© : Pre-Calculus - Chapter 1A Chapter 1A - Real Numbers Properties of Real Numbers Real numbers are used in almost every human endeavor. Whenever we need to quantify objects, we use numbers. Cooking recipes, prices, interest rates, blood pressure, height, age, voltage, and wind velocity are a few of the everyday objects that are quantified by real numbers. As we know, the two operations of addition and multiplication are defined for real numbers. In other words, for any two real numbers a and b, the sum a b and the product a b are uniquely defined real numbers. Two special real numbers are zero ( 0 ) and one ( 1 ). These operations satisfy: Properties of real numbers: Commutative: ab ba ab ba Example: 7 3 3 7 10 5 6 6 5 30 Associative: a b c a b c abc abc Example: 3 4 7 3 4 7 2 5 3 2 5 3 3 11 7 7 14 2 15 10 3 30 Identity: a0 0a a a1 1a a Example: 80 08 8 11 1 1 11 11 For each real number a, there is a real number, denoted by −a, called the negative of a, for which Inverse: a −a 0 −a a Example: 7 −7 0 Subtraction, denoted by a − b, is defined as follows: a − b a −b −1 For each a ≠ 0, there is a real number, denoted by 1 a or 1/a or a , called the reciprocal of a, for which 1 Inverse: a 1 a 1 a a Example: 7 1 1 7 1 7 7 Division, denoted by a b or a or a/b, where b ≠ 0, is defined as follows: b a b a a/b a 1b ab −1 b Finally, there is a property which relates addition and multiplication: Distributive: ab c ab ac Example: −4 3 2 −4 3 −4 2 −4 5 −12 −8 −20 © : Pre-Calculus a bc ac bc © : Pre-Calculus - Chapter 1A Types of Real Numbers Integers The real numbers are classified into several categories. The positive integers, also called counting numbers, are 1, 2, 3, 4, … The negative integers, … , −4, −3, −2, −1 are the negatives (or additive inverses) of the positive integers. An integer is either a positive integer, a negative integer, or zero. Rationals. The ratio a of any two integers a and b, where b ≠ 0, is called a rational number. Common fractions b are rationals. If b 1 then the number is also an integer, so integers are also rational numbers. Examples: 4 , 121 , − 16 , −7 5 31 54 Irrationals. Real numbers that are not rational are called irrational. Examples of irrational numbers include , 2 , and so on. One distinction between rational numbers and irrational numbers is that rational numbers have repeating or terminating decimal expansions, whereas irrational numbers do not. Summary The diagram below shows the relationship among these types of numbers, all of which are a part of the set of real numbers: That is, whole numbers are a subset of the integers, which are a subset of the rational numbers. The irrational numbers are disjoint from all of these and the real numbers are comprised of the union of the set of irrational numbers and rational numbers. Real Numbers Irrational Numbers l Numb ers tiona Ra ge Inte rs Whole Numbers © : Pre-Calculus © : Pre-Calculus - Chapter 1A Number Lines and Absolute Value A number line is a method of picturing the set of real numbers. Each point on the number line corresponds to exactly one real number, as in the picture below: P -5 -4 -3 -3/2 O -1 0 -2 3/2 5/2 π 1 2 3 Q 4 5 The absolute value of a number x, denoted by |x|, refers to the distance from that number to the origin, or zero (point O in the picture above). Note that, since distance is always positive or zero, the absolute value of a number will always be positive or zero. A more precise definition is as follows: if x is positive or zero, then |x| will be the same as x, but if x is negative, we must change the sign to positive, indicated by −x. Therefore, |x| x if x ≥ 0 −x if x 0 For instance, |−7| 7 |10| 10 |0| 0 |3 − 5| |−2| 2. Example: Evaluate |6 − 9|. Solution: |6 − 9| |−3| 3. Notice that in this example, we did not simply change the subtraction to addition. The absolute value bars act as parentheses, so what is inside must be evaluated first. Therefore, in general, it is incorrect to say that |x − 3| |x| 3, or |x − 3| x 3!! Also notice that |6 − 9| gave us 3, the distance between the numbers 6 and 9 on the number line. In general, |a − b| gives us the distance between the numbers a and b on the number line. Example: What is the distance between the numbers 2 and −24? Solution: The distance between any two numbers is the absolute value of their difference. Thus, distance between 2 and −24 equals |2 − −24| |2 24| |26| 26. © : Pre-Calculus © : Pre-Calculus - Chapter 1A Exercises for Chapter 1A - Real Numbers For problems 1-6, determine which property of the real numbers is used. 1. −6x 6x 0 2. 