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  1. #1
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    theorem

    x^3 + 2x^2 + px -3 is divided by x+1 the remainder is the same as when it is divided by x-2. find value of p


    thanks
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  2. #2
    Moo
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    Hello,

    Let y be the common remainder. y will be a constant, because you divide by a polynom of degree 1.

    <br />
x^3+2x^2+px-3=(x+1)(ax^2+bx+c)+y
    <br />
x^3+2x^2+px-3=(x-2)(a'x^2+b'x+c')+y<br />

    If you develop (x+1)(ax^2+bx+c), you get, by identification :

    a=1
    b=1
    c=p-1 (1)
    c+y=-3 \Longleftrightarrow y+3=-c

    (i leave you the right to do the calculus )

    In the same way,
    a'=1
    b'=4
    c'=p+8 (2)
    -2c'+y=-3 \Longleftrightarrow y+3=2c'

    Hence 2c'=-c.

    With (1) and (2), we can state that 2(p+8)=-(p-1)

    And then, just solve for p
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  3. #3
    Super Member wingless's Avatar
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    Let P(x) and Q(x) be polynomials and k be the root of Q(x)=0.

    Then the remainder of the division P(x)/Q(x) is equal to P(k).


    In your question,
    P(x) = x^3 + 2x^2 + px - 3 and Q(x)=x+1
    The root of Q(x) = 0 is,
    x+1=0, x=-1

    So find P(-1),
    P(x) = x^3 + 2x^2 + px - 3
    P(-1) = -1 + 2 -p - 3 = -p-2


    Now find the remainder for Q(x) = x-2
    The root of Q(x) = 0 is,
    x-2 =0, x=2

    So P(2) = 8 + 8 + 2p -3 = 2p + 13

    Since the remainders are the same,
    -p-2 = 2p + 13
    p=-5
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  4. #4
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    Quote Originally Posted by gracey View Post
    x^3 + 2x^2 + px -3 is divided by x+1 the remainder is the same as when it is divided by x-2. find value of p


    thanks
    To streamline things a bit ......

    Use the remainder theorem:

    Let the remainder be r. Then:

    -1 + 2 - p - 3 = r => -2 - p = r .... (1)

    8 + 8 + 2p - 3 = r => 13 + 2p = r .... (2)

    Equate (1) and (2): -2 - p = 13 + 2p => p = -5.
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