# Polynomial Roots

## Factor Theorem

**The polynomial P(x) is divisible by a polynomial of the form (x − a) if and only if P(x = a) = 0. **

The value** x = a** is called the ** root** or** zero** of P(x).

### Roots of a Polynomial

These are the values to nullify the polynomial.

#### Calculate the Roots of the Polynomial:

P(x) = x^{2} − 5x + 6

P(2) = 2^{2} − 5 · 2 + 6 = 4 − 10 + 6 = 0

P(3) = 3^{2} − 5 · 3 + 6 = 9 − 15 + 6 = 0

**x = 2 and x = 3 are roots or zeros of the polynomial**: P(x) = x^{2} − 5x + 6, because** P(2) = 0 and P(3) = 0. **

## Properties of the Roots and Factors of a Polynomial

1The **zeros or roots** are divisors of the independent term of the polynomial.

2For each **root** type **x = a** corresponds to it by a binomial of the type **(x − a)**.

3 A polynomial can be expressed in factors by writing it as a product of all the **binomials** of type **(x − a)**, which will correspond to the **roots**, **x = a**.

x^{2} − 5x + 6 = (x − 2) · (x − 3)

4The sum of the exponents of the binomial must be equal to the degree of the polynomial.

5All polynomials that do not have an independent term accept x = 0 as a root.

x^{2} + x = x · (x + 1)

Roots: x = 0, and x = − 1

6A polynomial is called irreducible or prime when it cannot be decomposed into factors.

P(x) = x^{2} + x + 1

#### Find the Roots and ** Factor** the Following Polynomial:

Q(x) = x^{2} + x + 1

The divisors of the independent term are: ±1, ±2, ±3.

Q(1) = 1^{2} − 1 − 6 ≠ 0

Q(−1) = (−1)^{2} − (−1) − 6 ≠ 0

Q(2) = 2^{2} − 2 − 6 ≠ 0

Q(−2) = (−2)^{2} − (−2) − 6 = 4 +2 − 6 = 0

Q(3) = 3^{2} − 3 − 6 = 9 − 3 − 6 = 0

The roots are: x = −2 and x = 3.

Q(x) = (x + 2) · (x − 3)