17 x x 17 3. 28 x − 7 28 x − 7 4. 14 x − 3 x − 3 14 5. 13x 1x 13 6. x 3 − y −y x 3 For problems 7-10, simplify and name the properties used. 7. x 7 − x 8. 4x − 3 12 9. 3x 4 4x 10. x 7x − 3 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. © For problems 11-20, evaluate the given expression. |−18 − 3| |7 − 14| −|15 − 4| −2 − 4 − |−6| −334 − 25 − |10| −48 −37 48 38 999 |−24 −24| 75 −49 − |74| 21 − 5 1 12 4 3 2 2 2 2 −51 5 3 8 1 1 1 1 2 2 2 3 3 Choose , , or : −|2| ? |−2| Describe the following in absolute-value notation: The distance between x and 3 is at most 2. Describe in words: |y 2| a (Hint: y 2 y − −2 ) Order the following list of numbers from smallest to largest: |3 − 5|, |−3 − 5|, |−3 5|, − |3 − −5|, − |3 − 5|, |− |−3 − 5|| : Pre-Calculus © : Pre-Calculus - Chapter 1A Answers to Exercises for Chapter 1A - Real Numbers 1. 2. 3. 4. 5. 6. 7. Additive Inverse Commutative Property of Addition Associative Property of Addition Commutative Property of Multiplication Multiplicative Inverse and Multiplicative Identity (Since 13x 1x 131 13 ) Commutative Property of Addition (remember that x 3 − y x 3 −y ) x 7 − x Commutative Property 7 x − x Associative Property 7 x − x Additive Inverse Property 7 0 Identity for Addition 7 NOTE: While you can simply say that x 7 − x 7, the properties of real numbers help explain why this simplifying is allowed. 8. 4x − 3 12 Distributive Property 4x − 12 12 Associative Property 4x −12 12 Additive Inverse* 4x 0 Identity for Addition 4x *-remember that 4x − 12 4x −12 9. 3x 4 4x Commutative Property-last 2 terms 3x 4x 4 Distributive Property (in reverse) 3 4x 4 7x 4 Again you could simplify immediately: 3x 4 4x 7x 4. The properties allow you to see why this works. 10. Later we will learn another method (FOIL) for doing this. It is based on the distributive property as shown below: x 7x − 3 x 7x −3 x 7x x 7−3 x 2 7x x 7−3 x 2 7x − 3x − 21 then simplify as #9 x 2 4x − 21 11. 21 © : Pre-Calculus © : Pre-Calculus - Chapter 1A 12. 7 13. −11 14. − 2 − 4 − |−6| (work parentheses from the inside out) −2 − 4 − 6 −2 − −2 −2 2 −4 15. − 334 − 25 − |10| (do multiplication before subtraction) −334 − 10 − 10 −314 −42 16. − 48 −37 48 38 999 ( Here it may be best to rearrange the terms.) −48 48 38 −37 999 0 1 999 1000 Note that the numbers may be added in any order (associative and commutative properties), but the above is the most expediant way. 17. |−24 −24| 75 −49 − |74| (Again rearranging terms may help.) |−48| 75 −49 − 74 48 −49 75 − 74 18. 19. 20. 21. 22. 23. 24. © −1 1 0 2 1 − 5 1 Find a common denominator 12 3 4 3 5 2 − 4 Write the mixed number as an improper fraction 12 12 12 27 − 9 18 3 or 1 1 12 12 12 2 2 2 2 2 2 − 5 1 Start by writing improper fractions: 5 3 8 8 − 41 Common denominator inside absolute value 12 5 3 8 64 12 − 123 5 24 24 59 12 cross cancel the 12 and the 24 5 24 59 or 5 9 or 5. 9 10 10 1 Work inside the parentheses first 1 1 1 2 2 2 3 3 1 3 2 3 4 6 6 6 6 1 5 7 6 6 1 36 35 35 36 −|2| |−2| Since −|2| −2 and |−2| 2 |x − 3| ≤ 2 y is more than a units from −2 − |3 − −5| −8 − |3 − 5| −2 |3 − 5| |−3 5| 2 : Pre-Calculus © : Pre-Calculus - Chapter 1B Chapter 1B - Exponents and Radicals Properties of Exponents Exponents can be used to represent repeated multiplication of a single factor. More formally, if n represents a positive integer, the expression x n means multiply x times itself n times. If n 0, then we define x n x 0 1, and for n 0 we define xn x x x x n times Examples: 4 3 4 4 4 64 2 −15 −15 −15 225 Note that −15 2 −15 2 −225 since exponents are computed before multipliying (unless there are