INDEX
Numbers and Sets
Complex Numbers
Exponentiation
Algebraic Equations and Inequalities
Functions
Discrete Algebra
Basic Formulas
Graphics of Basic Functions
Algebraic Transformations
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Algebraic Expressions

Basic Definitions

Polynomials

Algebraic Transformations

Outline

Factoring

Factoring Quadratic Polynomials

Factoring Cubic Polynomials

Factor Theorems

Other Transformations

Expanding

Completing Perfect Square

Rationalizing Denominators


Factoring
Key Topics Remaining:   Factoring Quadratic Polynomials » Factoring Cubic Polynomials » Factor Theorens » Expanding » Completing Perfect Square » Rationalizing Denominators

To factor a mathematical quantity means to express it as a product of two or more quantities.

Let us recall that the process of factoring a natural number involves expressing the number as a product of prime numbers, each of which has only two factors, the unity and the prime number itself.

Similarly, any polynomial can be represented by a product of irreducible polynomials, that is, polynomials such that cannot be further reduced to other factors aside from the unity and the polynomial itself.
Transformations of expressions by factoring are always correct, whatever values the symbolic variables in the expressions may have.

Let us consider a few simple examples to illustrate a practical importance of factoring in mathematics.

  1. The procedure of factoring often gives more simple expressions. For instance, the following factored expression,

    (x + y2) (2 x – y)3(x2 + 3 y)4,

    is a polynomial involving 36 terms.



  2. Some fractions can be simplified by factoring the numerator or denominator (or both). Consider, for example, the following fraction,

    .

    By direct multiplication and combining similar terms, one can easily verify that

    .

    Therefore,

    .



  3. Many equations can be easily solved by factoring. Consider, for example, the following polynomial equation of the third order,

    x3 – 2 x2 – 5 x + 6 = 0.

    Like above, one can verify that

    (x + 2) (x – 1) (x – 3) = x3 – 2 x 2 – 5 x + 6.

    Therefore,

    (x + 2) (x – 1) (x – 3) = 0,

    which gives the following solution set:

    x1 = –2,   x2 = 1,   and   x3 =3.

The technique of factoring will be discussed in the following sections.


Examples
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