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Math Help - Quadratic factors of a cubic polynomial

  1. #1
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    Quadratic factors of a cubic polynomial

    If x + bx -2 is factor of xcubed + (2b-1)x - p, find two possible values of p.


    Im not sure what method to use to solve this. do i divide them?
    Last edited by Detanon; December 9th 2009 at 07:42 AM. Reason: Moved from another thread
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  2. #2
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    Hello Detanon
    Quote Originally Posted by Detanon View Post
    If x + bx -2 is factor of xcubed + (2b-1)x - p, find two possible values of p.


    Im not sure what method to use to solve this. do i divide them?
    Suppose the other factor is (rx+s). Then:
    (x^2+bx-2)(rx+s) = x^3+(2b-1)x^2 -p
    Comparing coefficients:
    x^3: r = 1
    So:
    (x^2+bx-2)(x+s) = x^3+(2b-1)x^2 -p
    Comparing coefficients:
    x^2: (s+b) = (2b-1) \Rightarrow s = b-1 (1)

    x: -2+sb = 0 \Rightarrow sb = 2 (2)
    So, from (1) and (2):
    b(b-1)=2

    \Rightarrow b^2-b-2=0

    \Rightarrow (b-2)(b+1)=0

    \Rightarrow b = 2, -1

    \Rightarrow s = 1, -2
    Constant term:
    p = 2s

    \Rightarrow p = 2, -4
    Grandad
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  3. #3
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    Sorry, but I really dont get that.
    I dont get why you suppose the other facotr is (rx+s)
    and what does <br />
x^3: r <br />
    mean?

    Is there a different method to do it?
    Last edited by Detanon; December 9th 2009 at 09:51 AM.
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  4. #4
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    Hello Detanon
    Quote Originally Posted by Detanon View Post
    Sorry, but I really dont get that.
    I dont get why you suppose the other facotr is (rx+s)
    Because when one factor of a cubic is a quadratic, the other is linear; i.e. it is of the form (rx + s) for some values of r and s.
    and what does <br />
x^3: r <br />
mean?
    It doesn't mean anything. You have taken it out of context. Read the whole thing. I am comparing coefficients. (Do you know what thay means?) First I am looking at the coefficient of x^3. Later on, I look at the coefficients of x^2, x and the constant. I have used a colon (':') each time to show which coefficient I am looking at.
    Is there a different method to do it?
    No, not a method that is as sensible and straightforward as the one I have shown you.

    Read it through again - it's really very simple.

    Grandad
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  5. #5
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    I still dont get it.. Is there another method where dividing is involved? I thnik thats the one I get.
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  6. #6
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    Hello Detanon

    I really don't know what your problem is here. Multiply out the brackets from
    (x^2+bx-2)(rx+s)
    and then note that the result has to be identically equal to
    x^3 +(2b-1)x^2 -p
    (Can you do that?)

    In this way you get the equations for r, s, b and p that I have shown you. I repeat: if you understand the basic rules of algebra, then trust me, this is really very simple. (And certainly much easier than trying to do an algebraic long division.)

    Grandad
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  7. #7
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    Quote Originally Posted by Detanon View Post
    If x + bx -2 is factor of xcubed + (2b-1)x - p, find two possible values of p.


    Im not sure what method to use to solve this. do i divide them?
    If x^2 + bx - 2 is a factor then it should be obvious, after a little thought, that the other factor has to be x + \frac{p}{2}. This is the only way that you can get the x^3 term and the constant term of -p.

    So x^3 + (2b - 1) x^2 - p = (x^2 + bx - 2)\left(x + \frac{p}{2}\right). Expand this out. You require the coefficient of x^2 to be 2b - 1 and the coeffcient of x to be zero:

    \frac{p}{2} + b = 2b - 1 \Rightarrow p = 2b - 2 .... (1)

    \frac{bp}{2} - 2 = 0 \Rightarrow bp = 4 .... (2)

    Solve equations (1) and (2) simultaneously for b and p.

    You need to go back now and have a good hard think about all that's been said about this question.

    For checking purposes:

    Spoiler:
    b = 2, -1 and p = 2, -4
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  8. #8
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    Where do you get

    Constant term:

    p = 2s

    from??
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  9. #9
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    Hello Detanon
    Quote Originally Posted by Detanon View Post
    Where do you get

    Constant term:

    p = 2s

    from??
    The constant term when we expand:
    (x^2+bx-2)(rx+s) is -2s, and this is equal to the constant term in x^3 +(2b-1)x^2 -p

    Hence p = 2s

    Grandad

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