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Thread: factorization

  1. #1
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    factorization

    Given $\displaystyle p(x)=x^3+2x^2-x-2 $

    If $\displaystyle q(y)=y^6+2y^5-4y^4-6y^3+4y^2+2y-1$ , find the real number k such that
    $\displaystyle

    q(y)=y^3[(y-\frac{1}{y})^3+2(y-\frac{1}{y})^2-(y-\frac{1}{y})+k]
    $, y is not 0 .

    By using the substitution $\displaystyle x=y-\frac{1}{y}$ , show that the equation $\displaystyle q(y)=0$ can be transformed into the equation $\displaystyle p(x)=0$ .

    THanks ..
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  2. #2
    Senior Member TheAbstractionist's Avatar
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    Quote Originally Posted by thereddevils View Post
    Given $\displaystyle p(x)=x^3+2x^2-x-2 $

    If $\displaystyle q(y)=y^6+2y^5-4y^4-6y^3+4y^2+2y-1$ , find the real number k such that
    $\displaystyle

    q(y)=y^3[(y-\frac{1}{y})^3+2(y-\frac{1}{y})^2-(y-\frac{1}{y})+k]
    $, y is not 0 .

    By using the substitution $\displaystyle x=y-\frac{1}{y}$ , show that the equation $\displaystyle q(y)=0$ can be transformed into the equation $\displaystyle p(x)=0$ .

    THanks ..
    The constant term in $\displaystyle \left(y-\frac{1}{y}\right)^3+2\left(y-\frac{1}{y}\right)^2-\left(y-\frac{1}{y}\right)+k$ is $\displaystyle -4+k$ and this has to equal the coefficient of $\displaystyle y^3$ in $\displaystyle q(y).$ Hence $\displaystyle -4+k=-6\ \implies\ k=-2.$
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  3. #3
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    Quote Originally Posted by TheAbstractionist View Post
    The constant term in $\displaystyle \left(y-\frac{1}{y}\right)^3+2\left(y-\frac{1}{y}\right)^2-\left(y-\frac{1}{y}\right)+k$ is $\displaystyle -4+k$ and this has to equal the coefficient of $\displaystyle y^3$ in $\displaystyle q(y).$ Hence $\displaystyle -4+k=-6\ \implies\ k=-2.$

    THanks .But i don understand how did u get the constant term ?
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  4. #4
    Senior Member TheAbstractionist's Avatar
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    $\displaystyle \left(y-\frac{1}{y}\right)^3$ has no constant term (the terms are in $\displaystyle y^3,\,y,\,y^{-1}$ and $\displaystyle y^{-3}).$ Neither does $\displaystyle -\left(y-\frac{1}{y}\right)$ have any constant term. Hence the only constant term besides the $\displaystyle k$ comes from $\displaystyle 2\left(y-\frac{1}{y}\right)^2$ and that is -4.

    As a last resort, you can expand out $\displaystyle \left(y-\frac{1}{y}\right)^3+2\left(y-\frac{1}{y}\right)^2-\left(y-\frac{1}{y}\right)+k$ completely and see what happens.
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  5. #5
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    Smile

    hi
    i have the same problem with 3 constants to find
    Could you please show me how to find them with the same method

    $\displaystyle f : x \rightarrow \frac{x^{2}+x-2}{x^{2}+x}$
    find a,b and c such :
    $\displaystyle f(x) = a + \frac{b}{x}+\frac{c}{x+1}$
    $\displaystyle x$ is not 0 nor -1
    thanks a lot.
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  6. #6
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by Raoh View Post
    hi
    i have the same problem with 3 constants to find
    Could you please show me how to find them with the same method

    $\displaystyle f : x \rightarrow \frac{x^{2}+x-2}{x^{2}+x}$
    find a,b and c such :
    $\displaystyle f(x) = a + \frac{b}{x}+\frac{c}{x+1}$
    $\displaystyle x$ is not 0 nor -1
    thanks a lot.

    $\displaystyle f:\rightarrow \frac{x^2+x-2}{x^2+x} $

    so

    $\displaystyle f(x) = a+\frac{b}{x} + \frac{c}{x+1} = \frac{x^2+x-2}{x^2+x} $

    $\displaystyle a+\frac{b}{x} + \frac{c}{x+1} = \frac{x^2+x}{x^2+x} + \frac{-2}{x^2+x} $

    $\displaystyle a+\frac{b}{x} + \frac{c}{x^2+x} = 1 + \frac{-2}{x^2+x} $

    $\displaystyle \frac{-2}{x^2+x} =\frac{A}{x} + \frac{B}{x+1} $

    $\displaystyle -2= A(x+1) + Bx $

    $\displaystyle A=-2 , B=2 $

    $\displaystyle a+\frac{b}{x} + \frac{c}{x+1}=1+\frac{-2}{x} + \frac{2}{x+1} $

    it is clear
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  7. #7
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    thanks a lot
    i liked how "TheAbstractionist" answered the first problem,is there any way we can do this one with the same method.
    thanks again
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