## Interactive Algebra Review

Based on the algebra reviews in Applied Calculus and Finite Mathematics and Applied Calculus

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## 3. Multiplying and Factoring Algebraic Expressions Part B: Factoring Algebraic Expressions

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,

2x2 + x = x(2x + 1),

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 2x2 + x, since it is a factor of both 2x2 and x. On the other hand, x2 is not a common factor, since it is not a factor of the second term, x.

 Q1 x2
is
is not
 a common factor in 3x2 - x4 + 4x3
Q2         x
is
is not
 a common factor in x3/2 - x5/2 - 12 x1/2
 Q3 x3
is
is not
 a common factor in 2x3 - 5x3
Q4         3x x
is
is not
a common factor in        6x2 - 3x2 x

Once we have located a common factor, we can "factor it out" by applying the distributive law.

 Examples of Factoring Algebraic Expressions 1  The expression 2x3 - x2 + x has x as a common factor, so 2x3 -x2 + x = x(2x2 - x + 1). 2.  2x2 + 4x has 2x as a common factor, so 2x2 + 4x = 2x(x + 2). 3.  2x2y + xy2 - x2y2 has xy as a common factor, so 2x2y + xy2 - x2y2 = xy(2x + y - xy).

Here are some for you.

Q1
 2x2 - 6x3 + 8x4 has 2x2 as a common factor, so
 2x2 - 6x3 + 8x4 =
Q2
 (x2+1)(x2+5) - (x2+1)(2x2) has x2+1 as a common factor, so
 (x2+1)(x2+5) - (x2+1)(2x2) =
Q3
 (x- 3)5(x + 2)6 + (x -3)6(x + 2)5 has (x -3)5(x + 2)5 as a common factor, so
 (x - 3)5(x + 2)6 + (x -3)6(x + 2)5 =
Q4
 x3/2 + x - x has x1/2 as a common factor, and so
 x3/2 + x - x
=
Q5
 6x5 - 4x3 + 2x2   can be factored as
Q6
 10x(x2+1)4(x3+1)5 + 15x2(x2+1)5(x3+1)4   can be factored as

We would also like to be able to reverse calculations such as (x + 2)(2x - 5) = 2x2 - x - 10. That is, starting with the expression 2x2 - x - 10, we would like to factor it and get back the original expression (x + 2)(2x - 5).

An expression of the form ax2 + bx + c, where a, b, and c are real numbers, is called a quadratic expression in x. Thus, given a quadratic expression ax2 + 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 x2 + 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 "trial-and-error" approach, which we illustrate by means of an example and some exercises for you.

 Factoring Quadratics by "Trial & Error" Example Factor x2 - 6x + 5. Solution Concentrate on the first and last terms. x2 has factors x and x           Since x.x = x2 5 has factors 5 and 1. Group them together and make an attempt. (x + 5)(x + 1) = x2 + 6x + 5 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 (x - 5)(x - 1) = x2 - 6x + 5, so we have found the correct factorization.

Factor each of the following.
 Q1 x2 + 6x + 8 = Q2 x2 - 4x - 12 = Q3 2x2 - 5x + 2 = Q4 2x2 + 4x - 6 = Q5 4x2 - 5x - 6 = Q6 4x2 - 25 = Q7 x4 - 5x2 + 6 =

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.

 Solving a Quadratic Equation by Factoring Example Solve the equation 3x2 + 4x - 4 = 0. Solution We first factor the left-hand side, getting (3x - 2)(x + 2) = 0. 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, Either 3x - 2 = 0, giving x = 2/3, or x + 2 = 0, giving x = -2. Thus, there are two solutions: x = 2/3, and x = -2.

Solve the following equations. If there are two or more solutions, separate them by commas (for example, if the solution is x = -1 or x = 2, enter -1, 2 (spaces optional).
 Q1 x2 + x - 12 = 0 Solution: Q2 2x2 - 3x - 2 = 0 Solution: Q3 6x2 + 13x + 6 = 0 Solution:

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).

Last Updated: March, 2006
Copyright © 2001 Stefan Waner and Steven R. Costenoble