Arithmetic Operations. The real numbers have the following properties: In particular, putting a 1 in the Distributive Law, we get

Transcription

1 Review of Algebra

2 REVIEW OF ALGEBRA Review of Algebra Here we review the basic rules and procedures of algebra that you need to know in order to be successful in calculus. Arithmetic Operations The real numbers have the following properties: a b b a ab ba a b c a b c ab c ab ac abc abc (Commutative Law) (Associative Law) (Distributive law) In particular, putting a in the Distributive Law, we get b c b c b c and so b c b c EXAMPLE (a) (b) (c) 3y4 34 y y t7 t 4t 4t t If we use the Distributive Law three times, we get a bc d a bc a bd ac bc ad bd This says that we multiply two factors by multiplying each term in one factor by each term in the other factor and adding the products. Schematically, we have a bc d In the case where c a and d b, we have or a b a ba ab b a b a ab b Similarly, we obtain a b a ab b

3 REVIEW OF ALGEBRA 3 EXAMPLE (a) (b) (c) Fractions To add two fractions with the same denominator, we use the Distributive Law: Thus, it is true that a b c b b a b c a c a c b b a c b a b c b But remember to avoid the following common error: a b c a b a c (For instance, take a b c to see the error.) To add two fractions with different denominators, we use a common denominator: a b c ad bc d bd We multiply such fractions as follows: a b c ac d bd In particular, it is true that a b a b a b To divide two fractions, we invert and multiply: a b c d a b d c ad bc

4 4 REVIEW OF ALGEBRA EXAMPLE 3 3 (a) 3 3 (b) (c) (d) 3 y y s t u ut s t u u s t y y y y y y y y y y y y Factoring We have used the Distributive Law to epand certain algebraic epressions. We sometimes need to reverse this process (again using the Distributive Law) by factoring an epression as a product of simpler ones. The easiest situation occurs when the epression has a common factor as follows: Epanding 3(-)=3@-6 Factoring To factor a quadratic of the form b c we note that r s r s rs so we need to choose numbers r and s so that r s b and rs c. EXAMPLE 4 Factor 5 4. SOLUTION The two integers that add to give 5 and multiply to give 4 are 3 and 8. Therefore EXAMPLE 5 Factor 7 4. SOLUTION Even though the coefficient of is not, we can still look for factors of the form r and s, where rs 4. Eperimentation reveals that Some special quadratics can be factored by using Equations or (from right to left) or by using the formula for a difference of squares: 3 a b a ba b

5 REVIEW OF ALGEBRA 5 The analogous formula for a difference of cubes is 4 a 3 b 3 a ba ab b which you can verify by epanding the right side. For a sum of cubes we have 5 a 3 b 3 a ba ab b EXAMPLE 6 (a) (Equation ; a, b 3) (b) (Equation 3; a, b 5) (c) (Equation 5; a, b ) 6 EXAMPLE 7 Simplify. 8 SOLUTION Factoring numerator and denominator, we have To factor polynomials of degree 3 or more, we sometimes use the following fact. 6 The Factor Theorem If P is a polynomial and Pb 0, then b is a factor of P. EXAMPLE 8 Factor SOLUTION Let P If Pb 0, where b is an integer, then b is a factor of 4. Thus, the possibilities for b are,, 3, 4, 6, 8,, and 4. We find that P, P 30, P 0. By the Factor Theorem, is a factor. Instead of substituting further, we use long division as follows: Therefore Completing the Square Completing the square is a useful technique for graphing parabolas or integrating rational functions. Completing the square means rewriting a quadratic a b c

6 6 REVIEW OF ALGEBRA in the form a p q and can be accomplished by:. Factoring the number a from the terms involving.. Adding and subtracting the square of half the coefficient of. In general, we have a b c a b a c a b a a b a b a c 4a b a b c EXAMPLE 9 Rewrite by completing the square. SOLUTION The square of half the coefficient of is. Thus 4 4 ( ) 3 4 EXAMPLE Quadratic Formula By completing the square as above we can obtain the following formula for the roots of a quadratic equation. 7 The Quadratic Formula are The roots of the quadratic equation a b c 0 b sb 4ac a EXAMPLE Solve the equation SOLUTION With a 5, b 3, c 3, the quadratic formula gives the solutions 3 s s69 0 The quantity b 4ac that appears in the quadratic formula is called the discriminant. There are three possibilities:. If b 4ac 0, the equation has two real roots.. If b 4ac 0, the roots are equal. 3. If b 4ac 0, the equation has no real root. (The roots are comple.)

