A.2 Working with Expressions 1061 Now subtract: (x + 5)(2x + 3) (2x)(2x + 3) − x 2 (2x)(2x + 3) = (x + 5)(2x + 3) − x 2 (2x)(2x + 3) . We can multiply out the numerator (top) in hopes that we can add and then factor, but there is no reason to multiply out the denominator. (Note that after addition/subtraction the numerator is not in factored form.) (x + 5)(2x + 3) − x 2 (2x)(2x + 3) = 2x 2 + 13x + 15 − x 2 (2x)(2x + 3) = x 2 + 13x + 15 (2x)(2x + 3) The numerator does not factor, so this cannot be simplified. ◆ Advice: One common pitfall for a novice is the loss of direction. Keep sight of your goal. Just because a certain operation is “legal” doesn’t mean you benefit from doing it; make sure that what you are doing is leading you in the direction of your goal. For example, many times a student will correctly determine the LCD, express each term using the LCD, and then, after multiplying out the numerator of each term, proceed to simplify each term. If you do this accurately, then you are right back at the beginning of the problem, feeling a little dizzy from having completed a tight circle It is important to have a strategy. Break the problem solving process into a sequence of manageable steps and keep track of where you are in your game plan. Another fairly common pitfall occurs when you’re working with an expression once you’ve written something along the lines of x x+1 = (x)(x+2) (x+1)(x+2) . You are still working with an expression, so the rules for expressions hold. You can’t, for instance, simply decide to wipe out the denominator by “multiplying both sides by (x + 1)(x + 2)” because you really don’t have two “sides.” Clarifying your goal and game plan will help you avoid these mistakes. Division and Complex Fractions Arithmetic Underpinnings To divide a quantity by 2 is to take half of it. Q ÷ 2 = Q · 1 2 = Q 2 To divide a quantity by 10 is to take a tenth of it. Q ÷ 10 = Q · 1 10 = Q 10 To divide a quantity by 2 3 is Q ÷ 2 3 or Q 2/3 . We can multiply this expression by 1 in the form 3 2 3 2 in order to eliminate the 2 3 from the denominator: Q 2 3 = Q · 3 2 2 3 · 3 2 = Q · 3 2 6 6 = Q · 3 2 . 1062 APPENDIX A Algebra So, to divide by 2 3 is to multiply by 3 2 . To divide Q by R is to multiply Q by 1 R . To divide Q by r s is to multiply Q by s r . Q r s · s r s r = Q · s r 1 = Q · s r ◆ EXAMPLE A.11 Simplify. (a) 1/3 2/9 (b) 1 3 +1 1 3 + 1 2 SOLUTION (a) 1 3 2 9 = 1 3 · 9 2 = 3 2 (b) 1 3 + 1 1 3 + 1 2 = 1 3 + 3 3 2 6 + 3 6 = 4 3 5 6 = 4 3 · 6 5 = 8 5 ◆ Extracting the Fundamentals A B C D = A B · D C Why? A B C D = A B C D · D C D C = A B · D C 1 = A B · D C . Algebra ◆ EXAMPLE A.12 Simplify 1 x −x 1− 1 x 2 . SOLUTION Get the complex fraction into the form A B C D . You can work on each part, 1 x − x and 1 − 1 x 2 , individually or work both at once. We’ve done the latter. 