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Math Help - cubics

  1. #1
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    cubics

    how guys, i haven't done cubics for quite some time.
    how would you go about finding the x intercepts of a cubic such as x^3 - 3x -2? Also how would u move such a cubic in the negative direction of the x axis.
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  2. #2
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    Hello, chaneliman!

    How would you find the x-intercepts of: . y \:=\:x^3 - 3x -2
    You might note that x = 2 is a zero of the polynomial.

    Using long division to factor: . x-2)(x^2+2x+1) \;=\;(x-2)(x+1)^2" alt="x^3-3x-2 \:=\x-2)(x^2+2x+1) \;=\;(x-2)(x+1)^2" />

    The x-interecepts are: . 2\text{ and }-1




    How would u move such a cubic in the negative direction of the x axis?
    To move it a units to the left, replace x with (x+a).


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  3. #3
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    Quote Originally Posted by chaneliman View Post
    how guys, i haven't done cubics for quite some time.
    how would you go about finding the x intercepts of a cubic such as x^3 - 3x -2? Also how would u move such a cubic in the negative direction of the x axis.
    Solve x^3 - 3x - 2 = 0.

    f(x) ---> f(x + a) where a > 0 will move the graph of y = f(x) a units to the left.


    Oh ..... now you want to konw how to solve x^3 - 3x - 2 = 0 ......

    Trial and error. x = -1 seems to work. So x + 1 is a factor. So now you can factorise: (x + 1)(x^2 - .... -2).

    Or you could do a clever grouping: x^3 - 3x - 2 = (x^3 + 2x^2 + x) - (2x^2 + 4x + 2) = x(x^2 + 2x + 1) - 2(x^2 + 2x + 1) = ........

    Oh yes ..... In advance - you're welcome ........
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  4. #4
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    i understand everything except "You might note that is a zero of the polynomial."
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  5. #5
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    Quote Originally Posted by chaneliman View Post
    i understand everything except "You might note that is a zero of the polynomial."
    this is from the rational roots theorem (look it up). to find the possible rational roots here (that is, the x-values that cause the function to give a value of zero) we try the factors of 2. so we try +/- 1, and +/- 2. and it turns out that x = 2 and x = -1 work
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