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Thread: [SOLVED] polynomial quesstion

  1. #1
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    [SOLVED] polynomial quesstion

    hello all,
    i hope this question is in the right board, anyhow, here is my problem:

    if $\displaystyle x^3+px+q$ is divisible by both $\displaystyle x-3$ and $\displaystyle x+2$, find the values of p and q


    thanks
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  2. #2
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    don't worry guys, after some more work i got the answer myself, here is the working out for anybody interested:

    $\displaystyle P(x) = x^3+px+q$
    if both x-3 and x+2 are factors of P(x), then P(3) = 0 = P(-2)

    therefore
    $\displaystyle (3)^3+p3+q = (-2)^3+p2+q$
    simplifying $\displaystyle 27+3p = -8-2p$
    $\displaystyle -35 = 5p$
    $\displaystyle p = -7$

    therefore $\displaystyle P(x)=(x-3)(x+2)(x+k)$

    now, $\displaystyle (k*-1) + (k*-6) =7$
    $\displaystyle k=1$

    therefore $\displaystyle P(x)=(x-3)(x+2)(x+1)$
    expanding $\displaystyle P(x)=x^3-7x-6$

    therefore $\displaystyle q=-6$
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  3. #3
    Super Member dhiab's Avatar
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    Quote Originally Posted by str33tl0rd View Post
    hello all,
    i hope this question is in the right board, anyhow, here is my problem:

    if $\displaystyle x^3+px+q$ is divisible by both $\displaystyle x-3$ and $\displaystyle x+2$, find the values of p and q


    thanks
    Hello :There is the solution
    I'have :

    $\displaystyle \begin{array}{l}
    x^2 + px + q = \left( {x - 3} \right)\left( {x + 2} \right)\left( {ax + b} \right) \\
    x^2 + px + q = ax^3 + \left( {b - a} \right)x^2 + \left( { - b - 6a} \right)x - 6b \\
    \end{array}$
    I'HAVE THE SYSTEM :
    $\displaystyle \left\{ \begin{array}{l}
    a = 1 \\
    b - a = 0 \\
    - b - 6a = p \\
    - 6b = q \\
    \end{array} \right.
    $


    $\displaystyle a = 1,b =1$
    Subst :
    CONCLUSION
    $\displaystyle \left\{ \begin{array}{l}
    q = - 6 \\
    p = - 7 \\
    \end{array} \right.
    $
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  4. #4
    Senior Member pacman's Avatar
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    "if is divisible by both and , find the values of p and q."

    Not so ELEGANT SOLUTION:

    Notice that x^3 + px + q = (x - a)(x - b)(x - c) = (x - 3)(x + 2)(x - k)

    that the coefficient of x^2 is ZERO, that means the sum of its roots is ZERO.

    then, -(-3 + 2 - k) = 0, or

    3 - 2 + k = 0,

    k = -1; thus the 3 roots are, -2, -1 and 3.

    p = (-2)(-1) + (-2)(3) + (-1)(3) = 2 - 6 - 3 = 2 - 9 = -7

    q = -(-2)(-1)(3) = -6

    x^3 -7x - 6 = (x - 3)(x + 2)(x + 1).





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