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## 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:

**Comments:**

- 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
- 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} |