What are the possible number of positive real, negative real, and complex zeros of
f(x) = –7x4 – 12x3 + 9x2 – 17x + 3?

Question

What are the possible number of positive real, negative real, and complex zeros of 
f(x) = –7x4 – 12x3 + 9x2 – 17x + 3?

anonymous · Answer

The fundamental theorem of algebra says for any polynomial:
$$f(x)=a_1x^n+a_2x^{n-1}+...a_{n-1}x^{1}+a_nx^0$$
There are exactly n roots.
So you have:
$$f(x)=-7x^4-12x^3+9x^2-17x+3$$
Since its a fourth order it has 4 roots.
Now, for each one individually, you would have to factor it.
It happens to factor as:
$$(x-0.19023) (x+2.59494) (x-(0.345212+0.86546 i)) (x-(0.345212-0.86546 i))$$
So it has 2 complex roots (if it has one, then the complex conjugate is also one)
And one negative real. One positive real.