug -4 for x and 0 for the output, got an expression = 0
did the same thing for 2
since both equal 0, i set them equal to each other, simplified and got: 2p-q=13.3
Don't know what to do next.
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rocomath
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#2
Jan26-08, 06:09 PM
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Rather then setting them equal to each other. Solve for p with one of your x-intercepts then plug it in your other set with the other x-intercept.
AFG34
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#3
Jan26-08, 07:17 PM
P: 128
but there are two unknowns in each expression

Hootenanny
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#4
Jan26-08, 07:21 PM
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[SOLVED] polynomial with two unknowns


Quote Quote by AFG34 View Post
but there are two unknowns in each expression
What roco is getting at is that you can create a system of two simultaneous equations thus;

[tex]f(-4) = 0[/tex]

[tex]f(2) = 0[/tex]
AFG34
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#5
Jan26-08, 07:28 PM
P: 128
yes i know that, that is what i initially did. So you get 2 equations, both equal to 0, both have q and p in them. But i don't know what to do next.
Hootenanny
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#6
Jan26-08, 07:29 PM
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Quote Quote by AFG34 View Post
yes i know that, that is what i initially did. So you get 2 equations, both equal to 0, both have q and p in them. But i don't know what to do next.
Have you never solved simultaneous equations before?
AFG34
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#7
Jan26-08, 07:32 PM
P: 128
no, i haven't.
Hootenanny
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#8
Jan26-08, 07:33 PM
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Quote Quote by AFG34 View Post
i don't think so.
Okay, in that case if you post the two equations you obtain I shall walk you through the process.
AFG34
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#9
Jan26-08, 07:36 PM
P: 128
1) 0 = 16p - 4q - 104
2) 0 = 4p +2q + 184

then i divided both by 2:
1) 0 = 8p - 2q - 52
2) 0 = 2p + q + 92
Hootenanny
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#10
Jan26-08, 07:43 PM
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Good, so now multiply (2) by 2 and then add the two equations.
rocomath
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#11
Jan26-08, 07:47 PM
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Did you ever learn to solve matrices in Algebra class?
AFG34
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#12
Jan26-08, 07:47 PM
P: 128
8p + 132

ok so p = -11, q = -70

thnx
AFG34
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#13
Jan26-08, 07:49 PM
P: 128
Quote Quote by rocophysics View Post
Did you ever learn to solve matrices in Algebra class?
nope
rocomath
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#14
Jan26-08, 07:52 PM
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Quote Quote by AFG34 View Post
nope
Well either way, your answer isn't right b/c your answer should have ended up being in terms of a solution.

I'll go step by step. Let me type this up.
Hootenanny
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#15
Jan26-08, 07:55 PM
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Quote Quote by rocophysics View Post
Did you ever learn to solve matrices in Algebra class?
If the OP hasn't met simultaneous equations it's pretty safe to say that they haven't been introduced to linear algebra (which IMHO is over-kill for a question such as this).
Quote Quote by AFG34 View Post
8p + 132

ok so p = -11
Firstly, what you have written is an expression, not an equation. Secondly, you might want to check your coefficent of p.
AFG34
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#16
Jan26-08, 08:36 PM
P: 128
0 = 12p + 132 is an equation
ya i typed it wrong
checked the back of the book, got the right answer (f(x)= 3x^4 + 14x^3 - 11x^2 - 70x + 24). ok i know how to do them now, thnx

hootenanny, funny name
rocomath
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#17
Jan26-08, 08:40 PM
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Quote Quote by AFG34 View Post
0 = 12p + 132 is an equation
ya i typed it wrong
checked the back of the book, got the right answer. ok i know how to do them now, thnx
Want to learn the fast way?
AFG34
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#18
Jan26-08, 09:22 PM
P: 128
sure

got a quick question:
is the graph of y = log3^(x+4) the same as the graph of y = log3^x+4?
*the base is 3 not 10

I think the second one is y=log3^x moved up by 4 units.


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