1 x − x 1 − 1 x 2 = 1 x − x 2 x x 2 x 2 − 1 x 2 = 1−x 2 x x 2 −1 x 2 = 1 − x 2 x · x 2 x 2 − 1 A.2 Working with Expressions 1063 = (1 − x)(1 +x) x · x 2 (x − 1)(x + 1) = (1 − x)x x − 1 = (−1)(x − 1)x (x − 1) =−x. ◆ Talking the Talk—A Bit of Terminology Most of our examples have dealt with polynomials or quotients of polynomials. Polynomials are sums of products of constants and nonnegative integer powers of variables. A polynomial can be written in the form a 0 + a 1 x + a 2 x 2 + a 3 x 3 +···+a n x n , where a 0 , a 1 , a 2 , ,a n are constants. Since the powers to which the variable is raised are nonnegative integers (0,1,2, ),thevariable is never in the denominator or under a radical. 4x 3 − 3x +8 is a polynomial, as are 4 11 x 3 − √ 3x + 8 √ 5 and abx √ c 2 +1 for a, b, and c constants. −3 √ x + 5 is not a polynomial. The degree of a polynomial is the highest power to which x is raised. A polynomial of degree 2 is called a quadratic. We call a k the coefficient of the term x k . If a polynomial is of degree n, a n is called the leading coefficient. In the term 4 11 x 3 we call 4 11 the coefficient of the term. 3 is the exponent, or power, of x. Integers, Rationals, Irrationals, and Real Numbers Real Numbers: R = all x such that x corresponds to a point on the number line Integers: Z = { ,−3, −2, −1, 0, 1, 2, 3, } Positive Integers: Z + = { 1, 2, 3, 4, } Rational Numbers: Q, all numbers that can be written in the form a b , where a and b are integers, b = 0. Equivalently, the rationals are all numbers that either repeat a pattern infinitely or terminate when written in decimal notation. For example: 13 20 = 0.65 1 3 = 0.333 =0. ¯ 3 2 7 = 0.285714285714 Irrational Numbers: Real numbers that cannot be written in the form a b . Decimals that are nonterminating and nonrepeating are irrational numbers, and conversely. There are infinitely 1064 APPENDIX A Algebra many irrational (and rational) numbers between any two rational numbers. Examples of irrational numbers are √ 2, √ 10, and π. EXERCISE A.2 Let f(x)= x x+1 and let g(x) = 1 x 2 −1 . Find the following. If your answer involves the sum or difference of fractions, add the fractions; do not leave complex fractions in your answers. (a) g(x) −f(x) (b) f(x) g(x) (c) f(x+1) x+1 (d) x f(x) + 1 g(x) (e) Find f g(x) . (f) For what values of x is f g(x) =1? Answers: (a) −x 2 +x+1 (x+1)(x−1) (b) x(x −1) (c) 1 x+2 (d) x 2 + x (e) 1 x 2 (f) x =±1 Factoring Why factor? From what we have already seen, factoring permits us to simplify expressions. We can find least common denominators only if we can factor. We will also see that factoring helps us solve equations. What is factoring? To factor an expression is to write it as a product. (x + 3)(x − 1), x(−1)(+7)(x − 5), and 8 x (x + 4)(x) are all factored. (x 2 + 1)(x 2 − 1) is factored, but it can be further factored into (x 2 + 1)(x − 1)(x + 1). (x + 1)(x + 2) + 2isnot factored, nor is (x + 1)(x + 2) + 3(x + 4). (x + 3)(x − 1) factored form = x 2 + 2x − 3 multiplied out ◆ EXAMPLE A.13 Multiply out. (x + 3) 2 = (x + 3)(x + 3) = x 2 + 6x + 9 Note: A common error is to write x 2 + 9 as an answer and forget about the 6x. Perhaps the picture in Figure A.1 can steer you away from this. 3 x 3 x 3 3 3 3 3 3 xxx x x x (x + 3) 2 = x 2 + 2(3x) + 9 Figure A.1 ◆ A.2 Working with Expressions 1065 ◆ EXAMPLE A.14 Multiply out. (x − 2)(x + 2) = x 2 + 2x − 2x − 4 = x 2 − 4 ◆ Factoring and expanding, or multiplying out, are complementary