Question

How many complex roots does \[6{x^4} + {x^3} - 5 = 0\] have?

Answer 1

Hint: Let us consider the above polynomial as function \[f(x) = 6{x^4} + {x^3} - 5\] and then use Descartes’ rule of signs to get the number of complex roots. Also keep in mind that the occurrence of complex roots is always in pairs.

Complete answer:
Let us take the function \[f(x) = 6{x^4} + {x^3} - 5\]and use Descartes’ rule of signs for getting the complex roots.
Let us check the change of signs in the function \[f(x) = 6{x^4} + {x^3} - 5\]
That is, \[f(x)\]has \[ + \] \[ + \] \[ - \] as its signs, then the change of sign are once and so the function \[f(x)\]has a single positive root.
For\[x{\text{ }} \to {\text{ }}-x\],\[f( - x) = 6{( - x)^4} + {( - x)^3} - 5\]\[ = 6{x^4} - {x^3} - 5\]
Now we check the changes of sign in \[f( - x)\]that is, \[ + \] \[ - \] \[ - \]
We notice that the function changes its sign once and hence there is single negative root.
As we see that the function is a fourth degree equation and that complex roots occur in pairs.
Therefore, there are two complex roots of the function \[f(x) = 6{x^4} + {x^3} - 5\].

Additional information:
In mathematics, Descartes' rule of signs, first described by René Descartes in his work La Géométrie, is a technique for getting information on the number of positive real roots of a polynomial. It states that the count of positive roots is at most to the number of sign it changes in the sequence of polynomial's coefficients leaving the zero coefficients, and implies the difference between these two numbers is said to be even. It also states, in particular that if the count of sign changes is zero or one, then there are exactly zero or one positive roots, respectively. One may use Descartes' rule of signs for getting similar information on the number of roots in any interval. Descartes used the transformation \[x{\text{ }} \to {\text{ }}-x\] for using his rule for getting information of the number of negative roots.

Note:
We should keep in mind that graphing the function may be a first approach for a student but the Descartes’ rule of signs gives us an easier solution. Some may also try to approach with taking the derivative but then may find difficulty while searching the intervals.