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Thread: last polynomial

  1. #1
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    last polynomial

    P(x)=3.P(-x)+4x-1
    Find remainder when P(x-1) is divided by (x+2).
    :
    My work:

    P(-3)=3.P(3)-13

    there is no relation between P(x) and P(-x),middle term of both equations are different.One has a - sign,the other has not..
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  2. #2
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    Re: last polynomial

    well.. one can always go the brute force route

    $p(x) = 3p(-x) + 4x -1$

    let $p(x) = c_0 + c_1 x + c_2 x^2 + c_3 x^3 \dots$

    $p(-x) = c_0 - c_1 x + c_2 x^2 - c_3 x^3 \dots$

    $c_0 = 3 c_0 -1 \Rightarrow c_0 = \dfrac 1 2$

    $c_1 = -3c_1 + 4 \Rightarrow c_1 =1$

    $c_2 = 3 c_2 \Rightarrow c_2 = 0$ and similarly for $c_k,~k>2$

    $p(x) = x + \dfrac 1 2$

    $p(x-1) = (x-1) + \dfrac 1 2 = x - \dfrac 1 2$

    $\dfrac{x-\frac 1 2}{x+2} = -\dfrac 5 2$
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    Re: last polynomial

    How did you get 1 and 1/2,3c00[/S]?
    Last edited by kastamonu; Jun 11th 2017 at 02:23 PM.
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    Re: last polynomial

    How did you find C_{0},C_{1},C_{2}?
    Last edited by kastamonu; Jun 11th 2017 at 02:48 PM.
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  5. #5
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    Re: last polynomial

    $$P(x)=3P(-x)+4x-1$$
    $$P(-x)=3P(-(-x))+4(-x)-1 = 3P(x)-4x-1$$
    Substitution yields
    $$P(x)=3(3P(x)-4x-1)+4x-1$$
    $$P(x)=9P(x)-12x-3+4x-1$$
    $$-8P(x)=-8x-4$$
    $$P(x)=x+\frac{1}{2}$$
    Last edited by Ahri; Jun 11th 2017 at 02:57 PM.
    Thanks from Plato
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    Re: last polynomial

    Ahri,Many Thanks. I undersood.But I am curious about thesolution above.

    How did he find C_{0},C_{1},C_{2}?
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    Re: last polynomial

    Many Thanks . Ok I understood the solution but how did you find c0,c1 and c2?
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  8. #8
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    Re: last polynomial

    By comparing coefficients of the constant term, $x$ term, $x^2$ term, and so on.
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    Re: last polynomial

    Quote Originally Posted by romsek View Post
    well.. one can always go the brute force route

    $p(x) = 3p(-x) + 4x -1$

    let $p(x) = c_0 + c_1 x + c_2 x^2 + c_3 x^3 \dots$

    $p(-x) = c_0 - c_1 x + c_2 x^2 - c_3 x^3 \dots$

    $c_0 = 3 c_0 -1 \Rightarrow c_0 = \dfrac 1 2$

    $c_1 = -3c_1 + 4 \Rightarrow c_1 =1$

    $c_2 = 3 c_2 \Rightarrow c_2 = 0$ and similarly for $c_k,~k>2$

    $p(x) = x + \dfrac 1 2$

    $p(x-1) = (x-1) + \dfrac 1 2 = x - \dfrac 1 2$

    $\dfrac{x-\frac 1 2}{x+2} = -\dfrac 5 2$
    the last line should be

    $x-\frac 1 2 \pmod{x+2} = -\dfrac 5 2$
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  10. #10
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    Re: last polynomial

    Quote Originally Posted by romsek View Post
    the last line should be
    $x-\frac 1 2 \pmod{x+2} = -\dfrac 5 2$
    Reply #5 is more beautiful solution.
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