one way would be to factor
I can't factor it so we will have to use Descartes' Rule of Signs which is helpful for finding how many real roots you have
it goes like this:
for a polynomial with real coefients, consider [tex]f(x)=-3x^6-2x^4+5x+6[/tex].
after arranging the terms in decending order in terms of degree, count how many times the signs of the coeffients change direction and minus 2 from that number until you get to 1 or 0. that will be the number of even roots the function can have
We have (-, -, +, +). the signs changed 1 times, so it has 1 real positive root
to get the negative roots, we evaluate f(-x) and see how many times the root changes
[tex]f(-x)=-3x^6-2x^4-5x+6[/tex]
signs are (-, -, -, +). there was 1 change in sign
so the function has 1 real negative root
a total of 2 real roots
a function of degree [tex]n[/tex] can have at most, [tex] n[/tex] roots
our function is degree 6 so it has 6 roots
if 2 are real, then the others must be complex
6-2=4 so there are 4 complex roots
you can also show that there are only 2 real roots by using a graphing utility to see that there are only 2 real roots
