What is the number of complex zeros for the polynomial function?

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Answer 1 · 2017-11-23T21:01:23

#f(x)# has two complex zeros.

I'm not sure what methods you are familiar with, so I will present a couple...

Given:

#f(x) = x^3-96x+400#

Note that a cubic (with real coefficients) will always have at least one real zero and will have #0# or #2# non-real complex zeros.

Discriminant

The discriminant #Delta# of a cubic polynomial in the form #ax^3+bx^2+cx+d# is given by the formula:

#Delta = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd#

In our example, #a=1#, #b=0#, #c=-96# and #d=400#, so we find:

#Delta = 0+3538944+0-4320000+0 = -781056#

Since #Delta < 0# this cubic has #1# Real zero and #2# non-Real Complex zeros, which are Complex conjugates of one another.

Turning points

Differentiating, we find:

#f'(x) = 3x^2-96 = 3(x^2-32) = 3(x-4sqrt(2))(x+4sqrt(2))#

So #f(x)# has local maximum at #x = -4sqrt(2)# and local minimum at #x = 4sqrt(2)#.

Putting #x=4sqrt(2)# into the equation of #f(x)#, we find:

#f(4sqrt(2)) = (4sqrt(2))^3-96(4sqrt(2))+400 = 128sqrt(2)-384sqrt(2)+400 = 400-256sqrt(2) ~~ 37.96 > 0#

Since this local minimum is positive, #f(x)# has only one real zero.