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Math Help - seperating higher power variables

  1. #1
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    seperating higher power variables

    I believe this is a high school level algebra question but if its not please move it.

    I am having trouble expanding higher power equations, such as, what steps do i take to solve this?

    2x^3-4x^2-2x+4=0
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  2. #2
    Senior Member yeKciM's Avatar
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     2x^3-4x^2 -2x +4 = 0

     2x^2 (x-2 ) -2 (x-2) =0

     (2x^2 -2 ) (x-2) = 0

     2x^2 - 2 = 0 \Rightarrow x^2 = 1 \Rightarrow x_1 = 1 \; x_2 = -1

     x-2 = 0 \Rightarrow x_3 = 2

     (x-1)(x+1)(x-2) =0
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  3. #3
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    Thanks a lot, didn't think about breaking up the first 2 and the 2nd 2
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  4. #4
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    Another way to look at it: the "rational root theorem" says that any rational number roots to a polynomial equation with integer coefficients must be of the form \frac{m}{n} with m, the numerator, a divisor of the leading coefficient, and n, the denominator, a divisor of the constant term. Here, the leading coefficent is 2 which has divisors, 1, -1, 2, and -2, and the constant term is 4 which has divisors 1, -1, 2, -2, 4, -4. The only possible roots of the equation 2x^3- 4x^2- 2x+ 4= 0 are 1, -1, 2, -2, 4, -4, 1/2, -1/2, 1/4, and -1/4. Putting those numbers into the equation, it is easy to see, immediately, that x= 1 is a root and so x- 1 is a factor. We can continue checking or, by "long division" or "synthetic" division, see that x^2- 4x^2- 2x+ 4= (x- 1)(2x^2- 2x- 4)= 2(x- 1)(x^2- x- 2) and then see that the last factors as (x- 2)(x+ 1)

    (I accidently entered "x^^2" rather than "x^2" but instead of getting a usual error message I got "blackisted command". Hope I didn't offend anyone!)
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