parentheses as in the previous example) y3 y y y What does 5 4 equal? Answer: 5 4 5 5 5 5 625 We define x to a negative power as follows (we assume that x ≠ 0) x −1 1x x −2 12 x −n x 1n x Examples: 1 1 4 −3 64 43 1 −5 2 5 1 32 2 The following list of properties of exponentiation should be memorized, as you need to be able to manipulate expressions involving exponents. Property 1 2 3 4 5 6 7 Example a m a n a mn a m a m−n an a −n 1n a 2 3 2 4 2 34 2 7 x 4 x 4−2 x 2 x2 3 −4 14 1 81 3 0 2 0 a 1 for a ≠ 0 14 − x 1 iff 14 − x 2 ≠ 0 ab n a n b n a n an b bn 2y 3 2 3 y 3 8y 3 5 2 5 2 25 x x2 x2 a m n a mn x 3 −4 x 3−4 x −12 112 x Although these properties work for any real exponents m and n, the properties are easy to see when m © : Pre-Calculus © : Pre-Calculus - Chapter 1B and n are integers. "Proof" of Example 1(from previous page): 23 24 2 2 2 2 2 2 2 2222222 27 We can use these properties to simplify expressions with exponents in them. Example 1: Simplify the expression −z 3 3z 4 . Solution: −z 3 3z 4 −1 3 z 3 3 4 z 4 −181z 34 −81z 7 Example 2: Simplify the expression x −3 y 4 5y −1 −3 x −3 y 4 5y −1 −3 . Solution: x 9 y −12 5 −3 y 3 9 3 x3 512 y y 9 125x 15 y Example 3: Simplify the expression 15 4 5 −6 . Solution: 15 4 5 −6 3 5 4 5 −6 3 4 5 4 5 −6 3 4 5 4 5 −6 3 4 5 4−6 One could have stopped at 3 4 5 −2 3 4 5 −2 4 32 5 and not bothered about rewriting 5 −2 as 1/5 2 . Example 4: Simplify the expression 5 3 4 −2 10 2 7 3 . Solution: Before starting, scan the expression to find common terms. Here we note that 4 2 2, and 10 5 2. This means that we should think to combine the common integer factors of the terms 5 3 , 10 2 , and 4 −2 . © : Pre-Calculus © : Pre-Calculus - Chapter 1B 5 3 4 −2 10 2 7 3 5 3 2 2 −2 5 2 2 7 3 5 3 2 −4 5 2 2 2 7 3 5 32 2 2−4 7 3 5 5 2 −2 7 3 5 3 5 72 . 2 5 3 We could have stopped at the line 5 5 2 −2 7 3 , and not rewritten this as 5 72 . 2 If we have a complicated expression involving integers raised to a power, then the easiest first step would be to write the integers as a product of their prime factors. Remember that a positive integer (greater than 1) is said to be prime if its only integer divisors are itself and 1. Thus, 2, 3, 7, 23, and 31 are prime, while 4, 9, 28, and 32 are not. Example 5: The following example is fairly complicated. If you can follow all of the steps and understand the reasoning, you are well on your way to understanding and being able to use the exponential properties. 198 3 700 −4 Simplify the expression 5 4 6 −3 Solution: As we noted at the end of Example 4, when simplifying a complicated expression involving integers, factor each integer into its prime factors. 2 3 2 11 3 2 2 5 2 7 −4 198 3 700 −4 5 4 6 −3 5 4 2 3 −3 2 3 3 6 11 3 2 −8 5 −8 7 −4 5 4 2 −3 3 −3 2 3 3 6 11 3 2 −8 5 −8 7 −4 5 −4 2 3 3 3 2 3−83 3 63 5 −8−4 7 −4 11 3 2 −2 3 9 5 −12 7 −4 11 3 9 3 32 11 12 4 2 5 7 © : Pre-Calculus © : Pre-Calculus - Chapter 1B Radicals and Properties of Radicals Radicals (or roots) are, in effect, the opposite of exponents. In other words, the n th root of a number a is a number b such that b n a a 1/n b n a The number b is called an n th root of a. The number n is referred to as the index of the radical (if no index appears, n is understood to be 2). The principal n th root of a number is the n th root of a which has the same sign as a. For example both 2 and −2 satisfy x 2 4, but 2 is the (principal) square root of 4. Examples: 3 27 3 since 3 3 27 4 4 16 2 since 2 4 16 (Note −2 16 also, but 2 is the principal 4 th root 3 3 −64 −4 since −4 −64 4 −81 is not a real number and we will say that it does not exist. (In this course we won’t learn how to take an eventh power of a negative number.) Radicals are used to define rational exponents: 1 an m a n n n a am The notation a 1/n is extremely useful, and we encourage you to use it whenever you have to simplify expressions involving radicals. Examples: 1 125 3 3 125 5 2 −64 3 32 − 5 3 −64 2 1 3 32 5 3 2 −64 −4 2 16 1 13 1 3 8 2 5 32 3 Since radicals are nothing more than rational exponents, many of the properties of