7 REVIEW OF ALGEBRA 7 These three cases correspond to the fact that the number of times the parabola y a b c crosses the -ais is,, or 0 (see Figure ). In case (3) the quadratic a b c can t be factored and is called irreducible. y y y FIGURE Possible graphs of y=a@+b+c (a) b@-4ac>0 (b) b@-4ac=0 (c) b@-4ac<0 EXAMPLE The quadratic is irreducible because its discriminant is negative: b 4ac Therefore, it is impossible to factor. The Binomial Theorem Recall the binomial epression from Equation : a b a ab b If we multiply both sides by a b and simplify, we get the binomial epansion 8 a b 3 a 3 3a b 3ab b 3 Repeating this procedure, we get a b 4 a 4 4a 3 b 6a b 4ab 3 b 4 In general, we have the following formula. 9 The Binomial Theorem If k is a positive integer, then a b k a k ka k b kk kk k 3 a k b a k3 b 3 kk k n 3 n a kn b n kab k b k

8 8 REVIEW OF ALGEBRA EXAMPLE 3 Epand 5. SOLUTION Using the Binomial Theorem with a, b, k 5, we have Radicals The most commonly occurring radicals are square roots. The symbol s means the positive square root of. Thus sa means a and 0 Since a 0, the symbol sa makes sense only when a 0. Here are two rules for working with square roots: 0 sab sa sb a sa b sb However, there is no similar rule for the square root of a sum. In fact, you should remember to avoid the following common error: sa b sa sb (For instance, take a 9 and b 6 to see the error.) EXAMPLE 4 s8 (a) s 8 s9 3 (b) s y s sy sy Notice that s because s indicates the positive square root. (See Appendi A.) In general, if n is a positive integer, s n a means n a If n is even, then a 0 and 0. Thus s 3 8 because 3 8, but s 4 8 and s 6 8 are not defined. The following rules are valid: a s n a s n ab s n a s n n b b s n b EXAMPLE 5 s 3 4 s 3 3 s 3 3 s 3 s 3

9 REVIEW OF ALGEBRA 9 To rationalize a numerator or denominator that contains an epression such as sa sb, we multiply both the numerator and the denominator by the conjugate radical sa sb. Then we can take advantage of the formula for a difference of squares: (sa sb)(sa sb) (sa) (sb) a b s 4 EXAMPLE 6 Rationalize the numerator in the epression. SOLUTION We multiply the numerator and the denominator by the conjugate radical s 4 : s 4 s 4 s s 4 (s 4 ) (s 4 ) s 4 Eponents Let a be any positive number and let n be a positive integer. Then, by definition,. a n a a a. 3. a 0 a n a n n factors 4. a n s n a a mn s n a m (s n a) m m is any integer Laws of Eponents Let a and b be positive numbers and let r and s be any rational numbers (that is, ratios of integers). Then rs. a r a s a rs. a abr a r b r 5. a r a s b a r a r b b 0 r a r s a rs In words, these five laws can be stated as follows:. To multiply two powers of the same number, we add the eponents.. To divide two powers of the same number, we subtract the eponents. 3. To raise a power to a new power, we multiply the eponents. 4. To raise a product to a power, we raise each factor to the power. 5. To raise a quotient to a power, we raise both numerator and denominator to the power.

10 0 REVIEW OF ALGEBRA EXAMPLE 7 (a) (b) (c) (d) y y 4 3 s4 3 s64 8 s z (e) y3 y y y y y yy y y z 4 y y y y y y Alternative solution: 7 y 5 z 4 y y y y 4 3 (s4) Eercises A Click here for answers. 6 Epand and simplify.. 6ab0.5ac. yy a t 4 t tt y 4 6 y5 y t 5 t 38t Perform the indicated operations and simplify. 8 9b b u u. u y z 5. r s 6. s 6t a 3 ab 4 b yz a bc b ac c c 9 48 Factor the epression ab 8abc t t 9s 4. 4t t Simplify the epression

11 REVIEW OF ALGEBRA Complete the square Solve the equation Which of the quadratics are irreducible? Use the Binomial Theorem to epand the epression. 73. a b a b Simplify the radicals. s 3 s s3 s s 3 54 s sy s 3 y 8. s6a 4 b 3 8. s 5 96a 6 s 5 3a Use the Laws of Eponents to rewrite and simplify the epression a n a n a 3 b 4 y 88. a 5 b 5 y y y 3 z s 5 y (sa) s (st) 5 s 4 3 t sst 4 s Rationalize the epression. s s s5 07. s s s 09 6 State whether or not the equation is true for all values of the variable. 09. s 0. 6 a. a y y a 6 4 4a a n s 4 r n s 4 r (s) s h s h h s sy s 4 y y 4

12 ANSWERS Answers. 3a bc. 3 y a t t y 4 y 5 y t 56t b u 3u u. b 3ab 4a z rs a b yz y 3t 6. a c b c ab5 8c t t t 40. t 3st 3s 4. t ( 5 ) ( 3 ) , 0 6., 4 9 s s s33 s5 67., 4 68., s 69. Irreducible 70. Not irreducible 7. Not irreducible (two real roots) 7. Irreducible a 6 6a 5 b 5a 4 b 0a 3 b 3 5a b 4 6ab 5 b 6 a 7 7a 6 b a 5 b 35a 4 b 3 35a 3 b 4 a b 5 7ab 6 b y 8. 4a bsb 8. a a n3 87. a y b y s s s 3 y 6 3 y y 95 z 6 t r n s 4 s s 8 3 s s sy y a 34 t 5 s h s h s s 3 4 s s 09. False 0. False. True. False 3. False 4. False 5. False 6. True 56 5 s3 6 8