Find the number of complex roots, the possible number of real roots and the possible rational roots of the polynomial function

Question

Find the number of complex roots, the possible number of real roots and the possible rational roots of the polynomial function. Then, solve for all roots.

x^3 + 2x^2 + 3x + 6 = 0

Answer 1

A cubic polynomial always has three roots, With integer coefficients, the Conjugate Root Theorem tells us that imaginary and complex roots always come in conjugate pairs. Hence if there are no imaginary or complex roots, then there must be three real roots. If there are two imaginary or complex roots, then there can be only one real root. So when it comes to real roots, there will be either three or one.

To determine the number of possible rational roots, apply the Rational Root Theorem which states that the possible rational roots will be factors of the constant term (-6) over factors of the coefficient of the highest degree term (1).

Factors of -6 are ±1, ±2, ±3, ±6

Factors of 1 are ±1

List of possible rational roots: ±1/1, ±2/1, ±3/1, ±6/1 = ±1, ±2, ±3, ±6

So there are eight possible rational roots. As it turns out, only -2 is a rational root. Apply synthetic division then factor the quadratic quotient to find the other roots. It turns that the other two roots are imaginary numbers (±i√3).

Answer 2

Descartes rule of signs tells you that there are no positive real roots and a maximum of 3 real roots; of course, since the equation is a cubic there must be at least one real root, which will be negative.

Possible rational roots: -1, -2, -3, and -6.

Try them in order: -1 is not a root but -2 is; divide it out to get

x²+3 which has 2 imaginary roots, namely i√3 and -i√3.

To check your answer, graph the equation.