procedures. We discussed multiplying out and the distributive law in Section A.2. Below we give a couple of more involved illustrations before we take up factoring. ◆ EXAMPLE A.15 Expand the following products. (a) (x 3 − 3x 2 + 1)(−x + 2x 3 ) (b) (x + 3)(x − 1)(2x − 1) SOLUTION (a) (x 3 − 3x 2 + 1)(−x + 2x 3 ) = x 3 (−x + 2x 3 ) − 3x 2 (−x + 2x 3 ) + 1(−x + 2x 3 ) =−x 4 +2x 6 +3x 3 −6x 5 −x+2x 3 We’ve multiplied; now we get organized. = 2x 6 − 6x 5 − x 4 + 3x 3 + 2x 3 − x We add like terms, = 2x 6 − 6x 5 − x 4 + 5x 3 − x. (b) (x + 3)(x − 1)(2x − 1). Multiply any two of these and then multiply the result by the third. (x + 3)(x − 1)(2x − 1) = [(x + 3)(x − 1)](2x + 1) = [x 2 + 2x − 3](2x + 1) = 2x 3 + 4x 2 − 6x + x 2 + 2x − 3 = 2x 3 + 5x 2 − 4x − 3. ◆ Factoring an expression is to write it as a product. It is the reverse of multiplying out. You can always check your factoring by multiplying out. Factoring Out a Common Factor: The Distributive Law in Reverse ◆ EXAMPLE A.16 Factor the following as much as possible. (a) (λw) 2 + wλ 2 + λw 2 (λw) 2 + wλ 2 + λw 2 = λ 2 w 2 + wλ 2 + λw 2 = λw(λw + λ + w) (b) (λ + C) 3 − (λ + C) 2 + 4(λ + C) 7 (λ + C) 3 − (λ + C) 2 + 4(λ + C) 7 = (λ + C) 2 (λ + C) − 1 + 4(λ + C) 5 (c) (A + B)x 2 + 2(A + B)x = (A + B)(x)(x + 2) This final expression is factored completely. 3 (d) 2x(x + 1) + 3(x + 1) = (2x + 3)(x + 1) (e) 2x(x − 1) + 3(1 − x) = 2x(x − 1) − 3(x − 1) = (2x − 3)(x − 1) Notice that 1 − x could be rewritten as −(x − 1) so the common factor of x − 1 could be extracted. ◆ 3 An expression is said to be factored completely if it is factored and each factor cannot be further factored. 1066 APPENDIX A Algebra Factoring Quadratics of the Form x 2 + bx + c We want to express x 2 + bx + c in the form (x + )(x +),where and are constants. (x + )(x +)=x 2 + x +x+ =x 2 +( +)x + . This has to be equal to x 2 + bx + c, so we must find two numbers whose product is c and sum is b. Replace by one of them and by the other. ◆ EXAMPLE A.17 Factor the expressions below. (a) x 2 − x − 2 (b) x 2 − 14x + 24 (c) x 2 − 10x + 24 (d) x 2 − 9x + 24 SOLUTION (a) Look for two numbers whose product is −2 and whose sum is −1. Answer: −2 and 1. x 2 − x − 2 = (x + )(x + ) = (x − 2)(x + 1) (b) Look for two numbers whose product is 24 and whose sum is −14. If they are going to add to −14, they will both be negative. Our options are −1, −24, or −2, −12, or −3, −8, or −4, −6. Answer: −12 and −2. x 2 − 14x + 24 = (x + )(x + ) = (x − 12)(x − 2) (c) Look for two numbers whose product is 24 and whose sum is −10. Answer: −4 and −6. x 2 − 10x + 24 = (x + )(x + ) = (x − 4)(x − 6) (d) Look for two numbers whose product is 24 and whose sum is −9. Our options are as in part , but we can’t get a sum of 9. This expression can’t be factored; it’s irreducible. ◆ Factoring Quadratics of the Form ax 2 + bx + c Option 1: Guess and check: ax 2 + bx + c = (ex + f )(gx + h) = egx + (f g + eh)x + fh. Tohave equality we must find e, f , g, and h such that a = ef , c = fh,and b = fg + eh. This approach isn’t as formidable as it looks provided that a and c have few factors. ◆ EXAMPLE A.18 2x 2 + 5x − 3 = ( x + )( x + ) A.2 Working with Expressions 1067 SOLUTION We know that we must have (2x + )(x + ) and the only question is whether to use −1and3or1and−3and where to place these numbers. 