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Thread: Factor & Remainder Theorem

  1. #1
    Newbie s2951's Avatar
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    Question Factor & Remainder Theorem

    I have prob getting other two factors.
    Had solved that (x-1) is a factor of 2X^3+5X^2-11X+4.

    2X^3+5X^2-11X+4
    =2(1)^3+5(1)^2-11(1)+4
    =0
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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by s2951 View Post
    I have prob getting other two factors.
    Had solved that (x-1) is a factor of 2X^3+5X^2-11X+4.

    2X^3+5X^2-11X+4
    =2(1)^3+5(1)^2-11(1)+4
    =0
    Use long division to find the answer to $\displaystyle \frac{2x^3+5x^2-11x+4}{x-1}$

    This should give a quadratic from which you can find the other two solutions (remember that $\displaystyle x-1$ is also a factor)
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  3. #3
    Newbie s2951's Avatar
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    Question

    I prefer to do this method, but got stuck.. whats next?

    f(x)=(x-1)q(x)
    2x^3+5x^2-11x+4=(x-1)(2x^2+x-4)
    2x^3+5x^2-11x+4=x(2x^2+x-4)-1(2x^2+x-4)
    ??? = ???
    Last edited by s2951; Nov 29th 2009 at 04:11 AM. Reason: Correct me if im wrong
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  4. #4
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    Quote Originally Posted by s2951 View Post
    I prefer to do this method, but got stuck.. whats next?

    f(x)=(x-1)q(x)
    2x^3+5x^2-11x+4=(x-1)(2x^2+x-4)
    2x^3+5x^2-11x+4=x(2x^2+x-4)-1(2x^2+x-4)
    ??? = ???
    You appear to be assuming that the other factor is $\displaystyle 2x^2+ x- 4$. Why?

    If you insist upon that method (though ordinary division or synthetic division is simpler) you need to write $\displaystyle 2x^3+ 5x^3- 11x+ 4= (x-1)(ax^2+ bx+ c)$. Do the multiplication and combine like powers and you will get three equations for a, b, and c. (Actually, you get 4 equations but they are not "independent" since x-1 is a factor of $\displaystyle 2x^3+ 5x- 11x+ 4$.
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  5. #5
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by s2951 View Post
    I prefer to do this method, but got stuck.. whats next?

    f(x)=(x-1)q(x)
    2x^3+5x^2-11x+4=(x-1)(2x^2+x-4)
    2x^3+5x^2-11x+4=x(2x^2+x-4)-1(2x^2+x-4)
    ??? = ???
    Let $\displaystyle q(x)=Ax^2+Bx+C$ (it's a method I prefer)

    $\displaystyle (x-1)(Ax^2+Bx+C) = 2x^3+5x^2-11x+4$

    Compare coefficients

    • $\displaystyle x^3 \: \rightarrow \: 2 = A$
    • $\displaystyle x^2 \: \rightarrow \: 5 = -A+B \: \: \therefore \: B = 7$
    • $\displaystyle x^0 \: \rightarrow \: 4 = -C $


    $\displaystyle (x-1)(2x^2+7x-4) = 2x^3+5x^2-11x+4$

    $\displaystyle \Delta = 7^2-(4 \times 2 \times -4) = 81$. As this is a perfect square the quadratic has rational solutions and can be factorised.

    $\displaystyle 2x^2+7x-4 = (2x-1)(x+4)$

    Therefore $\displaystyle 2x^3+5x^2-11x+4 = (x-1)(x+4)(2x-1)$
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