# Thread: seperating higher power variables

1. ## 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?

$2x^3-4x^2-2x+4=0$

2. $2x^3-4x^2 -2x +4 = 0$

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

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

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

$x-2 = 0 \Rightarrow x_3 = 2$

$(x-1)(x+1)(x-2) =0$

3. Thanks a lot, didn't think about breaking up the first 2 and the 2nd 2

4. 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!)