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    Q1) Using algebraic division ONLY show that x^3 -4x^2 + 3x can be written in the form:
    (x-2)(x^2 + ax + b) + r

    where a, b and r and integer constants.


    Q2)

    Find the value of k for which the equation:
    x^2 - 2(k+1)x - 2x^2 - 7= 0
    has equal roots
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    (Original post by jsmith6131)
    Q1) Using algebraic division ONLY show that x^3 -4x^2 + 3x can be written in the form:
    (x-2)(x^2 + ax + b) + r

    where a, b and r and integer constants.


    Q2)

    Find the value of k for which the equation:
    x^2 - 2(k+1)x - 2x^2 - 7= 0
    has equal roots

    What do u need help with mate? U got any working?
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    (Original post by jsmith6131)
    Q1) Using algebraic division ONLY show that x^3 -4x^2 + 3x can be written in the form:
    (x-2)(x^2 + ax + b) + r

    where a, b and r and integer constants.


    Q2)

    Find the value of k for which the equation:
    x^2 - 2(k+1)x - 2x^2 - 7= 0
    has equal roots
    Do you know how to do algebraic long division? If so divide x^3 - 4x^2 + 3x by x-2 and this will give you the values for a, b and r.

    For the second question, use the discriminant of the quadratic formula.
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    O thanks, for Q2) I now got k = 4, -2 which is correct

    for q1) when i divided by x-2 I got x^2 - 2x with a remainder of -x
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    this is why I am stuck
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    (Original post by jsmith6131)
    Q1) Using algebraic division ONLY show that x^3 -4x^2 + 3x can be written in the form:
    (x-2)(x^2 + ax + b) + r

    where a, b and r and integer constants.


    Q2)

    Find the value of k for which the equation:
    x^2 - 2(k+1)x - 2x^2 - 7= 0
    has equal roots
    i think i've done a similar if not the same question and in that a previous part told you what r the remainder was and then you had to use long division to get the other bit
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    (Original post by jsmith6131)
    O thanks, for Q2) I now got k = 4, -2 which is correct

    for q1) when i divided by x-2 I got x^2 - 2x with a remainder of -x
    You haven't done the last step of the division. Remember that x^3 - 4x^2 + 3x is the same as x^3 - 4x^2 + 3x + 0 If you use this you should get an integer remainder after the division.
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    I'm assuming you're after how you'd do these.

    Fpr the first one you need to divide the original polynomial by (x-2). This will give another polynomial (which will be a quadratic with no constant) and a remainder (which is r).

    For the second one, you need to simplify the quadratic. Them you need to find a value which gives a discriminant of 0 (which is b^2-4ac)
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    The answer is (x-2)(x^2 - 2x - 1) - 2
 
 
 
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