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    Please can someone help me with this question, i don't even know where to start!

    Given that f(x) = (x-2) (x^3 + ax^2 + bx + c) find the values of the constants a, b and c.

    Thankyouu
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    (Original post by stanmoor)
    Please can someone help me with this question, i don't even know where to start!

    Given that f(x) = (x-2) (x^3 + ax^2 + bx + c) find the values of the constants a, b and c.

    Thankyouu
    You need to give us the rest of the question otherwise you are going to get some pretty unhelpful responses.
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    (Original post by Mr M)
    You need to give us the rest of the question otherwise you are going to get some pretty unhelpful responses.
    This.

    Though from what I can see, it will probably involve long division...
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    (Original post by stanmoor)
    Please can someone help me with this question, i don't even know where to start!

    Given that f(x) = (x-2) (x^3 + ax^2 + bx + c) find the values of the constants a, b and c.

    Thankyouu
    I'm sorry but what type of maths is this?
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    Throw in any values you want, YOU CAN'T BE WRONG.

    Now that's Maths for you.
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    I agree that it looks like long division, and surely there's more to the question..
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    I have a better one :

    a, b and c are constants

    find them
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    (Original post by kaosu_souzousha)
    I have a better one :

    a, b and c are constants

    find them
    Haha. I reckon they are 2, 26 and 1,203,123.
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    How accurate was my first post?

    I'm psychic me.
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    (Original post by stanmoor)
    Please can someone help me with this question, i don't even know where to start!

    Given that f(x) = (x-2) (x^3 + ax^2 + bx + c) find the values of the constants a, b and c.

    Thankyouu
    f(2) = 8+4a+2b+c
    you need another (x- something) bracket to find the second equation. After this has been done , you can then solve simultaneously to find A, B and C.... normally its just a and b.... are you sure this is the full question and you havent left anything out?
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    (Original post by StephenP91)
    Haha. I reckon they are 2, 26 and 1,203,123.
    :banana2:
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    (Original post by MissCandyFloss)
    f(2) = 8+4a+2b+c
    you need another (x- something) bracket to find the second equation. After this has been done , you can then solve simultaneously to find A, B and C.... normally its just a and b.... are you sure this is the full question and you havent left anything out?
    you do realise that

    f(2) = 0 ?

    f(2) = ( 8 + 4a + 2b + c ) * ( 0 ) = 0
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    I think you put a couple of values in for x and then solve simultaneously.
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    (Original post by kaosu_souzousha)
    you do realise that

    f(2) = 0 ?

    f(2) = ( 8 + 4a + 2b + c ) * ( 0 ) = 0
    ohhhhh hes multiplying both brackets! lol i thought (x-2) was a factor... oops!
 
 
 
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