1.3 Algebraic Expressions. Copyright Cengage Learning. All rights reserved.

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1 1.3 Algebraic Expressions Copyright Cengage Learning. All rights reserved.

2 Objectives Adding and Subtracting Polynomials Multiplying Algebraic Expressions Special Product Formulas Factoring Common Factors 2

3 Objectives Factoring Trinomials Special Factoring Formulas Factoring by Grouping Terms 3

4 Algebraic Expressions A variable is a letter that can represent any number from a given set of numbers. If we start with variables, such as x, y, and z and some real numbers, and combine them using addition, subtraction, multiplication, division, powers, and roots, we obtain an algebraic expression. Here are some examples: 2x 2 + 3x + 4 A monomial is an expression of the form ax k, where a is a real number and k is a nonnegative integer. 4

5 Algebraic Expressions A binomial is a sum of two monomials and a trinomial is a sum of three monomials. In general, a sum of monomials is called a polynomial. For example, the first expression listed above is a polynomial, but the other two are not. 5

6 Algebraic Expressions Note that the degree of a polynomial is the highest power of the variable that appears in the polynomial. 6

7 Adding and Subtracting Polynomials 7

8 Adding and Subtracting Polynomials We add and subtract polynomials using the properties of real numbers. The idea is to combine like terms (that is, terms with the same variables raised to the same powers) using the Distributive Property. For instance, 5x 7 + 3x 7 = (5 + 3)x 7 = 8x 7 8

9 Adding and Subtracting Polynomials In subtracting polynomials, we have to remember that if a minus sign precedes an expression in parentheses, then the sign of every term within the parentheses is changed when we remove the parentheses: (b + c) = b c [This is simply a case of the Distributive Property, a(b + c) = ab + ac, with a = 1.] 9

10 Example 1 Adding and Subtracting Polynomials (a) Find the sum (x 3 6x 2 + 2x + 4) + (x 3 + 5x 2 7x). (b) Find the difference (x 3 6x 2 + 2x + 4) (x 3 + 5x 2 7x). Solution: (a) (x 3 6x 2 + 2x + 4) + (x 3 + 5x 2 7x) = (x 3 + x 3 ) + ( 6x 2 + 5x 2 ) + (2x 7x) + 4 Group like terms = 2x 3 x 2 5x + 4 Combine like terms 10

11 Example 1 Solution cont d (b) (x 3 6x 2 + 2x + 4) (x 3 + 5x 2 7x) = x 3 6x 2 + 2x + 4 x 3 5x 2 + 7x Distributive Property = (x 3 x 3 ) + ( 6x 2 5x 2 ) + (2x + 7x) + 4 Group like terms = 11x 2 + 9x + 4 Combine like terms 11

12 Multiplying Algebraic Expressions 12

13 Multiplying Algebraic Expressions To find the product of polynomials or other algebraic expressions, we need to use the Distributive Property repeatedly. In particular, using it two times on the product of two binomials, we get (a + b)(c + d) = a(c + d) + b(c + d) = ac + ad + bc + bd This says that we multiply the two factors by multiplying each term in one factor by each term in the other factor and adding these products. Schematically, we have (a + b)(c + d) = ac + ad + bc + bd 13

14 Multiplying Algebraic Expressions In general, we can multiply two algebraic expressions by using the Distributive Property and the Laws of Exponents. When we multiply trinomials or other polynomials with more terms, we use the Distributive Property. It is also helpful to arrange our work in table form. The next example illustrates both methods. 14

15 Example 3 Multiplying Polynomials Find the product: (2x + 3) (x 2 5x + 4) Solution 1: Using the Distributive Property (2x + 3)(x 2 5x + 4) = 2x(x 2 5x + 4) + 3(x 2 5x + 4) Distributive Property = (2x x 2 2x 5x + 2x 4) + (3 x 2 3 5x + 3 4) Distributive Property = (2x 3 10x 2 + 8x) + (3x 2 15x + 12) Laws of Exponents 15

16 Example 3 Solution = 2x 3 7x 2 7x + 12 cont d Combine like terms Solution 2: Using Table Form x 2 5x + 4 2x + 3 3x 2 15x x 3 10x 2 + 8x 2x 3 7x 2 7x + 12 Multiply x 2 5x + 4 by 3 Multiply x 2 5x + 4 by 2x Add like terms 16

17 Special Product Formulas 17

18 Special Product Formulas 18

19 Special Product Formulas The key idea in using these formulas (or any other formula in algebra) is the Principle of Substitution: We may substitute any algebraic expression for any letter in a formula. For example, to find(x 2 + y 3 ) 2 we use Product Formula 2, substituting x 2 for A and y 3 for B, to get 19

20 Example 4 Using the Special Product Formulas Use the Special Product Formulas to find each product. (a) (3x + 5) 2 (b) (x 2 2) 3 Solution: (a) Substituting A = 3x and B = 5 in Product Formula 2, we get (3x + 5) 2 = (3x) 2 + 2(3x)(5) = 9x x + 25 (b) Substituting A = x 2 and B = 2 in Product Formula 5, we get (x 2 2) 3 = (x 2 ) 3 3(x 2 ) 2 (2) + 3(x 2 )(2) = x 6 6x x

21 Factoring Common Factors 21

22 Factoring Common Factors We use the Distributive Property to expand algebraic expressions. We sometimes need to reverse this process (again using the Distributive Property) by factoring an expression as a product of simpler ones. For example, we can write x 2 4 = (x 2)(x + 2) We say that x 2 and x + 2 are factors of x 2 4. The easiest type of factoring occurs when the terms have a common factor. 22

