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Math Help - Remainder theorem

  1. #1
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    Exclamation Remainder theorem

    The expression f(x) = ax^3 - ( a + 3b )x^2 + 2bx + c is excatly divisible by x^2 - 2x. When f(x) is divided by x - 1 , the remainder is 8 more than when it is divided by x + 1. Factorise f(x) completely.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Ilsa View Post
    The expression f(x) = ax^3 - ( a + 3b )x^2 + 2bx + c is excatly divisible by x^2 - 2x. When f(x) is divided by x - 1 , the remainder is 8 more than when it is divided by x + 1. Factorise f(x) completely.
    Well, f(x) is divisible by x^2 - 2x = x(x - 2) so we know f(x) is divisible by x. Thus c = 0 immediately. Consider then
    f(x) = ax^3 - (a + 3b)x^2 + 2bx

    From the original statement (divisible by x(x - 2)), we get
    f(2) = a(2)^3 - (a + 3b)(2)^2 + 2b(2) = 0

    Call
    f(1) = a(1)^3 - (a + 3b)(1)^2 + 2b(1) = d

    Then we know that
    f(-1) = a(-1)^3 - (a + 3b)(-1)^2 + 2b(-1) = d - 8

    Three equations, three unknowns.

    -Dan
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  3. #3
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    Quote Originally Posted by topsquark View Post
    Well, f(x) is divisible by x^2 - 2x = x(x - 2) so we know f(x) is divisible by x. Thus c = 0 immediately. Consider then
    f(x) = ax^3 - (a + 3b)x^2 + 2bx

    From the original statement (divisible by x(x - 2)), we get
    f(2) = a(2)^3 - (a + 3b)(2)^2 + 2b(2) = 0

    Call
    f(1) = a(1)^3 - (a + 3b)(1)^2 + 2b(1) = d

    Then we know that
    f(-1) = a(-1)^3 - (a + 3b)(-1)^2 + 2b(-1) = d - 8

    Three equations, three unknowns.

    -Dan
    If f(x) is divisible by x, how is c=0?
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  4. #4
    Forum Admin topsquark's Avatar
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    f(x) = ax^3 - ( a + 3b )x^2 + 2bx + c is excatly divisible by x^2 - 2x.
    If x divides f(x) then
    \displaystyle \frac{f(x)}{x} leaves no remainder. Thus

    \displaystyle \frac{ax^3 - (a + 3b)x^2 + 2bx + c}{x} = ax^2 - (a + 3b)x + 2b + \frac{c}{x}

    This gives a remainder of c, but f(x) is divisible by x, thus c = 0.

    You can do this quicker by using the fact that if f(x) is divisible by x - r then f(r) = 0. In this case r = 0, so f(0) = 0.

    -Dan
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