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Math Help - [SOLVED] Algebraic Division

  1. #1
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    [SOLVED] Algebraic Division

    Normal algebraic division using remainder or factor theorem is usually fine however this exam question asks to find values in a different way:

    The polynomial
     f(x) = px^3 - x^2 + qx - 6
    has (x - 3) as a factor. When  f(x) is divided by  (x - 2) the remainder is -20.
    Show that p = 2 and find the value of q.

    I can find these values easily by division however the question is worth 6 marks and requires a different approach.
    How would you work this question out with that in mind?

    Thanks if you can help.
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  2. #2
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    Find f(3) and f(2) in terms of p and q. Then solve that resulting system of equations.
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  3. #3
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    Quote Originally Posted by db5vry View Post
    Normal algebraic division using remainder or factor theorem is usually fine however this exam question asks to find values in a different way:

    The polynomial
     f(x) = px^3 - x^2 + qx - 6
    has (x - 3) as a factor. When  f(x) is divided by  (x - 2) the remainder is -20.
    Show that p = 2 and find the value of q.

    I can find these values easily by division however the question is worth 6 marks and requires a different approach.
    How would you work this question out with that in mind?

    Thanks if you can help.
    If x - 3 is a factor, then the remainder when f(x) is divided by x - 3 is 0. Therefore f(3) = 0.

    If f(x) is divided by x - 2, the remainder is -20. So f(2) = -20.


    Using these pieces of information...

    f(3) = p(3)^3 - 3^2 + q(3) - 6 = 0

    27p - 9 + 3q - 6 = 0

    27p + 3q = 15


    and


    f(2) = p(2)^3 - 2^2 + q(2) - 6 = -20

    8p - 4 + 2q - 6 = -20

    8p + 2q = -10.


    So now you have 2 equations in 2 unknowns which you can solve simultaneously.


    27p + 3q = 15
    8p + 2q = -10


    54p + 6q = 30
    24p + 6q = -30


    So 30p = 60

    p = 2.


    If 8p + 2q = -10

    8(2) + 2q = -10

    16 + 2q = -10

    2q = -26

    q = -13.
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  4. #4
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    Quote Originally Posted by Prove It View Post
    If x - 3 is a factor, then the remainder when f(x) is divided by x - 3 is 0. Therefore f(3) = 0.

    If f(x) is divided by x - 2, the remainder is -20. So f(2) = -20.


    Using these pieces of information...

    f(3) = p(3)^3 - 3^2 + q(3) - 6 = 0

    27p - 9 + 3q - 6 = 0

    27p + 3q = 15


    and


    f(2) = p(2)^3 - 2^2 + q(2) - 6 = -20

    8p - 4 + 2q - 6 = -20

    8p + 2q = -10.


    So now you have 2 equations in 2 unknowns which you can solve simultaneously.


    27p + 3q = 15
    8p + 2q = -10


    54p + 6q = 30
    24p + 6q = -30


    So 30p = 60

    p = 2.


    If 8p + 2q = -10

    8(2) + 2q = -10

    16 + 2q = -10

    2q = -26

    q = -13.
    Thank you for explaining that, it was really clear and it is very greatly appreciated!
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