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1 Copyright © Cengage Learning. All rights reserved.
Precalculus Review Copyright © Cengage Learning. All rights reserved.

2 Copyright © Cengage Learning. All rights reserved.
0.3 Multiplying and Factoring Algebraic Expressions Copyright © Cengage Learning. All rights reserved.

3 Multiplying Algebraic Expressions

4 Multiplying Algebraic Expressions
Distributive Law The distributive law for real numbers states that a(b ± c) = ab ± ac (a ± b)c = ac ± bc for any real numbers a, b, and c. Quick Example 2(x – 3) is not equal to 2x – 3 but is equal to x – 2(3) = 2x – 6.

5 Multiplying Algebraic Expressions
There is a quicker way of expanding expressions, called the “FOIL” method (First, Outer, Inner, Last). Consider, for instance, the expression (x + 1)(x – 2). The FOIL method says: Take the product of the first terms: x  x = x2, the product of the outer terms: x  (–2) = –2x, the product of the inner terms: 1  x = x, and the product of the last terms: 1  (–2) = –2, and then add them all up, getting x2 – 2x + x – 2 = x2 – x – 2.

6 Example 1 – FOIL a. (x – 2)(2x + 5) = 2x2 + 5x – 4x – 10
b. (x2 + 1)(x – 4) = x3 – 4x2 + x – 4 c. (a – b)(a + b) = a2 + ab – ab – b2 = a2 – b2

7 Example 1 – FOIL d. (a + b)2 = (a + b)(a + b) = a2 + ab + ab + b2
cont’d d. (a + b)2 = (a + b)(a + b) = a2 + ab + ab + b2 = a2 + 2ab + b2 e. (a – b)2 = (a – b)(a – b) = a2 – ab – ab + b2 = a2 – 2ab + b2

8 Multiplying Algebraic Expressions
Special Formulas (a – b)(a + b) = a2 – b2 (a + b)2 = a2 + 2ab + b2 (a – b)2 = a2 – 2ab + b2 Quick Examples 1. (2 – x)(2 + x) = 4 – x2 2. (1 + a)(1 – a) = 1 – a2 3. (x + 3)2 = x2 + 6x + 9 4. (4 – x)2 = 16 – 8x + x2 Difference of two squares Square of a sum Square of a difference

9 Factoring Algebraic Expressions

10 Factoring Algebraic Expressions
Factoring Using a Common Factor To use this technique, locate a common factor—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, because it is a factor of both 2x2 and x). Once you have located a common factor, “factor it out” by applying the distributive law. Quick Examples 2x3 – x2 + x has x as a common factor, so 2x3 – x2 + x = x(2x2 – x + 1)

11 Factoring Algebraic Expressions
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 to get the 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)(f x + g) for some real numbers d, e, f, and g.

12 Factoring Algebraic Expressions
There are some quadratics, such as x2 + x + 1, that cannot be factored in this form at all. Here, we consider only quadratics that do factor, and in such a way that the numbers d, e, f, and g are integers. The usual technique of factoring such quadratics is a “trial and error” approach.

13 Factoring Algebraic Expressions
Factoring Quadratics by Trial and Error To factor the quadratic ax2 + bx + c, factor ax2 as (a1x)(a2x) (with a1 positive) and c as c1c2, and then check whether or not ax2 + bx + c = (a1x ± c1)(a2x ± c2). If not, try other factorizations of ax2 and c. Quick Example To factor x2 – 6x + 5, first factor x2 as (x)(x), and 5 as(5)(1): (x + 5)(x + 1) = x2 + 6x + 5. (x – 5)(x – 1) = x2 – 6x + 5. No good Desired factorization

14 Example 3 – Factoring Quadratics
Factor the following: a. 4x2 – 5x – 6 b. x4 – 5x2 + 6 Solution: a. Possible factorizations of 4x2 are (2x)(2x) or (x)(4x). Possible factorizations of –6 are (1)(–6), (2)(–3).

15 Example 3 – Solution cont’d We now systematically try out all the possibilities until we come up with the correct one. (2x)(2x) and (1)(–6): (2x + 1)(2x – 6) = 4x2 – 10x – 6 (2x)(2x) and (2)(–3): (2x + 2)(2x – 3) = 4x2 – 2x – 6 (x)(4x) and (1)(–6): (x + 1)(4x – 6) = 4x2 – 2x – 6 (x)(4x) and (2)(–3): (x + 2)(4x – 3) = 4x2 + 5x – 6 Change signs: (x – 2)(4x + 3) = 4x2 – 5x – 6 No good No good No good Almost! Correct

16 Example 3 – Solution cont’d b. The expression x4 – 5x2 + 6 is not a quadratic, you say? Correct. It’s a quartic (a fourth degree expression). However, it looks rather like a quadratic. In fact, it is quadratic in x2, meaning that it is (x2)2 – 5(x2) + 6 = y2 – 5y + 6 where y = x2.

17 Example 3 – Solution The quadratic y2 – 5y + 6 factors as
cont’d The quadratic y2 – 5y + 6 factors as y2 – 5y + 6 = (y – 3)(y – 2) so x4 – 5x2 + 6 = (x2 – 3)(x2 – 2) This is a sometimes useful technique.


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