Algebra 2! enter the number of complex zeros for the polynomial function in the box. f(x)=x^5+4x^3−5x?

2 complex roots

Step-by-step explanation:

The function f(x)=x^5+4x^3−5x can be factored as follows:

f(x)=x(x^4+4x^2−5).  One root, a real root, is zero.

That leaves g(x) = x^4+4x^2−5.  Substitute p = x^2, obtaining p^2 + 4p - 5 = 0.  This factors as follows:  (p+5)(p-1) = 0.  Thus, p = -5 and p = 1.

Recalling that p = x^2, we have -5 = x^2 and +1 = x^2.  The latter yields x = 1 and x = -1.  The former yields +i√5 and =i√5.  

Thus, the given poly has 3 real zeros:  -1, 1 and 0.  Due to the imaginary roots shown above, this means that this poly has 2 complex roots.

thats not a system of equations, that is a solution