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Thread: Remainder theorem problem

  1. #1
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    Remainder theorem problem

    Im currently stuck on this question:

    When P(z) is divided by z+1 the remainder is -8 and when divided by z-3 the remainder is 4. Find the remainder when P(z) is divided by (z-3)(z+1)

    I know that P(-1) = -8 and P(3) = 4 but i have no clue what to do from here.

    Thanks for any help
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  2. #2
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    We have $\displaystyle P(z)=(z+1)Q(z)-8.$

    Let $\displaystyle Q(z)=(z-3)R(z)+k.$

    Thus $\displaystyle 4=P(3)=4Q(3)-8=4k-8$ $\displaystyle \implies$ $\displaystyle k=3.$

    Hence $\displaystyle P(z)=(z+1)(z-3)R(z)+3(z+1)-8=(z+1)(z-3)R(z)+3z-5$ so the remainder when $\displaystyle P(z)$ is divided by $\displaystyle (z+1)(z-3)$ is $\displaystyle 3z-5.$
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  3. #3
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    Im not sure what youve done here

    Thus

    And whats R(z)?
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  4. #4
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    I am basically using the result that given any polynomials $\displaystyle f(x),g(x)$ with $\displaystyle g(x)\not\equiv0,$ there exist polynomials $\displaystyle q(x),r(x)$ with either $\displaystyle r(x)\equiv0$ or $\displaystyle \deg r(x)<\deg g(x)$ such that $\displaystyle f(x)=q(x)g(x)+r(x).$ (This applies to polynomials over a field such as the field of rationals, reals or complex numbers.) The polynomial $\displaystyle r(x)$ is the remainder when $\displaystyle f(x)$ is divided by $\displaystyle g(x).$

    Actually, I didn’t do a great job in my post above. Let me re-do the proof in a clearer way.

    Given polynomials $\displaystyle P(z)$ and $\displaystyle (z-1)(z+3)$ we have that there exist a polynomial $\displaystyle R(z)$ and constants $\displaystyle k,h$ such that

    $\displaystyle P(z)\ =\ (z-1)(z+3)R(z)+kx+h$

    As $\displaystyle P(-1)=-8$ and $\displaystyle P(3)=4$ we have $\displaystyle -k+h=-8$ and $\displaystyle 3k+h=4.$ Solving the two simultaneous equations should yield $\displaystyle k=3,\,h=-5.$ Hence the remainder is $\displaystyle 3x-5.$
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  5. #5
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    Ahh i see :d

    Thanks very much for your help
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