Question

What are the possible number of positive real, negative real, and complex zeros of
f(x) = 8x^4 + 13x^3 - 11x^2 + x + 9?

Answer 1

http://www.wolframalpha.com/input/?i=+8x%5E4+%2B+13x%5E3+-+11x%5E2+%2B+x+%2B+9

Answer 2

-2.17 and -0.71

Answer 3

@Jonask that's not what he's looking for. Use descartes rule of signs. Number of sign changes in the function gives the possible number of positive zeroes, then insert (-x) to find negative zeroes.

Answer 4

f(x) = 8x^4 + 13x^3 - 11x^2 + x + 9

changes sign from positive to negative twice, so two (or zero) possible positive zeroes.

f(-x) = 8(-x)^4 + 13(-x)^3 - 11(-x)^2 + (-x) + 9
becomes
f(-x) = 8x^4 - 13x^3 - 11x^2 - x + 9

Two changes in sign, so two (or zero) possible negative zeroes.

So I think there can be...

2 real pos, 2 real neg
2 real pos, 2 complex
2 real neg, 2 complex
4 complex

Answer 5

http://turner.faculty.swau.edu/mathematics/math110de/materials/descartes/

I only know this crap because I've been teaching it to a pre-calc student :P

Answer 6

As wolfram alpha showed, there are 2 real neg. and 2 complex zeroes.

Note: There has to be 4 total roots since it's a degree 4 polynomial. It can be any one of those combinations of zeroes:

2 real pos, 2 real neg
2 real pos, 2 complex
2 real neg, 2 complex
4 complex

Answer 7

Thanks for the help!!

Answer 8

No prob. The rule of signs is easy once you have an idea of how to use it :)