Raices enteras de un polinomio

Fundamental Theorem of Algebra and Finding Roots

Overview of the Fundamental Theorem

• The first part of the Fundamental Theorem of Algebra states that every polynomial p(x) has at least one real or complex root.

• The second part asserts that a polynomial of degree n will have exactly n roots.

Structure of Polynomials

• A polynomial p(x) = a_n x^n + a_n-1 x^n-1 + ... + a_1 x + a_0 , where coefficients can be real or complex, decreases in exponent from n to 0.

Integer Roots and Divisors

• If a polynomial has integer roots, these roots must be divisors of the constant term a_0 .

• This concept is formalized: if p(x)=a_0 , then integer roots are divisors of the independent term.

Finding Roots Example with Quadratic Polynomial

Evaluating Possible Roots

• For the quadratic polynomial, we need to find integer divisors of 12 (the constant term).

• Divisors include both positive and negative values: ±1, ±2, ±3, ±4, ±6, ±12.

Testing Potential Roots

• Testing x = -1 : yields 20; not a root.

• Testing x = 1 : results in -4; not a root.

Successful Root Findings

• Testing x = -3 : gives 42; not a root.

• Testing x = 3 : results in 0; thus, it is confirmed as a root.

Finalizing Roots for Quadratic Polynomial

Additional Root Discovery

• Testing x = 4 : also results in 0; confirming it as another root.

Expressing Polynomial Form

• The polynomial can now be expressed as p(x)= (x - 3)(x - 4).

Factoring Quadratic Equations

Factorization Process Explained

• To factor an equation like x^2 - 7x + 12 = 0, identify two numbers that multiply to give the constant term (12), while summing to give the linear coefficient (-7).

Exploring Cubic Polynomials

Finding Integer Roots for Cubic Polynomial

• For cubic equations like x^3 -5x^2 +2x +8, begin by identifying divisors of the constant term (8).

Evaluating Further Potential Roots

• Testing potential roots such as -1 reveals it is indeed a root since substituting yields zero.

Finding Roots of a Polynomial

Identifying Potential Roots

• The process begins by testing potential roots for the polynomial p(x) . Initially, substituting x = 1 yields p(1) = 6 , indicating that 1 is not a root.

• Next, testing x = -2 results in p(-2) = 24 , confirming it is also not a root.

Confirming Valid Roots

• When substituting x = 2 , the result is p(2) = 0 . This indicates that x = 2 is indeed a root of the polynomial.

• The polynomial can now be expressed as (x + 1)(x - 2)(m(x)) , where m(x) represents an additional factor yet to be determined.

Further Root Analysis

• Utilizing GeoGebra, further evaluations are conducted to find additional roots of the polynomial.

Discovering Additional Roots

• Testing with x = 4 , we find that p(4) = 0. Thus, it confirms that another root is at x = 4.

• The polynomial can now be fully expressed as:

• First root: x + 1

• Second root: x - 2

• Third root: x - 4

Summary of Found Roots

• The identified roots of this cubic polynomial are:

• First root: ** x = -1**

• Second root: ** x = 2**

• Third root: ** x = 4**