Polynomial with two unknowns

  1. [SOLVED] polynomial with two unknowns

    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.
     
  2. jcsd
  3. 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.
     
  4. but there are two unknowns in each expression
     
  5. Hootenanny

    Hootenanny 9,681
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    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]
     
  6. 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.
     
  7. Hootenanny

    Hootenanny 9,681
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    Have you never solved simultaneous equations before?
     
  8. no, i haven't.
     
  9. Hootenanny

    Hootenanny 9,681
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    Okay, in that case if you post the two equations you obtain I shall walk you through the process.
     
  10. 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
     
  11. Hootenanny

    Hootenanny 9,681
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    Good, so now multiply (2) by 2 and then add the two equations.
     
  12. Did you ever learn to solve matrices in Algebra class?
     
  13. 8p + 132

    ok so p = -11, q = -70

    thnx
     
    Last edited: Jan 26, 2008
  14. nope
     
  15. 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.
     
  16. Hootenanny

    Hootenanny 9,681
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    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).
    Firstly, what you have written is an expression, not an equation. Secondly, you might want to check your coefficent of p.
     
  17. 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
     
    Last edited: Jan 26, 2008
  18. Want to learn the fast way?
     
  19. 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.
     
    Last edited: Jan 26, 2008
  20. This will only apply to linear equations.

    [tex]a_1 x+b_1 y=c_1[/tex]
    [tex]a_2 x+b_2 y=c_2[/tex]

    [tex]x=\frac{\left|\begin{array}{cc}c_1 & b_1 \\ c_2 & b_2\end{array}\right|}{\left|\begin{array}{cc}a_1 & b_1 \\ a_2 & b_2\end{array}\right|}=\frac{c_1 b_2 - b_1 c_2}{a_1 b_2 - b_1 a_2}[/tex]

    [tex]y=\frac{\left|\begin{array}{cc}a_1 & c_1 \\ a_2 & c_2\end{array}\right|}{\left|\begin{array}{cc}a_1 & b_1 \\ a_2 & b_2\end{array}\right|}=\frac{a_1 c_2 - c_1 a_2}{a_1 b_2 - b_1 a_2}[/tex]

    Notice that the denominator for both x & y are the same (determinant). While the numerators vary.

    You can also solve for 3 unknowns but that takes too long to type.
     
    Last edited: Jan 26, 2008
  21. I'm confused to as what your problem is. Is it ...

    [tex]y=\log_3{(x+4)}[/tex]

    ?
     
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