exponents also apply to radicals. Property Example 1 n am n a m 2 n a nb ab n 3 n 4 m a b n a n n a ,b ≠ 0 b mn 5 32 3 5 32 27 3 3 a 27 8 4 2x 3 8 3 23 8 81 9 27 3 3 2 8 2x 5a If n is odd n a n a 3 −127 3 −127 5b If n is even n a n |a| 4 −127 4 |−127| 127 The following list is a restatement of these properties, but in exponential notation. You need to be © : Pre-Calculus © : Pre-Calculus - Chapter 1B familiar with both radical and exponential notation, and be able to convert between the two. Property Example 3 32 3 1/5 32 1/5 2 3 8 a m/n 1 a 2 a 1/n b 1/n ab 1/n 3 a 1/n b 1/n 4 a 1/m m 1/n 1/n a b 1/n ,b ≠ 0 a 1/mn 5a If n is odd a n 1/n a 5b If n is even a n 1/n |a| 27 1/2 3 1/2 27 3 1/2 81 1/2 9 27 8 1/3 2x 1/4 1/3 271/3 3 2 8 1/2 2x 1/8 −127 3 1/3 −127 −127 4 1/4 |−127| 127 Examples: © −3 2 3 (refer to Property 5b) 3 16 3/2 16 3/2 16 1/2 4 3 64 (refer to property 1-given the right hand side) 3 1 3 3 3/2 (refer to property 1) −16 −16 −16 2 −16 2 There is no answer as we cannot take the square root of −16. 32 1/5 27 1/3 32 1/5 27 1/3 2 3 6 8 x 8 |x| : Pre-Calculus © : Pre-Calculus - Chapter 1B Simplifying Radicals Properties 2 and 3 in the previous section can be used to simplify radical expressions. A radical expression is simplified when the following conditions hold: 1. All possible factors (”perfect roots”) have been removed from the radical. 2. The index of the radical is as small as possible. Remember, the index of 3 10 is 3. 3. No radicals appear in the denominator. For example, to simplify the radical 40 , we factor the number into prime factors. Since the index is 2, any square factors are ”pulled out” of the radical: 40 23 5 22 2 5 2 2 2 5 (Using Property 2) 2 10 , or 2 10 . Notes for the above example: 1 1 1 Recall that Property 2 states that n ab n a n b or, in exponential notation ab n a n b n Since 2 3 2 2 2 , the perfect square is pulled out of the radical, while the remaining 2 stays inside the radical. The previous example could also be done directly by finding a factor of 40 which is a perfect square: 40 4 10 4 10 (by Property 2 again-see previous note) 2 10 This method works quite well for square roots. Example1: Simplify 72x 3 Solution: 72x 3 3 2 2 2 2 x 2 x 3 2 2 2 x 2 2x 3 2 |x| 2 x 6|x| 2x Example 2: Simplify Solution: 3 3 54xy 4 54xy 4 3 2 33 x y4 3y 32xy 3y 3 2xy © : Pre-Calculus © : Pre-Calculus - Chapter 1B Example 3: Simplify 4 80x 2 y4 Solution: 4 80x 2 y4 4 24 5 x2 y4 2 4 5x 2 (since 4 y 4 |y| ) |y| Finally, we can use the fact that radicals can be written as fractional exponents to be sure the radical is in ”lowest terms”. Example 4: Simplify 4 9 4 Solution: 9 4 32 2 34 1 32 3 Example 5: Simplify 6 81x 4 6 Solution: 81x 4 6 3 4 x 4 4 4 36x6 2 2 33x3 3 32x2 3 9x 2 These last two examples demonstrated the utility of exponential notation. © : Pre-Calculus © : Pre-Calculus - Chapter 1B Rationalizing Denominators Rationalizing the Denominator is a technique used whenever a radical appears in the denominator of an expression. To begin with, make sure the radical is simplified; that is, perfect roots are pulled out. The next step depends on how the radical appears in the denominator: I. The Radical is a single term Multiply both the numerator and the denominator of the expression by something which will produce a perfect root in the denominator. Example 1: Rationalize the denominator: 8 3 54 Solution: First simplify the radical in the denominator: 8 8 8 (Recall that 3 3 3 3) 3 3 3 54 332 23 Note that if the 2 was a cube, we would no longer need a radical in the denominator. To rationalize this denominator, we multiply both the numerator and denominator by 3 2 2 as follows: 8 3 22 834 434 22 32 3 3 22 3 3 23 3 8 332 Question: Answer: Why do we pick 3 2 2 ? We want to multiply the denominator of 8 by a term which will enable us to remove 332 any radicals. Thus, we need to figure out what to multiply 3 2 by. Well, if we write this in exponential notation, we have 3 2 2 1/3 . Clearly if we multiply by 2 2/3 we get something nice. 2 1/3 2 2/3 2 1 2 . Moreover 2 2/3 3 22 . 