23 Example 6 Solution cont d Factor each expression: (b) 8x 4 y 2 + 6x 3 y 3 2xy 4 ; The greatest common factor of the three terms in the polynomial is 2xy 2, and we have 8x 4 y 2 + 6x 3 y 3 2xy 4 = (2xy 2 )(4x 3 ) + (2xy 2 )(3x 2 y) + (2xy 2 )( y 2 ) = 2xy 2 (4x 3 + 3x 2 y y 2 ) (c) (2x + 4)(x 3) 5(x 3). x 3 is the common factor. So, (2x + 4)(x 3) 5(x 3) = [(2x + 4) 5](x 3) = (2x 1)(x 3) Distributive Property Simplify 23

24 Factoring Trinomials 24

25 Factoring Trinomials To factor a trinomial of the form x 2 + bx + c, we note that (x + r)(x + s) = x 2 + (r + s)x + rs so we need to choose numbers r and s so that r + s = b and rs = c. 25

26 Example 7 Factoring x 2 + bx + c by Trial and Error Factor: x 2 + 7x + 12 Solution: We need to find two integers whose product is 12 and whose sum is 7. By trial and error we find that the two integers are 3 and 4. Thus, the factorization is x 2 + 7x + 12 = (x + 3)(x + 4) 26

27 Factoring Trinomials To factor a trinomial of the form ax 2 + bx + c with a 1, we look for factors of the form px + r and qx + s: ax 2 + bx + c = (px + r)(qx + s) = pqx 2 + (ps + qr)x + rs Therefore, we try to find numbers p, q, r, and s such that pq = a, rs = c, ps + qr = b. If these numbers are all integers, then we will have a limited number of possibilities to try for p, q, r, and s. 27

28 Example 9 Recognizing the Form of an Expression Factor each expression. (a) x 2 2x 3 (b) (5a + 1) 2 2(5a + 1) 3 Solution: (a) x 2 2x 3 = (x 3)(x + 1) Trial and error (b) This expression is of the form where represents 5a + 1. This is the same form as the expression in part (a), so it will factor as 28

29 Example 9 Solution cont d = (5a 2)(5a + 2) 29

30 Special Factoring Formulas 30

31 Special Factoring Formulas Some special algebraic expressions can be factored using the following formulas. The first three are simply Special Product Formulas written backward. 31

32 Example 11 Factoring Differences and Sums of Cubes Factor each polynomial. (a) 27x 3 1 (b) x Solution: (a) Using the Difference of Cubes Formula with A = 3x and B = 1, we get 27x 3 1 = (3x) = (3x 1)[(3x) 2 + (3x)(1) ] = (3x 1) (9x 2 + 3x + 1) 32

33 Example 11 Solution cont d (b) Using the Sum of Cubes Formula with A = x 2 and B = 2, we have x = (x 2 ) = (x 2 + 2)(x 4 2x 2 + 4) 33

34 Special Factoring Formulas A trinomial is a perfect square if it is of the form A 2 + 2AB + B 2 or A 2 2AB + B 2 So, we recognize a perfect square if the middle term (2AB or 2AB) is plus or minus twice the product of the square roots of the outer two terms. 34

35 Example 12 Recognizing Perfect Squares Factor each trinomial. (a) x 2 + 6x + 9 (b) 4x 2 4xy + y 2 Solution: (a) Here A = x and B = 3, so 2AB = 2 x 3 = 6x. Since the middle term is 6x, the trinomial is a perfect square. By the Perfect Square Formula we have x 2 + 6x + 9 = (x + 3) 2 35

36 Example 12 Solution cont d (b) Here A = 2x and B = y, so 2AB = 2 2x y = 4xy. Since the middle term is 4xy, the trinomial is a perfect square. By the Perfect Square Formula we have 4x 2 4xy + y 2 = (2x y) 2 36

37 Special Factoring Formulas When we factor an expression, the result can sometimes be factored further. In general, we first factor out common factors, then inspect the result to see whether it can be factored by any of the other methods of this section. We repeat this process until we have factored the expression completely. 37

38 Example 13 Factoring an Expression Completely Factor each expression completely. (a) 2x 4 8x 2 (b) x 5 y 2 xy 6 Solution: (a) We first factor out the power of x with the smallest exponent. 2x 4 8x 2 = 2x 2 (x 2 4) = 2x 2 (x 2)(x + 2) Common factor is 2x 2 Factor x 2 4 as a difference of squares 38

39 Example 13 Solution cont d (b) We first factor out the powers of x and y with the smallest exponents. x 5 y 2 xy 6 = xy 2 (x 4 y 4 ) Common factor is xy 2 = xy 2 (x 2 + y 2 )(x 2 y 2 ) = xy 2 (x 2 + y 2 )(x + y)(x y) Factor x 4 y 4 as a difference of squares Factor x 2 y 2 as a difference of squares 39

40 Factoring by Grouping Terms 40

41 Factoring by Grouping Terms Polynomials with at least four terms can sometimes be factored by grouping terms. The following example illustrates the idea. 41

42 Example 15 Factoring by Grouping Factor each polynomial. (a) x 3 + x 2 + 4x + 4 (b) x 3 2x 2 3x + 6 Solution: (a) x 3 + x 2 + 4x + 4 = (x 3 + x 2 ) + (4x + 4) Group terms = x 2 (x + 1) + 4(x + 1) Factor out common factors = (x 2 + 4)(x + 1) Factor out x + 1 from each term 42

43 Example 15 Solution cont d (b) x 3 2x 2 3x + 6 = (x 3 2x 2 ) (3x 6) Group terms = x 2 (x 2) 3(x 2) = (x 2 3)(x 2) Factor out common factors Factor out x 2 from each term = (x V3)(x + V3)(x 2) 43