After using the rule of the difference of cubes, how many x's can yo.:Resolved:.

[voidflux]

Hello everyone, I had a question, I need to solve for all the x values, to find the x-intercepts.
If i had an equation like the following:
(x^4-x) = 0;
so i factor out an x
x(x^3 - 1) = 0;
now i have to use the differences of cubes formula:
x^3-a^3 = (x-a)(x^2+ax+a^2);
so u would have
x(x-1)(x^2 + x + 1);
so u have
x = 0, x = 1, but what do u do about the x^2+x+1?
I tried the quadratic equation and i get a - in the square root so what does this mean?
Thanks for listening,

[NotLKH]

i is imaginary, but
i² = -1

-Lou

[TheAlchemist]

hey guys,
voidflux.

Since the original equation is a quartic there will be 4 roots or values of x. You have already found 2 but these are real and imply that the graph of the function crosses the x axis at these points. The other 2 appear to be complex as there is a negative quantity under the square root sogn.
This is where what NotLKH has said comes in. By using the imaginary number i such that i^2 = -1 we can find the other roots(imaginary) so there is nothing worng with what you have done so far.
solving x^2 + x + 1 = 0 we get x = [-1 +/- sqrt(-3)] /2

now sqrt(-3) can be written as sqrt(3)i. therefor the four roots of the equation are:
x = 0 ,1, [-1 + sqrt(3)i]/2, [-1 - sqrt(3)i]/2

[voidflux]

Thanks everyone for your help!! All these questions were going to be on a Gateway exam and I got a 50/50 on it so now I can enter calc...I have no idea why they even made us take this test considering I took calc in highschool and it didn't invovle hardly any of the stuff the test was on! Anyways! thanks again!