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Math Help - The remainder and factor theorems 2 (what did I do wrong?)

  1. #1
    Member Mr Rayon's Avatar
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    Exclamation The remainder and factor theorems 2 (what did I do wrong?)

    The remainder when x^3 + ax^2 + bx + 1 is divided by (x - 5) is -14. When the cubic polynomial is divided by (x + 1), the remainder is -2. Find a and b.

    To solve a and b, I did the following:

    P(5) = 5^3 + a(5)^2 + b(5) + 1
    = 125 + 25a + 5b + 1
    = 25a + 5b + 126

    25a + 5b + 126 = -14

    P(-1) = (-1)^3 + a(-1)^2 + b(-1) + 1
    = -1 + a - b + 1
    = a - b

    a - b = -2

    Using the method of elimination:

    25a + 5b = -140
    25a - 25b = -50
    30b = -190
    b= -19/3

    Substituting the value of b into: a - b = -2
    a - (-19/3) = -2
    a + 19/3 = -2
    a = -25/3


    But the answers are supposed to be:

    a = -5, b= -3


    What did I do wrong? Please show me complete working out. All help will be appreciated.
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  2. #2
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    Quote Originally Posted by Mr Rayon View Post
    The remainder when x^3 + ax^2 + bx + 1 is divided by (x - 5) is -14. When the cubic polynomial is divided by (x + 1), the remainder is -2. Find a and b.

    To solve a and b, I did the following:

    P(5) = 5^3 + a(5)^2 + b(5) + 1
    = 125 + 25a + 5b + 1
    = 25a + 5b + 126

    25a + 5b + 126 = -14

    P(-1) = (-1)^3 + a(-1)^2 + b(-1) + 1
    = -1 + a - b + 1
    = a - b

    a - b = -2

    Using the method of elimination:

    25a + 5b = -140
    25a - 25b = -50
    30b = -190 Mr F says: -140 - (-50) = -90. So you should have 30b = -90.

    b= -19/3

    Substituting the value of b into: a - b = -2
    a - (-19/3) = -2
    a + 19/3 = -2
    a = -25/3


    But the answers are supposed to be:

    a = -5, b= -3


    What did I do wrong? Please show me complete working out. All help will be appreciated.
    Note: Do yourself a favor and simplify your equations before trying to solve them. By dividing equation 1 through by 5 you get 5a + b = -28. Much easier to work with.
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  3. #3
    Member Mr Rayon's Avatar
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    Oh wait I think I understand now! From above I was assuming that they were simulataneous equations. So to solve a and b I just need to use the quadaratic formulat on both separately?
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  4. #4
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    Quote Originally Posted by Mr Rayon View Post
    Oh wait I think I understand now! From above I was assuming that they were simulataneous equations. So to solve a and b I just need to use the quadaratic formulat on both separately?
    No. Read my post again - the red stuff (I made an edit to make my point more obvious).
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  5. #5
    Member Mr Rayon's Avatar
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    Quote Originally Posted by mr fantastic View Post
    Note: Do yourself a favor and simplify your equations before trying to solve them. By dividing equation 1 through by 5 you get 5a + b = -28. Much easier to work with.
    Oh...yes I understand completely now!
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