How do you find the number of complex zeros for the function f(x)=3x^3+2x^2-x?

1 Answer
Oct 5, 2016

x_1=0; x_2=-1; x_3=1/3

Explanation:

f(x) is a polynomial with degree 3, therefore we have 3 solutions real or complex.

To simplify the computation we could factorize f(x)

f(x)=3x^3+2x^2-x=x(3x^2+2x-1)

We have to find the zeros, then:

x(3x^2+2x-1)=0

A product is zero when the factors are zero:

x=0=>x_1=0

3x^2+2x-1=0

Using the The Quadratic Formula:

x_(2,3)=(-b+-sqrt(b^2-4ac))/(2a)

with: a=3; b=2; c=-1

:.x_(2,3)=(-2+-sqrt(4+12))/6=(-2+-sqrt(16))/6=(-2+-4)/6

=>x_2=-cancel(6)^1/cancel(6)_1=-1
=>x_3=cancel(2)^1/cancel(6)_3=1/3