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Dividing Polynomials ~ Root/Factor Theroem

  1. Oct 18, 2007 #1
    1. The problem statement, all variables and given/known data

    A cubic polynomial gives remainders [tex](5x+4)[/tex] and [tex] (12x-1) [/tex] when divided by [tex] x^{2} - x + 2 [/tex] and [tex] x^{2} + x - 1 [/tex] respectively. Find the polynomial

    2. Relevant equations
    :S Well, im using the root theorem, the factor theorem, and possibly just basics on long division..

    We know that:

    [tex]\frac{P(x)}{D(x)}=Q(x) + \frac {R(x)}{D(x)}[/tex]

    3. The attempt at a solution

    [tex] P(x)=ax^{3} + bx^{2} + cx + d [/tex] which is our standard for a cubic polynomial.

    **Please note I will use Q(x1) and Q(x2) for the tw different quotients of the two divisions.
    Then we know that

    [tex]\frac {P(x)}{x^{2} - x + 2} = Q (x_{1}) + \frac{5x+4}{x^{2}-x+2}[/tex]

    Therefore, our P (x) is the following for the first division.

    [tex]P(x)=Q(x_{1})(x^{2} - x + 2) + 5x + 4[/tex]

    If we do the same for the next division, we obtain the following (using the same procedure)

    [tex]P(x)=Q(x_{2})(x^{2} + x + 1) + 12x - 1[/tex]

    I have no idea, I can possibly make them equal each other, and sort of solve:
    [tex]Q(x_{2})(x^{2} + x + 1) + 12x - 1=Q(x_{1})(x^{2} - x + 2) + 5x + 4[/tex]

    [tex]Q(x_{2})(x^{2} + x + 1) =Q(x_{1})(x^{2} - x + 2) - 7x + 5[/tex]

    [tex]Q(x_{2})(x^{2} + x + 1) =Q(x_{1})(x^{2} - x + 2) - 7x + 5[/tex]

    ...any advice to lead me into the right path, I feel im going the wrong way.
     
    Last edited: Oct 18, 2007
  2. jcsd
  3. Oct 19, 2007 #2
    Q(x) has to be linear. So put Q(x) = px+q, say.

    Let the cubic be ax^3 + bx^2 +cx +d
    ax^3 + bx^2 +cx +d = (px+q)(x^2 - x + 2) + 5x +4

    ax^3 + bx^2 + (c-5)x + (d-4) = (px + q)(x^2 - x + 2)

    And similarly for the other one. Then it is solvable by equation like powers yadda yadda yadda.
     
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