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Math Help - Polynomials: Synthetic Division. Help needed ASAP!!

  1. #1
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    Polynomials: Synthetic Division. Help needed ASAP!!

    Hello, my cousin has a math problem and needs this problem solved soon. I myself am not sure so any help would be great!

    f(x)= x cubed - x squared -5x -3

    a) i) Show that (x+1) is a factor of f(x)
    ii) Hence or otherwise factorise f(x) fully

    b) One of the turning points of the graph of y=f(x) lies on the x-axis. Write down the coordinates of this turning point

    It is part b which she needs help with

    I typed part a out incase it is needed.
    Thanks in advance
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  2. #2
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    You know that it has x+1 as a factor. Completed, the factorization is

    (x-3)(x+1)^{2}

    The turning point that lies on the x-axis is where it has a root of multiplicity 2.

    See which one that is?.
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  3. #3
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    I'm sorry but I am not sure what you mean

    purely an educated guess but is it (1,-3)?

    Possibly? Really not sure thanks for the reply
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  4. #4
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    f(x) = x^3 - x^2 - 5x - 3

    A)
    i. To show that x+1 is a factor of f(x), you can just use long division and show that it divides without leaving any remainder.
    ii. So you can factorize it as (x+1)(quotient). If it needs, you can factorize the quotient too.

    B) It says that one of the turning points lies on x axis. First, find the turning points using f'(x) = 0. You should find two turning points. One of them is on the x axis, plug them in f(x) and see which one is.
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  5. #5
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    Ahh!
    Theres hope for me and my cousin yet!!

    so you get, x= 5/3 and x=-1 , plug it so you get f(x)=0 ??

    Thanks!
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  6. #6
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    Hello, Scotland!

    f(x)\:=\: x^3 -x^2-5x - 3

    a) i) Show that (x+1) is a factor of f(x)
    . .ii) Factorise f(x) fully.
    We can use the Remainder/Factor Theorem.

    . . \text{-}1)^3 - (\text{-}1)^2 - 5(\text{-}1) - 3 \;=\;0" alt="f(\text{-}1) \:=\\text{-}1)^3 - (\text{-}1)^2 - 5(\text{-}1) - 3 \;=\;0" />

    Since f(\text{-}1) = 0, then (x+1) is a factor of f(x).


    Then we have: . f(x) \;=\;(x+1)(x^2-2x-3) \;=\;(x+1)^2(x-3)



    b) One of the turning points of the graph of f(x)
    lies on the x-axis. .Find the coordinates of this turning point.
    We can answer this without Calculus . . .

    In part (a), we found that: . x+1)^2(x-3)" alt="f(x) \:=\x+1)^2(x-3)" />

    The curve has x-intercepts at (-1,\,0)\text{ and }(3,\,0).

    The intercept (-1,\,0) has multiplicity 2.
    . . Hence, the curve is tangent to the x-axis at x = -1.

    Therefore, there is a turning point at (-1,\,0).


    The graph looks like this:
    Code:
                 |
                 |          *
                 |
                 |
                 |         *
                 |
    --------o----+--------o----
         * -1  * |       * 3
       *         *     *
      *          |  *
                 |
     *           |
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  7. #7
    Eater of Worlds
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    You don't necessarily need calculus. See where it touches but does not cross the x-axis?. Which root is that?. It's in your factorization. Which one has exponent 2?.
    Last edited by galactus; November 24th 2008 at 05:38 AM.
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  8. #8
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    Ah right

    A question, why is (-1,0) a tangent because it has multiplicity of 2?

    Don't really understand that part...

    Thanks again
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  9. #9
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    f(x) = (x+1)^2(x-3)

    Think about the graph y = x+1. The root x = -1 has multiplicity one because it can be written y = (x+1)^1. It crosses the x-axis there because it is negative on one side and positive on the other.

    If we have a root of multiplicity 2, we get something like  y = (x+1)^2. But (x+1)^2 is never negative so instead of crossing the x-axis at x = -1, it just touches it, then moves off in the direction it came from. So when we have roots of multiplicity 2, they have a tangent at the axis.

    This can be generalised to all even and odd multiplicities.
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  10. #10
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    Oh ok, I understand now.

    Thanks for all your help!
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