4 If b is to be 5, we’ll want to multiply the 2 by 3, so we have 2x 2 + 5x − 3 = (2x − 1)(x + 3). Check this by multiplying out. ◆ Try the next exercise on your own. EXERCISE A.3 Factor. (a) 3x 2 − 14x − 5 (b) 3x 2 − 2x − 5 ANSWER (a) We know that we must have (3x + )(x + ). 3x 2 − 14x − 5 = (3x + 1)(x − 5) (b) 3x 2 − 2x − 5 = (3x − 5)(x + 1) Option 2: A systematic method for factoring ax 2 + bx + c: Look for two numbers whose product is ac and sum is b. If you can find such numbers, you can factor. Otherwise, you can’t. We’ll work an example to show you what to do with these two numbers and then describe the general procedure. ◆ EXAMPLE A.19 Factor 2x 2 + 5x − 3. SOLUTION Look for two numbers whose product is −6 and sum is 5. Answer: 6 and −1. These numbers add to 5, so we can use them to split the 5x term. 2x 2 + 5x − 3 = 2x 2 + 6x − x − 3 Now we’ll factor the first two terms and the last two terms independently. We will be able to arrange this so that the resulting two terms have a common factor. 2x 2 + 6x − x − 3 = 2x(x + 3) − (x + 3) Lo and behold, we can factor out a common factor of (x + 3). 2x(x + 3) − (x + 3) = (x + 3)(2x − 1) You might wonder what would happen if we split 5x into −x + 6x instead of 6x − x. No problem. 2x 2 + 5x − 3 = 2x 2 − x + 6x − 3 = x(2x − 1) + 3(2x − 1). We can factor out the common factor of (2x − 1). 2x 2 + 5x − 3 = (2x − 1)(x + 3), as before. ◆ 4 Starting with (−2x + )(−x + ) is equivalent (factor −1 out of both factors) so we don’t deal with it separately. 1068 APPENDIX A Algebra General Procedure Given ax 2 + bx + c: 1. Look for two numbers whose product is ac and sum is b. Let’s suppose q and r work. 2. Using these two numbers, split the bx term. Then group the first two terms together and the last two terms together. (ax 2 + qx) + (rx + c) 3. Factor each expression in parentheses so that a factor of the second group looks the same as a factor of the first group. 4. You will now have an expression out of which you can factor a common factor. ◆ EXAMPLE A.20 Factor. (a) −6x 2 + 19x − 15 (b) 6x 2 + 5x − 6 SOLUTION (a) Look for two numbers that multiply to 90 and add to 19. Answers: 10 and 9. −6x 2 + 19x − 15 =−6x 2 +10x + 9x − 15 = (−6x 2 + 10x) + (9x − 15) = 2x(−3x + 5) − 3(−3x + 5) Note: We factored out −3 instead of 3 from the second group in order to make it match a factor from the first group. = (−3x + 5)(2x − 3) (b) Look for two numbers that multiply to −36 and add to 5. Answers: 9 and −4. 6x 2 + 5x − 6 = 6x 2 + 9x − 4x − 6 = 3x(2x + 3) − 2(2x + 3) = (2x + 3)(3x − 2) ◆ Factoring the Difference of Two Perfect Squares (A 2 − B 2 ) = (A − B)(A + B) Be on the lookout for the difference of perfect squares that might be “in disguise.” An expression like A 2 − B 2 should jump out at you, screaming to be factored, but there can be less obvious examples, as illustrated below. ◆ EXAMPLE A.21 Factor. (a) 9x 2 − 4z 6 (b) 8x 8 − 18(x − y) 2 (c) x 4 − 1 SOLUTION (a) 9x 2 − 4z 6 = (3x) 2 − (2z 3 ) 2 = (3x − 2z 3 )(3x + 2z 3 ) (b) 8x 8 − 18(x − y) 2 = 2 4x 8 − 9(x − y) 2 = 2 (2x 4 ) 2 − 3(x − y) 2 = 2[2x 4 − 3(x − y)][2x 4 + 3(x − y)] = 2(2x 4 − 3x + 3y)(2x 4 + 3x − 3y) A.2 Working with Expressions 1069 (c) x 4 − 1 = (x 2 − 1)(x 2 + 1) = (x − 1)(x + 1)(x 2 + 1) The sum of two perfect squares cannot be factored. ◆ Variations on the Theme Once you’ve learned about factoring out common factors and factoring quadratics (including the difference of two perfect squares), you can in practice do a bit more. Let’s take the example 2x 2 − x − 1 = (2x + 1)(x − 1) and see how it can be dressed up a little. ◆ EXAMPLE A.22 Factor 2x 6 − x 4 − x 2 completely. SOLUTION First factor out any factors common to all terms. 2x 6 − x 4 − x 2 = x 2 (2x 4 − x 2 − 1) Now look for a quadratic in masquerade. x 2 2(x 2 ) 2 − (x 2 ) − 1 The expression in brackets is of the form 2u 2 − u − 1, where u = x 2 . 2u 2 − u − 1 = (2u + 1)(u − 1). Therefore we have x 2 (2x 2 + 1)(x 2 − 1), but this is still not factored completely. 2x 6 − x 4 − x 2 = x 2 (2x 2 + 1)(x + 1)(x − 1) ◆ ◆ EXAMPLE A.23 Simplify 2x 6 −x 4 −x 2 x−2x 2 x−x 3 −2x 2 −x+1 . SOLUTION We’ll use the result of Example A.22 to factor the numerator. 2x 6 −x 4 −x 2 x−2x 2 x−x 3 −2x 2 −x+1 = x 2 (2x 2 + 1)(x + 1)(x − 1) x(1 − 2x) · (x + 1)(−2x + 1) x(1 − x)(1 + x) = (2x 2 + 1)(x − 1)(x + 1) 1-2x · (-2x+1) 1 − x = (2x 2 + 1)(x − 1)(x + 1) · −1 −1 + x = (2x 2 + 1)(−x − 1) ◆ We have by no means exhausted the topic of factoring—one can learn to factor the difference of two perfect cubes (and also the sum) and more, but for now we’re mod- erately well equipped to go on. As we discuss solving equations, we’ll touch again on factoring. 1070 APPENDIX A Algebra PROBLEMS FOR SECTION A.2 1. Compute the following sums and simplify your answer. (a) 1 x 2 + 3−x x + x 3+x (b) 1 x+w − 1 w w (c) 2 (y+z) 2 − 2 y 2 · 1 y 2. Simplify. (a) 2x 2 −8 x −x−2 (b) y+ 1 xy x y +x (c) 1 1 x + 1 y (d) ( √ x+ √ y) 2 2xy − √ xyx −1 y (e) x 2 y 3 −xy 4 y 3 x 2 −x 3 y 2 3. Factor as much as possible. (a) x 2 − x − 6 (b) 2x 2 − x − 3 (c) 6x 3 + 6x 2 − x (d) 16x 2 y 4 − 1 (e) (x − 1)xy + 3(1 − x) (f) x 4 − 3x 2 − 10 (g) x 4 + 3x 2 − 4x 3 4. Factor as much as possible. (a) b w+2 − b w (b) b 2w − b w (c) x 3 b x+2 − x 3 b x (d) b 2x − 4 (e) b 2w − b w − 6 5. Simplify as much as possible 3x 2 −x−2 9x 2 −4 x 4 −9 x 4 +x 2 . A.3 SOLVING EQUATIONS The strategy one adopts for solving an equation depends upon what type of equation one has. In particular, to solve an equation for x one has to determine what type of equation it is in x. . have by no means exhausted the topic of factoring—one can learn to factor the difference of two perfect cubes (and also the sum) and more, but for now we’re mod- erately well equipped to go on common factor of x − 1 could be extracted. ◆ 3 An expression is said to be factored completely if it is factored and each factor cannot be further factored. 1066 APPENDIX A Algebra Factoring Quadratics. expressions. We can find least common denominators only if we can factor. We will also see that factoring helps us solve equations. What is factoring? To factor an expression is to write it as a