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Math Help - Polynomial division (I think)

  1. #1
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    Polynomial division (I think)

    I'm absolutely stumped by this problem:

    If f(x) = (x + 5) (x - k) and the remainder is 28 when f(x) is divided by x - 2, find k.

    I figure that I'm supposed to use synthetic division and then have the remainder equal 28, but that isn't getting me the right answer (which is k = -2). Could someone please help me out?
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  2. #2
    MHF Contributor Unknown008's Avatar
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    You can use the remainder theorem, which states that if f(x) is divided by x-a, and the remainder is b, then f(a) = b.

    In your case,

    f(x) = (x+5)(x-k)

    f(2) = (2+5)(2-k) = 28

    Solve for k
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  3. #3
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    Thanks!
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  4. #4
    MHF Contributor Unknown008's Avatar
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    You're welcome.

    I tried using long division too, but it's easy to get the numbers messed up. Both methods work though.

    Through long division, I get as equation:

    14 - 7k = 28

    Then, you get k = -2.
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  5. #5
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    Hello, blackdragon190!

    f(x) \:=\: (x + 5) (x - k) and the remainder is 28
    when f(x) is divided by x - 2.

    \text}Find }k.

    If you must use long division . . .


    We have: . \bigg[x^2 + (5-k)x - 5k\bigg] \div \bigg[x-2\bigg]


    . . \begin{array}{ccccccccc} & & & & x & + & 7-k \\  & & -- & -- & --- & -- & --- \\ x-2 & | ^& x^2 & + & (5-k)x & - & 5k \\ & & x^2 & - & 2x \\ & & -- & -- & --- \\ &&&& (7-k)x & - & 5k \\ &&&& (7-k)x & - & 2(7-k) \\ & & & & --- & -- & --- \\ &&&&&& \text{-}7k +14 & \leftarrow\text{ remainder}\end{array}


    Hence: . \text{-}7k+14 \;=\:28 \quad\Rightarrow\quad \text{-}7k \:=\:14 \quad\Rightarrow\quad \boxed{k \:=\:-2}


    Edit: Too slow . . . again!
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  6. #6
    MHF Contributor Unknown008's Avatar
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    Quote Originally Posted by Soroban View Post
    Edit: Too slow . . . again!
    I was trying to post the long division method, but the code tabs weren't behaving and there were some sort of limited spaces... I gave up then, and posted only the final equation.
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  7. #7
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    Edit: Too slow . . . again!
    Thanks anyway!
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