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    Hi there

    I have a question

    x^2-3x-18=0 (Hint use the fact that x=3 is a root of this equation.)

    Now I have the answer to this and I know you basically divide x^2-3x-18 by the root (I think) I just dont have an explaination as to how this is done? Have looked at it for some times but doesnt mean much to me please could someone explain this to me or perhaps point me in the direction of a website or something that goes over this stuff? Thankyou ever so much

    Kind Regards
    S
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    Doesn't look like x=3 is a root of that equation to me...
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    i think he meant x = -3
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    x^2-3x-18=0
    If x = -3 is a root of this equation, then  (x+3) is a factor. (Why?)

    Proof
    x^2-3x-18=0
    (x+3)(some function of x) = 0
    Because if x = -3,(x+3) = 0
    0*(some function of x) = 0
    LHS = 0, RHS = 0. Left hand side = Right hand side. Therefore (x+3) is a factor.

    Now you need to find (some function of x)

    (x+3)(some function of x) = x^2-3x-18

    Deductions
    1)first term of (some function of x) must be 'x', because when you expand the bracket, the 'x' from (x+3) will multiply with the 'x' from (some function of x) to give x^2, which fits the above equation!

    So,
    (x+3)(x + something) = x^2-3x-18

    2)Look at '-18'. What multipled by '+3' from (x+3) gives you '-18'?

    Find out that number. Manually expand it out and see if it fits the expression on the right.


    Now that you've got the expression, replaced the given equation with the one you have found:
    (x+3)(x+ something) = 0

    Therefore:
    Either (x+3) = 0 (Which is given in the question) or (x + something) = 0

    Find the other root(solution) for the equation.
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    I might be talking joo-joo then but the equation doesn't seem to have any complex roots. Real roots at x=-3 and 6?
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    FYI: This is not complex numbers! (mathematically :P) Unless you're describing it as being complex!
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    (Original post by RedJimmy)
    I might be talking joo-joo then but the equation doesn't seem to have any complex roots. Real roots at x=-3 and 6?
    that is right.
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    Haha, the word complex should probably be banned when describing difficulty in mathematical problems.
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    (Original post by RedJimmy)
    Haha, the word complex should probably be banned when describing difficulty in mathematical problems.
    It should be used conservatively, or it can lead to a misconception such as this
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    thankyou
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    Technically they actually are complex roots, but they are also real roots.

    same thing as saying that 2 is an integer and a real number.
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    ^ Yeap, all numbers have their real and imaginary parts
 
 
 
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