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Math Help - a question about polynomials

  1. #1
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    another question about factor theorem/polynomials



    Question: please refer to the picture above.
    I think there is something wrong about my working as my remainder's answer is different from the choices, but i don't see anywhere went wrong. ="=
    Mind helping me to check whether is there anything wrong in my working?

    NOTE: I don't really know how to use [ math ] [ /math]. So, if you see some small p or big p, they're the same.

    <br />
(x-1)(x+3)(-x+p)
    =( x^2+2x-3)(-x+p)
    = -x^3  +x^2 p -2x^2 +2xp+3x-3p

    let x be -4,
    -(-4)^3 + (-4)^2 p  -2(-4)^2  + 2(-4)p +3(-4) -3p=10
    64 + 16p -32 -8p-12-3p=10
    5p=-10
    p=-2

    let x be 4 and substitute p=-2,
    -64 -32 -16 -16 +12 +6 = -110
    my remainder is -110, but according to the choices, the remainder should be -126.
    Last edited by mr fantastic; December 29th 2008 at 11:42 AM. Reason: Deletd new question added as an edit and moved it to a new thread
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  2. #2
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    Hello, wintersoltice!

    The roots of a cubic equationj, f(x) = 0, are 1,-3\text{ and }p.
    Given that f(x) is such that the coefficient of x^3 is -1,
    and that it has a remainder of 10 when divided by (x+4), find:

    (a) the value of p.

    (b) the remainder when divided by (x-4)
    The function is: . f(x) \:=\:-(x-1)(x+3)(x-p)


    (a) "The remainder is 10 when divided by (x+4)": . f(\text{-}4) = 10

    . . f(\text{-}4) \:=\:-(\text{-}4-1)(\text{-}4+3)(\text{-}4-p) \:=\:10  \quad\Rightarrow\quad 5(p+4) \:=\:10 \quad\Rightarrow\quad \boxed{p \:=\:-2}


    (b) The function is: . f(x) \:=\:-(x-1)(x+3)(x+2)

    "The remainder when f(x) is divided by (x-4)": .we want f(4)

    . . f(4) \:=\:-(4-1)(4+3)(4+2) \:=\:\boxed{-126}


    Answer: . (ii)

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  3. #3
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    Quote Originally Posted by Soroban View Post
    Hello, wintersoltice!

    The function is: . f(x) \:=\:-(x-1)(x+3)(x-p)


    (a) "The remainder is 10 when divided by (x+4)": . f(\text{-}4) = 10

    . . f(\text{-}4) \:=\:-(\text{-}4-1)(\text{-}4+3)(\text{-}4-p) \:=\:10  \quad\Rightarrow\quad 5(p+4) \:=\:10 \quad\Rightarrow\quad \boxed{p \:=\:-2}


    (b) The function is: . f(x) \:=\:-(x-1)(x+3)(x+2)

    "The remainder when f(x) is divided by (x-4)": .we want f(4)

    . . f(4) \:=\:-(4-1)(4+3)(4+2) \:=\:\boxed{-126}


    Answer: . (ii)


    o i see.
    so, my p is correct.
    but i still don't know where i went wrong in my working, making my remainder incorrect. =_=
    oh well, anyway, thanks. =)
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  4. #4
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    Quote Originally Posted by wintersoltice View Post
    [snip]
    let x be 4 and substitute p=-2,
    -64 -32 -16 -16 +12 +6 = -110
    my remainder is -110, but according to the choices, the remainder should be -126.
    Careless arithmetic. It should be -64 - 32 - 32 - 16 + 12 + 6.
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