# Zero, complex and rational roots

## Null roots

Any algebraic equation whose independent term is zero admits number zero as root whose multiplicity is equal to the smallest exponent of the unknown.

These roots are called null roots.

Examples: ## Complex roots

Let's solve the algebraic equation x² -2x + 2 = 0: It can be demonstrated that if a complex number whose imaginary part is not null is the root of an equation with real coefficients, its conjugate is also root of this equation.

Consequences:

• The number of complex roots of an algebraic equation of real coefficients is necessarily even;
• If an algebraic equation of real coefficients is of an odd degree, it will assume at least one real root.

## Rational roots

Given an algebraic equation of integer coefficients with and , if there are rational roots, they will be as follows. , with P and what cousins ​​among themselves, on what P is a divider of and what is divisor of .

For example, in the equation we have: • Not every number obtained is the root of the equation. After listing the rational root candidates we have to do the verification.
• This search for rational roots can only be done on integer coefficient equations.
• If = 1, the root candidates are the dividers of .
• If the sum of the coefficients of the equation is zero, the number 1 will be root of the equation.

Example 1

Solve the equation .

Resolution

Since the coefficient of the highest degree term is 1, the candidates for rational roots are the dividers of the independent term:

{-6, -3, -2, -1, 1, 2, 3, 6}

Let's check some of these values:

 P (-6) = 840 (- 6 not root) P (6) = 1260 (6 not root) P (1) = 0 → 1 it's root P (-1) = 0 → -1 it's root

Since we have a 4th degree equation and know two of its roots, applying the Briot-Ruffini device we get a 2nd degree equation: So the equation can be written as (x - 1) (x + 1) .Q (x) = 0, with Q (x) = x² + x - 6. The solutions of the equation are -1, 1 and the roots of Q (x): Therefore, the solution set of the equation é:

 S = {-3, -1, 1, 2}

Example 2

Solve the equation :

Resolution

Putting x in evidence, we have: So a root is 0 and the others are solutions of the equation .

Note that in all coefficients are integers. Since the coefficient of the highest degree term is 1, the rational root candidates are the dividers of the independent term:

{-3, -1, 1, 3}

Verifying:

 P (-3) = 0 → -3 it's root P (3) = 120 P (-1) = -8 P (1) = 0 → 1 it's root

We can write the equation this way: We already know that the quotient of per x é . Now let's split per (x + 3) and this quotient for (x - 1) to get Q (x): Q (x) = x² + 1 So the solution set of the equation é:

 S = {-3, 0.1, -1, i}
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