How do you determine the number of complex roots of a polynomial of 3 degrees?

1 Answer
Oct 30, 2015

Answer:

See explanation...

Explanation:

Suppose we are given #f(x) = ax^3+bx^2+cx+d# with #a > 0#

The slope of the tangent of the graph of #f(x)# at any #x# is given by the derivative:

#f'(x) = 3ax^2+2bx+c#

Using the quadratic formula, the cubic has turning points where #f'(x) = 0#, which is when

#x = (-2b+-sqrt(4b^2-12ac)) / (6a) = (-b+-sqrt(b^2-3ac))/(3a)#

If #b^2-3ac < 0# then #f'(x) = 0# only has Complex roots, so #f(x)# has no Real turning points and therefore #f(x) = 0# has exactly one Real root and two non-Real Complex roots.

If #b^2-3ac = 0# then #f'(x) = 0# has one repeated Real root, so #f(x)# has an inflection point, but is strictly monotonically increasing. So again #f(x) = 0# has exactly one Real root and two non-Real Complex roots, unless the inflection point is itself a root. If #f(-b/(3a)) = 0# then this is a threefold Real root and there are no non-Real Complex roots.

If #b^2-3ac > 0# then #f'(x) = 0# has two distinct Real roots:

#x_1 = (-b-sqrt(b^2-3ac))/(3a)#

#x_2 = (-b+sqrt(b^2-3ac))/(3a)#

We know that #f(x_2) < f(x_1)#, but there are still several possibilities:

(1) #f(x_1) < 0#, #f(x_2) < 0#

#f(x) = 0# has one Real root in #(x_2, oo)# and two non-Real Complex roots.

(2) #f(x_1) = 0#, #f(x_2) < 0#

#f(x) = 0# has a repeated Real root at #x = x_1#, and another Real root in #(x_2, oo)#.

(3) #f(x_1) > 0#, #f(x_2) < 0#

#f(x) = 0# has three distinct Real roots in #(-oo, x_1)#, #(x_1, x_2)# and #(x_2, oo)#.

(4) #f(x_1) > 0#, #f(x_2) = 0#

#f(x) = 0# has one Real root in #(-oo, x_1)# and a repeated Real root at #x = x_2#.

(5) #f(x_1) > 0#, #f(x_2) > 0#

#f(x) = 0# has one Real root in #(-oo, x_1)# and two non-Real Complex roots.