5 10 Example 2: Rationalize the denominator: Solution: 5 10 10 5 10 10 10 10 2 Example 3: Rationalize the denominator: 3 Solution: 1 2 3 5x 5x 3 5x 3 3 3 3 5x 5x 3 1 5x 2 (note that you only need one more 5x ) 5x 5x II. The Denominator is a Sum of Terms In this example the denominator has the form a b m (here we only concern ourselves with square roots). The conjugate of this expression is a − b m . Similarly, the conjugate of the expression a b is a − b . In the example on the next page, notice what happens when we multiply the © : Pre-Calculus © : Pre-Calculus - Chapter 1B numerator and denominator by the conjugate: 5 5− 7 5− 7 5 5 7 5− 7 5 7 5− 7 25 − 5 7 52 − 2 7 25 − 5 7 . 18 Here the multiplication in the denominator is done using a distributive property technique called FOIL (shown below). You can also use the special product a ba − b a 2 − b 2 which we will discuss in detail in another section. 5 7 5− 7 5 5 5 − 7 First 5 7 Outer Inner 52 − 5 7 5 7 − 52 − 2 7 7 7 − 7 Last 2 Note how the inner terms cancel. 25 − 7 18 Question: What should you multiply 5 − 3 by to rationalize it? Answer: Multiply 5 − 3 by 5 3 . 5− 3 5 3 25 − 9 16. Remember a − ba b a 2 − b 2 . Question: Answer: What should you multiply a b c by to rationalize it? Multiply a b c by a − b c . a b c a − b c a 2 − b 2 c a 2 − b 2 |c| . Example 4: Solution: 2x 5− 3 2x 5 3 Rationalize the denominator: 5 3 2x 5 3 5− 3 52 − 3 2x 5 3 22 x 5 3 11 © : Pre-Calculus 2 2 © : Pre-Calculus - Chapter 1B Example 5: 4 5 6 Rationalize the denominator: Solution: 4 5 6 © 5 − 6 5 − 6 : Pre-Calculus 4 5 5 − 6 2 − 6 2 4 5 − 6 5−6 −4 5 − 6 4 6 −4 5 © : Pre-Calculus - Chapter 1B Combining Radical Expressions Radical expressions can only be combined by addition or subtraction if there are like terms-terms with the same index and same radicand (the expression inside the radical). 2 2 3 4 2 −4 3 , In the above expression, the terms with 2 are alike and the terms with 3 are alike. Therefore, the above expressions simplifies as follows: 2 2 3 4 2 −4 3 2 4 2 2 3 −4 3 1 4 2 2 − 4 3 5 2 −2 3 Notice that the two remaining terms are not alike and, hence, cannot be simplified. Some terms which do not look alike at first glance may be alike after simplifying. Therefore, it is important that you simplify all radicals before combining like terms. Examples: 50 − 32 2 5 2 2 − 2 5 2 5 2 −4 2 2 5 − 4 1 2 2 2 3 3 7 80x − 2 270x 4 3 10 7 3 2 4 5x − 2 3 3 3 2 5x 4 3 10 7 2 3 2 5x − 2 3 3 2 5x 4 3 10 14 3 10x − 6 3 10x 4 3 10 8 3 10x 4 3 10 (note that these terms are NOT alike even though they both have a 10) 1 − 1 , we begin by using Property 3 ( n a n a ), then simplifying and In simplifying 2 8 n b b rationalizing the denominator before combining: 1 − 1 1 − 1 (Note that 8 4 2) 8 2 2 8 2 1 − 1 (Multiply both by ) 2 2 2 2 1 2 1 2 − 2 2 2 2 2 2 2 − (now get a common denominator) 2 4 2 2 2 − 4 4 2 2 − 2 (now combine radicals) 4 2 4 © : Pre-Calculus © : Pre-Calculus - Chapter 1B Exercises for Chapter 1B - Exponents and Radicals For problems 1-3, evaluate the expression. −3 4 5 −3 2 −2 3. 5 1. 2. 4. 5. 6. For problems 4-6, simplify the expression. −3ab 4 4ab −3 5a 3 a 2 −1 16a 10 0 −2 15x −2 y 5x 3 y −4 For problems 7-13, compute the following, if possible: 7. 3 −343 8. 4 81 16 9. −49 10. Write 3 y 4 using exponents 11. Write 1 12. −64 3 125 13. 27 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. © x 10 − 23 using radicals − 23 For problems 14-18, simplify the expression. 48 3 54x 4 3 56 4 162x 3 y 6 10 32x 5 For problems 19-26, rationalize the denominator. 3 2 2 18 6 3 16 9 4 8x 2 1 5 3 5 3 : Pre-Calculus © 25. 26. : Pre-Calculus - Chapter 1B 4 7 − 3 5 4 − 11 For problems 27-31, simplify the expression. 27 − 5 48 2 45 4 48 5 4 2 − 3 4 3 75x 3 − 27x 3x 5 3 3 30. 2 − 2 2 6 3 31. 45 − 6 125 32. 16x 3 3 16x 4 y 3 54x 4 y x 25x 27. 28. 29. © : Pre-Calculus © : Pre-Calculus - Chapter 1B Answers to Exercises for Chapter 1B - Exponents and Radicals 1. 2. 3. 4. 5. 6. 7. 