Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Polynomial with two unknowns

  1. Jan 26, 2008 #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. Jan 26, 2008 #2
    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. Jan 26, 2008 #3
    but there are two unknowns in each expression
     
  5. Jan 26, 2008 #4

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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. Jan 26, 2008 #5
    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. Jan 26, 2008 #6

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Have you never solved simultaneous equations before?
     
  8. Jan 26, 2008 #7
    no, i haven't.
     
  9. Jan 26, 2008 #8

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Okay, in that case if you post the two equations you obtain I shall walk you through the process.
     
  10. Jan 26, 2008 #9
    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. Jan 26, 2008 #10

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Good, so now multiply (2) by 2 and then add the two equations.
     
  12. Jan 26, 2008 #11
    Did you ever learn to solve matrices in Algebra class?
     
  13. Jan 26, 2008 #12
    8p + 132

    ok so p = -11, q = -70

    thnx
     
    Last edited: Jan 26, 2008
  14. Jan 26, 2008 #13
    nope
     
  15. Jan 26, 2008 #14
    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. Jan 26, 2008 #15

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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. Jan 26, 2008 #16
    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. Jan 26, 2008 #17
    Want to learn the fast way?
     
  19. Jan 26, 2008 #18
    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. Jan 26, 2008 #19
    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. Jan 26, 2008 #20
    I'm confused to as what your problem is. Is it ...

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

    ?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?