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

##### 1 Answer

#### Answer:

See explanation...

#### Explanation:

Suppose we are given

The slope of the tangent of the graph of

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

Using the quadratic formula, the cubic has turning points where

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

If

If

If

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

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

We know that

**(1)**

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

**(2)**

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

**(3)**

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

**(4)**

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

**(5)**

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