8. −3 −3 −3 −3 81 1 1 125 53 5 2 5 2 25 4 2 22 4 −3ab 4ab −3 −12a 11 b 4−3 −12a 2 b 5a 3 a 2 −1 16a 10 0 125a 3 12 1 125a a −2 −2 −2 4 15x y 3y y 3y 5 −2 x5 5x 3 y −4 x3 x2 (Other methods are possible) −7 , since −7 3 −343 4 81 3 4 2 16 x5 3y 5 2 10 x 10 9y 9. Not possible; there is no real number whose square is −49 4 10. y 3 11. x 10 3 −2 10 x 3 2 3 100 x2 12. −4 13. 14. 15. 16. 17. 18. 19. 20. 21. © 2 3 2 2 27 3 9 5 3 25 125 48 2 4 3 2 2 3 4 3 3 54x 4 3 2 3 3 x 4 3x 3 2x 3 56 3 2 3 7 2 3 7 4 162x 3 y 6 4 2 3 4 x 3 y 6 3y 4 2x 3 y 2 (Since y 6 y 4 y 2 ) 27 125 3 5 5 1 1 32x 5 10 2 5 x 5 2 10 x 10 2 2 x 2 2x 3 2 3 2 2 2 2 2 2 (simplify radical first) 18 2 32 2 2 3 2 2 2 2 32 2 (Note the cancellation of the 2’s) 3 6 6 (simplify radical first) 3 3 16 24 6 (OK to cancel first) 232 3 22 3 3 3 2 22 10 : Pre-Calculus © : Pre-Calculus - Chapter 1B 3 3 22 3 23 334 2 9 9 (factor first to see what you need) 4 3 4 8x 2 x 4 2x 3 9 (to get 4th powers you need one 2 and 3 x’s) 4 4 23 x 2x 3 9 4 2x 3 4 24x4 9 4 2x 3 2x 2 1− 5 1− 5 2 2 1− 5 1 5 12 − 5 22. 23. 21 − 5 −4 1− 5 − 2 3 5 −3 5 −3 2 5 −3 5 − 32 24. 3 5 3 −3 5 − 9 3 5 −9 or −4 4 7 3 4 7 3 2 2 7 3 7 − 3 25. 4 7 − 3 7 3 4 7 3 (since the 4’s cancel) 5 4 11 4 5 55 5 4 11 2 5 2 4 − 11 4 11 4 − 11 26. 4 27 − 5 48 2 45 3 2 3 − 5 2 4 3 2 3 2 5 3 3 − 5 4 3 6 5 3 4 3 −1 6 5 7 3 5 5 4 4 4 4 4 28. 48 5 2 − 3 3 2 3 5 4 2 − 3 4 3 243 542 −343 5 4 2 2 − 3 4 3 542 − 43 29. 75x 3 − 27x 3x 5 5 2 3 x 3 − 3 3 x 3x 5 5x 3x − 3 3x x 2 3x x 2 5x − 3 3x 30. (rationalize the denominators first) 2 6 3 3 2 − 2 3 2 6 3 6 − 2 3 6 2 3 6 2 2 3 3 6 2 6 9 6 4 3 − 6 6 6 27. © : Pre-Calculus © : Pre-Calculus - Chapter 1B 2 6 9 6 −4 3 6 11 6 − 4 3 6 3 31. 45 − 6 125 3 2 5 − 6 5 3 3 5 − 5 6 3 5 − 5 2 5 32. 16x 3 3 16x 4 y 3 54x 4 y x 25x 4x x 2x 3 2xy 3x 3 2xy 5x x 9x x 5x 3 2xy © : Pre-Calculus © : Pre-Calculus - Chapter 1C Chapter 1C - Polynomials Definition of a Polynomial A polynomial is the most common algebraic expression. Polynomials are expressions which contain terms of the form ax k , where a is a real constant and k is a nonnegative integer. More formally, a polynomial expression is an expression of the form a n x n a n−1 x n−1 a 1 x a 0 In the above expression, a n is a nonzero real number, a 0 , a 1 , ... , a n−1 are real numbers, and n is a positive integer. n is called the degree of the polynomial, a n is called the leading coefficient, and a 0 is called the constant term. If n 2 the polynomial is called a binomial or quadratic. If n 3 the polynomial is called a trinomial or cubic. Example The height, s (in feet), of a ball thrown in the air at time t 0 seconds is given by the polynomial equation s −16t 2 80t 5 . Here the degree of the polynomial is 2 (the highest exponent), the leading coefficient is −16, and the constant is 5. Example The volume, V, of a cube with edge length x is given by V x 3 . Here the degree of the polynomial is 3, the leading coefficient is 1 (remember that x 3 1x 3 ), and the constant is 0. Polynomial expressions do not contain negative exponents nor do they contain radicals. So an expression such as 9 − x 2 is not a polynomial. Question: Is x 3 1x a polynomial? Answer: No, the 1x term is the same as x −1 and the exponent is not a positive integer. © : Pre-Calculus © : Pre-Calculus - Chapter 1C Sums and Differences of Polynomials Addition and Subtraction of Polynomials is done by combining like terms, that is, terms which have the same variable and exponent. Example 1: Simplify the expression: 15x 2 − 6 8x 3 − 14x 2 17 Solution: 15x 2 − 6 8x 3 −14x 2 17 8x 3 15 − 14x 2 −6 17 8x 3 x 2 11 To subtract polynomials, recall that subtraction is defined as adding the opposite (inverse); that is, a − b a −b . It is important that you distribute the negative sign over the entire polynomial. Example 2: Simplify the expression: 15x 2 − 6 − 8x 3 − 14x 2 17 Solution: 15x 2 − 6 − 8x 3 − 14x 2 17 15x 2 − 6 −8x 3 14x 2 − 17 (NOT −8x 3 − 14x 2 17 !!) 