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

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Answer 1 · 2016-10-05T13:22:32

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

#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#

#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#