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Note You need to understand how to multiply algebraic expressions using the distributive law before starting work on this tutorial. If you feel you need to review this, go back to Part A: Multiplying Algebraic Expressions.
We can think of factoring as applying the distributive law in reverse. For example,
which can be checked by using the distributive law. The first technique of factoring is to locate a common factor  that is, a term that occurs as a factor in each of the expressions being added or subtracted. For example, x is a common factor in 2x^{2} + x, since it is a factor of both 2x^{2} and x. On the other hand, x^{2} is not a common factor, since it is not a factor of the second term, x.
Once we have located a common factor, we can "factor it out" by applying the distributive law.
1 The expression 2x^{3}  x^{2} + x has x as a common factor, so
2. 2x^{2} + 4x has 2x as a common factor, so
3. 2x^{2}y + xy^{2}  x^{2}y^{2} has xy as a common factor, so

Here are some for you.
We would also like to be able to reverse calculations such as (x + 2)(2x  5) = 2x^{2}  x  10. That is, starting with the expression 2x^{2}  x  10, we would like to factor it and get back the original expression (x + 2)(2x  5).
An expression of the form ax^{2} + bx + c, where a, b, and c are real numbers, is called a quadratic expression in x. Thus, given a quadratic expression ax^{2} + bx + c, we would like to write it in the form (dx + e)(fx + g) for some real numbers d, e, f, and g.
There are some quadratics, such as x^{2} + x + 1, that cannot be factored in this form at all. Here, we shall consider only quadratics that do factor, and in such a way that the numbers d, e, f and g are integers (whole numbers). (Other cases are fully discussed in Section 5.) The usual technique of factoring such quadratics is a "trialanderror" approach, which we illustrate by means of an example and some exercises for you.
Example Factor x^{2}  6x + 5. Solution Concentrate on the first and last terms.
5 has factors 5 and 1. Group them together and make an attempt.
This is fine, except for the sign of the middle term. But notice that we can also get the 5 by multiplying (5) and (1). In other words, 5 also has factors (5) and (1). Using these instead gives
so we have found the correct factorization. 
The next example is here to remind you why we should want to factor polynomials in the first place. We shall return to this in Section.5.
Example Solve the equation 3x^{2} + 4x  4 = 0. Solution We first factor the lefthand side, getting
Thus, the product of the two quantities (3x  2) and (x + 2) is zero. Now, if a product of two numbers is zero, it means that one or the other must be zero. In other words,
or x + 2 = 0, giving x = 2. Thus, there are two solutions: x = ^{2}/_{3}, and x = 2. 
Now go over the examples and try some of the exercises in Section A.3 of the Algebra Review in Applied Calculus and Finite Mathematics and Applied Calculus. Alternatively, go on to the next tutorial (Rational Expressions).