15x 2 −6 − 8x 3 14x 2 − 17 −8x 3 29x 2 − 23 Example 3: Simplify the expression: 3x 3 − 2x 2 8 3x 4 2x 2 − 5 Solution: 3x 3 − 2x 2 8 3x 4 2x 2 − 5 3x 3 − 2x 2 8 3x 4 2x 2 − 5 3x 4 3x 3 −2 2x 2 8 − 5 3x 4 3x 3 3 Example 4: Simplify the expression: −5x 2 − 1 − −3x 2 5 Solution: − 5x 2 − 1 − −3x 2 5 (distribute negative signs) −5x 2 1 3x 2 − 5 −2x 2 − 4 © : Pre-Calculus © : Pre-Calculus - Chapter 1C Products of Polynomials Mulitplication of Polynomials is based on the Distributive Property of Real Numbers as illustrated in the example below: Example 1: Expand the following product x − 68x 7. Solution: x − 68x 7 x8x 7 − 68x 7 (treat 8x 7 as a single number first) x8x x7 − 68x − 67 8x 2 7x − 48x − 42 8x 2 − 41x − 42 Notice, in the underlined step, that the distributive properties follow the familiar pattern FOIL, which is explained below. x8x x7 − 68x − 67 First Outer Inner Last terms terms terms terms Although FOIL only works with binomials, the distributive property method works on any polynomials. Example 2: Expand 5x − 6x 3 − 4x 2 2. Solution 1 5x − 6x 3 − 4x 2 2 5xx 3 − 4x 2 2 − 6x 3 − 4x 2 2 5x 4 − 20x 3 10x − 6x 3 24x 2 − 12 5x 4 − 26x 3 24x 2 10x − 12 (Don’t forget to distribute the negative sign as well!!!) Solution 2: The distributive method is easy to organize and picture using a rectangle, or box, whose lengths are the two factors as illustrated below: −4x 2 x3 5x 5x 4 −20x 2 3 10x −6 −6x 3 24x 2 −12 5x 4 − 20x 3 − 6x 3 24x 2 10x − 12 5x 4 − 26x 3 24x 2 10x − 12 Example 3: Expand x 2 9x 2 − 6x 9. Solution: x 2 9x 2 − 6x 9 x 2 x 2 − 6x 9 9x 2 − 6x 9 x 4 − 6x 3 9x 2 9x 2 − 54x 81 x 4 − 6x 3 18x 2 − 54x 81 Example 4: Expand a ba − b. Solution: a ba − b aa − b ba − b a 2 − ab ba − b 2 a 2 − ab ab − b 2 (By the commutative property) © : Pre-Calculus © : Pre-Calculus - Chapter 1C a2 − b2 The previous example a ba − b a 2 − b 2 is an example of a Special Product. Special products can not only make multiplication of certain polynomials easier, they are also useful when you want to factor a polynomial; that is, given a polynomial, find two or more polynomials whose product is the given polynomial. The most common special products are listed below (here a and b represent any numbers, variables, or algebraic expressions). a ba − b a 2 − b 2 a b 2 a 2 2ab b 2 a − b 2 a 2 − 2ab b 2 Sum of Cubes: a ba 2 − ab b 2 a 3 b 3 Difference of Cubes: a − ba 2 ab b 2 a 3 − b 3 Cube of a Binomial: a b 3 a 3 3a 2 b 3ab 2 b 3 a − b 3 a 3 − 3a 2 b 3ab 2 − b 3 NOTE that there is no special product for the sum of squares that involves only real numbers.. Difference of Squares: Square of a Binomial: Historical Note: The French mathematician Blaise Pascal (1623-1662) discovered an easy way to remember the coefficients of a binomial power using what is now called Pascal’s Triangle. After the first two rows, each succesive row is obtained by adding the two numbers immediately above it as shown below: 1 a b 0 1 1 1 a b 1 1a 1b 1 2 1 a b 2 1a 2 2ab 1b 2 1 3 3 1 etc. a b 3 1a 3 3a 2 b 3ab 2 1b 3 Using this triangle, we can see that a b 4 1a 4 4a 3 b 6a 2 b 2 4ab 3 1b 4 The numbers in Pascal’s Triangle are also useful in Probability. Example 5: Expand Solution: x − 2 3 . x − 2 3 x 3 − 3x 2 2 3x2 2 − 2 3 x 3 − 6x 2 12x − 8 Example 6: Expand x 2yx − 2y 2 . Solution: x 2yx − 2y 2 © : Pre-Calculus 2 x 2 − 2y 2 (work inside the parenthesis first) x 2 − 4y 2 2 x 2 2 − 2x 2 4y 2 4y 2 2 x 4 − 8x 2 y 2 16y 4 © : Pre-Calculus - Chapter 1C Quotients of Polynomials Division of Polynomials is done using standard long division techniques. Example 1: Review basic long division for numbers. Use long division to find 1350 18. Solution: The algorithm for long division is: Divide Multiply Subtract Bring Down. Repeat these steps as long as it is possible to divide. Note that 18 7 126, so 18 divides 135 seven times: 7 18 (divide) 1 3 5 0 1 2 6 (multiply) 9 (subtract) Now bring down the 0 and repeat, noting that 18 5 90: 18 7 5 0 1 3 5 1 2 6 9 0 9 0 0 So 1350 18 75. Example 2: Now use long division to divide polynomials: 8x 2 13x − 6 x 2 2 Solution: First, divide the highest terms. 8xx 8x. Then follow the steps discussed in the above example. 