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Math Help - X^3 + 2x^3 - 5x - 6 = (x-2)(x+3)(x-k), find k.

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    X^3 + 2x^3 - 5x - 6 = (x-2)(x+3)(x-k), find k.

    X^3 + 2x^3 - 5x - 6 = (x-2)(x+3)(x-k), i have to find the value of k,


    can you please help me with starting the question, i am completely confuse


    (ps of anyone is confuse ^ means power)
    Last edited by mr fantastic; January 7th 2012 at 05:05 PM. Reason: Re-titled.
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    Re: Advance level past paper question

    Expand the right-hand side and add like terms. Then use the fact that polynomials are equal iff their corresponding coefficients are equal.
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    Re: Advance level past paper question

    Quote Originally Posted by mathkid12 View Post
    X^3 + 2x^3 - 5x - 6 = (x-2)(x+3)(x-k), i have to find the value of k,


    can you please help me with starting the question, i am completely confuse


    (ps of anyone is confuse ^ means power)
    Expand the RHS and equate the coefficients. A shortcut can be gained by considering the constant term: -6 = -6 \cdot -k

    If you're unsure where -6 on the right comes from then what is the constant term of (x-2)(x+3)
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    Re: Advance level past paper question

    I am assuming you mean:

    x^3+2x^2-5x-6=(x-2)(x+3)(x-k)

    In the factored form, look at the constant terms in each factor and what their product must be by looking at the expanded form.
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    Thumbs up Re: Advance level past paper question

    Quote Originally Posted by emakarov View Post
    Expand the right-hand side and add like terms. Then use the fact that polynomials are equal iff their corresponding coefficients are equal.

    so wait please
    (x-2)(x+3)(x-k) = x^2 + 3x - 2x - 6 (x-k)
    =x^3 + 3x^2 -6x^2 - 6x- kX^2 - 3xk + 2xk +6k

    AM i goinG correct so far, and what advice can you give me to proceed, AND THANKS ALOT
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    Re: Advance level past paper question

    Another way: \frac{x^3+2x^2-5x+6}{(x-2)(x+3)}=\ldots= x+1 so, k=-1 . You can quickly divide using Ruffini's rule.
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    Re: Advance level past paper question

    ohh ok thanks but i still a bit confuse
    Last edited by mathkid12; January 3rd 2012 at 12:10 PM. Reason: tried back the question using this method
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    Re: Advance level past paper question

    If you are choosing to expand the RHS, then you will have:

    x^3+2x^2-5x-6=(x-2)(x+3)(x-k)

    x^3+2x^2-5x-6=\left(x^2+x-6\right)(x-k)

    x^3+2x^2-5x-6=\left(x^3+x^2-6x\right)-\left(kx^2+kx-6k\right)

    x^3+2x^2-5x-6=x^3+(1-k)x^2-(6+k)x+6k

    Equating coefficients involving k, we find:

    2=1-k

    -5=-(6+k)

    -6=6k

    Take your pick, and solve for k.
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    Re: Advance level past paper question

    Quote Originally Posted by mathkid12 View Post
    X^3 + 2x^3 - 5x - 6 = (x-2)(x+3)(x-k), i have to find the value of k,


    can you please help me with starting the question, i am completely confuse


    (ps of anyone is confuse ^ means power)

    Find the value of k in terms of X right?

    x^3 + 2x^3 - 5x -6 = (x-2)(x+3)(x-k)

    = x^3 + 2x^3 - 5x - 6 = (x^2 + x - 6)(x - k)
    = x^3 + 2x^3 - 5x - 6 = (x^3 - (k)x^2 + x^2 - (k)x -6x -(k)6
    Subtract x^3 from both sides to get

    2x^3 -5x - 6 = (k)x^2 + x^2 - (k)x - 6x - (k)6

    Add 6x to both sides to get

    2x^3 + x - 6 = (k)x^2 + x^2 - (k)x - (k)6

    Subtract x^2 from both sides to get

    2x^3 + x^2 + x - 6 = (k)x^2 + (k)x - (k)6

    Take k out from the right side to get

    2x^3 + x^2 + x - 6 = (k)(x^2 + x - 6) <--------- I may have messed up from this part, check your math too.

    Look, we have x^2 + x - 6 on both sides so divide

    2x^3 + 1 = (k)
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    Re: Advance level past paper question

    Quote Originally Posted by okokjae View Post
    Find the value of k in terms of X right?

    x^3 + 2x^3 - 5x -6 = (x-2)(x+3)(x-k)

    = x^3 + 2x^3 - 5x - 6 = (x^2 + x - 6)(x - k)
    = x^3 + 2x^3 - 5x - 6 = (x^3 - (k)x^2 + x^2 - (k)x -6x -(k)6
    Subtract x^3 from both sides to get

    2x^3 -5x - 6 = (k)x^2 + x^2 - (k)x - 6x - (k)6

    Add 6x to both sides to get

    2x^3 + x - 6 = (k)x^2 + x^2 - (k)x - (k)6

    Subtract x^2 from both sides to get

    2x^3 + x^2 + x - 6 = (k)x^2 + (k)x - (k)6

    Take k out from the right side to get

    2x^3 + x^2 + x - 6 = (k)(x^2 + x - 6) <--------- I may have messed up from this part, check your math too.

    Look, we have x^2 + x - 6 on both sides so divide

    2x^3 + 1 = (k)
    Scratch this post if you meant x^3 + 2x^2 - 5x - 6 = (x-2)(x+3)(x-k) and not x^3 + 2x^3 - 5x - 6 = (x-2)(x+3)(x-k)
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