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Math Help - A cubic question

  1. #1
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    A cubic question

    The polynomial P(x) has a remainder of 2 when divided by x-1 and a remainder of 3, when divided by x-2. The remainder when P(x) is divided by  (x-1)(x-2) is ax + b, i.e. P(x) can be written as P(x) = (x-1)(x-2)Q(x)+ax+b

    Find a and b.

    Well I got this already a= 1 and b=1

    but the second question I don't know what to do...

    Given that  P(x) is a cubic polynomial with coefficient of x^3 being 1, and −1 is a solution of the equation P(x) = 0, find P(x).

    I'm not too sure how to start... If -1 is a solution of the question when P(x) = 0 that means (x+1) must be a factor right? I don't know what to do from here...

    All replies are appreciated
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  2. #2
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    Is it the same \displaystyle P(x) as it is in Part 1? If so, \displaystyle P(1) = 2 and \displaystyle P(2) = 3 by the Remainder Theorem.
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  3. #3
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    Oh wow, I really should think a bit before I start posting.

    Using P(x) = (x-1)(x-2)Q(x)+ax+b

    Let Q(x) = x+1 because -1 is a solution of x

    Then multiply out and sub  a=b=1
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  4. #4
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    Quote Originally Posted by jgv115 View Post
    The polynomial P(x) has a remainder of 2 when divided by x-1 and a remainder of 3, when divided by x-2. The remainder when P(x) is divided by  (x-1)(x-2) is ax + b, i.e. P(x) can be written as P(x) = (x-1)(x-2)Q(x)+ax+b

    Find a and b.

    Well I got this already a= 1 and b=1

    but the second question I don't know what to do...

    Given that  P(x) is a cubic polynomial with coefficient of x^3 being 1, and −1 is a solution of the equation P(x) = 0, find P(x).

    I'm not too sure how to start... If -1 is a solution of the question when P(x) = 0 that means (x+1) must be a factor right? I don't know what to do from here...

    All replies are appreciated
    As P(-1)=-6Q(-1)-1+1=-6Q(-1)=0, this implies that $$-1 is a root of $$Q(x)..

    CB
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