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Challenging math questions involving polynomial functions

  1. Sep 18, 2005 #1
    Any help with the following is appreciated. Thank you!

    2. If 2x^3 - 9x^2 + 13x + k is divisible by x - 2, then it is also divisible by x -1.

    I don't understand why this statement is true

    8. An unknown polynomial f(x) of degree 37 yields a remainder of 1 when divided by x - 1, a remainder of 3 when divided by x - 3, a remainder of 21 when divided by x - 5.
    Find the remainder when f(x) is divided by (x-1)(x-3)(x-5)

    9. If ax^3 + bx + c, with a not equal 0, c not equal 0, has a factor of the form x^2 + px + 1, show that a^2 - c^2 = ab

    10. Given that the cubic equation x^3 - 3x^2 + ax + b = 0 has rational coefficients and has the root - 1 + sqrt(3)i, determine the values of a and b.
     
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  3. Sep 18, 2005 #2

    arildno

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    2. Hint:
    If P(x) is divisible by x-2, what must then P(2) be?
     
  4. Sep 18, 2005 #3

    Hurkyl

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    What have you tried on any of these?
     
  5. Sep 18, 2005 #4
    8.

    f (1) = ...a + b + c = 1
    f (3) = ...9a + 3b + c = 3
    f (5) = ...25a + 5b + c = 21

    I know degree of f(x) = 37, which means that there's more than just a, b and c. However, I have no idea how to do this kind of questions. Sorry i forgot to show the work that i did.
     
  6. Sep 18, 2005 #5
    9.

    ax^3 + bx + c = f(x)
    x^2 + px + 1 --> factor is of degree 2 so in order to get ax^3, I would need ax.
    hm... so the factored form would be ...
    (x^2 + px + 1)(x - h) = 0

    this is all i could come up with on my own
     
  7. Sep 18, 2005 #6
    10.

    x1 = - 1 + sqrt(3)i
    x2 = - 1 - sqrt(3)i
    x3 = x3

    so the factored form would be ...

    (x - x3)(x - (-1 + sqrt(3)i)(x - (-1 - sqrt(3)i) = 0
    (x - x3)(x^2 + 2x + 4) = x^3 - 3x^2 + ax + b
     
  8. Sep 18, 2005 #7

    Hurkyl

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    8. I think the division theorem would be much more fruitful. (Modular arithmetic better, if you've done arithmetic modulo a polynomial before)

    You want to write f(x) = q(x) (x-1)(x-3)(x-5) + r(x), and you don't care about the actual value of q(x), just that degree(r) < 3.


    9. the factored form would be ... (x^2 + px + 1)(x - h) = 0

    Why do you have "= 0"? You're not solving an equation, you're factoring a polynomial! You want f(x) = something, not 0 = something! (I won't say more yet, I'll give you a chance to figure it out)


    10.

    (x - x3)(x - (-1 + sqrt(3)i)(x - (-1 - sqrt(3)i) = 0

    No. Weren't you saying you were trying to factor f(x)? If so, then why are you introducing an "=0"?

    (x - x3)(x^2 + 2x + 4) = x^3 - 3x^2 + ax + b

    This one is correct, though, but why'd you stop there?
     
    Last edited: Sep 18, 2005
  9. Oct 2, 2005 #8
    Hi I solved 10. Could I get more hints on 8 and 9 though... I am still having troubles with them.
     
  10. Oct 3, 2005 #9

    AKG

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    9. If ax^3 + bx + c, with a not equal 0, c not equal 0, has a factor of the form x^2 + px + 1, show that a^2 - c^2 = ab

    Well if ax³ + bx + c has a factor of the form x² + px + 1, then you know there is some polynomial r(x) such that:

    r(x)(x² + px + 1) = ax³ + bx + c

    Now you should be able to say something about the degree of r, as well as the leading coefficient of r. Do you know what those facts are?

    8. An unknown polynomial f(x) of degree 37 yields a remainder of 1 when divided by x - 1, a remainder of 3 when divided by x - 3, a remainder of 21 when divided by x - 5.
    Find the remainder when f(x) is divided by (x-1)(x-3)(x-5)


    Well you know that you can write f(x) = q(x)[(x-1)(x-3)(x-5)] + r(x). This is simply a consequence of the division algorithm, and the division algorithm tells you something about the degree of r. Knowing the degree of r, you can write r as a polynomial with unknown coefficients. You know that dividing f(x) by (x-1) gives a remainder of 1, so you know that:

    f(x)/(x-1) = q(x)(x-3)(x-5) + r(x)/(x-1) has a remainder of 1, which simply means that:

    r(x)/(x-1) has a remainder of 1. Using division algorithm again, you can say:

    r(x) = q(x)(x - 1) + 1, since you know 1 to be the remainder. Since you should already know the degree of r, you know the degree of q. In fact, you also know how to express the leading coefficient of q in terms of the leading coefficient of r. You can make two more similar equations:

    r(x) = p(x)(x - 3) + 3
    r(x) = t(x)(x - 5) + 21

    There are a certain number of unknown coefficients in r, and a number of unknown coefficients in each of the p, q, and t. But given these equations, as well as the fact that:

    ax² + bx + c = a'x² + b'x + c' for all x if and only if a = a', b = b' and c = c'

    you should be able to come up with n equations with n unknowns, and solving for your unknowns will include solving for those unknown coefficients in r, and r is the remainder you're looking for.
     
  11. Oct 3, 2005 #10
    sorry i just realized i made a big mistake -_-! This isn't college math, this is just grade 12 calculus math. I don't know what the division theorem is! I am so sorry!
     
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