8x −3 x 2 8x 2 13x −6 8x 2 16x −3x −6 −3x −6 0 so 8x 2 13x − 6 x 2 8x − 3 Another way of writing this is to say that 8x 2 13x − 6 x 28x − 3. Example 3: Perform the following operation: x 4 − 3x 2 − 4 x 2 x Solution: We begin by writing out the long division, using 0’s for any missing powers of x: © : Pre-Calculus © : Pre-Calculus - Chapter 1C x 2 x x 4 0x 3 −3x 2 0x −4 4 We need to determine how many times x 2 will divide into x 4 . Since x 2 x 2 , the first term of our x quotient is x 2 : Divide x 2 x 2 x x 4 0x 3 −3x 2 0x −4 Now multiply and subtract: x2 x 2 x x 4 0x 3 −3x 2 0x −4 Multiply x 4 x 3 Subtract −x 3 Bring down the next term: x2 x 2 x x 4 0x 3 −3x 2 0x −4 x 3 x4 −x 3 −3x 2 Bring down The next term of the quotient will be −x2 −x, so we repeat the process: x 3 −x x2 Divide x 2 x x 4 0x 3 −3x 2 0x −4 x 3 x4 −x 3 −3x 2 x2 x x x 2 4 x4 −x 0x 3 −3x 2 0x −4 x 3 −x 3 −3x 2 −x 3 Multiply x2 x x x 2 4 x4 −x 2 −x 0x 3 −3x 2 0x −4 x 3 −x 3 −3x 2 −x 3 Subtract −x 2 −2x 2 0x Bring down 2 The next term of our quotient will be −2x2 −2, so bring down the next term and repeat: x © : Pre-Calculus © : Pre-Calculus - Chapter 1C −x x2 x x x 2 0x 4 −2 −3x 3 2 Divide 0x −4 x 3 x4 −x 3 −3x 2 −x 3 −x 2 −2x 2 0x x2 x x x 2 4 x4 −x 0x 3 −2 −3x 2 0x −4 x 3 −x 3 −3x 2 −x 3 −x 2 −2x 2 0x −2x 2 −2x Multiply x2 −x −2 x 2 x x 4 0x 3 −3x 2 0x −4 x4 x 3 −x 3 −3x 2 −x 3 −x 2 −2x 2 0x −2x 2 −2x Subtract 2x −4 Bring down Since x 2 does not evenly divide 2x, the long division is complete. We can say that x 4 − 3x 2 − 4 x 2 x x 2 − x − 2 with a remainder of 2x − 4. We could also say either of the following: x 4 − 3x 2 − 4 x 2 − x − 2 2x − 4 x2 x x2 x or: x 4 − 3x 2 − 4 x 2 xx 2 − x − 2 2x − 4 © : Pre-Calculus © : Pre-Calculus - Chapter 1C Factoring The process of factoring is a type of ”reverse-multiplication”, where you are given a polynomial and have to write it as a product of factors. A polynomial is completely factored when each factor is prime, or cannot be factored again. The idea is very similar to factoring numbers: 60 5 12 is one factoring of 60 562 5232 5 2 2 3 is the prime factorization of 60 There are several strategies to determine how to factor polynomials. I. Common Factors A common factor is a factor of every term of an expression. Common factors can be pulled out of an expression using the distributive property in reverse: ab ac ab c Examples: x 3 4x xx 2 4 −2x 2 6x −2xx − 3 (Note the negative in the second term) 6mn 2 15m 2 n − 30m 3 n 3 3mn2n 5m − 10m 2 n 2 Finding common factors should always be the first step in factoring an expression. II. Factoring by Grouping Factoring by Grouping is especially useful when you have more than three terms in the polynomial. The technique of factoring by grouping is really using common factors creatively, as shown in the following example: Example 1: Factor 2x 3 − x 2 6x − 3 Solution: Although there is no factor common to all terms (except 1, which we ignore), we can ”group” the polynomial by twos, each of which have a common factor: 2x 3 − x 2 6x − 3 x 2 2x − 1 32x − 1 Now notice that there are two terms, each of which has the factor 2x − 1. 2x − 1x 2 3 It is very easy to check your answer by mulitplying it out: 2x − 1x 2 3 2xx 2 3 − 1x 2 3 2x 3 6x − x 2 − 3 2x 3 − x 2 6x − 3 ✓ III. Factoring Using Special Products © : Pre-Calculus © : Pre-Calculus - Chapter 1C The special products we learned in the section ”Products of Polynomials” can be used to help factor expressions which have an appropriate form: Example 2: Solution: Factor completely x 3 − 9x As mentioned in the previous section, the first thing we do is look for a common factor: x 3 − 9x xx 2 − 9 Notice how the second factor can now be written as x 2 − 3 2 , a difference of two squares. Recall that a 2 − b 2 a ba − b. x 3 − 9x xx 2 − 9 xx 3x − 3 The expression is now completely factored. Example 3: Factor completely x 3 8y 3 Solution: Notice how the expression can be rewritten as x 3 2y 3 , a sum of two cubes. Recall that a ba 2 − ab b 2 a 3 b 3 x 3 8y 3 x 2y x 2 x2y 2y 2 x 2yx 2 2xy 4y 2 Example 4: Factor completely x 2 25 Solution: Recall that there is no special product for a sum of squares a 2 b 2 . In fact, this expression cannot be factored; it is prime over the real numbers. IV. Factoring Binomials (x 2 bx c ) If possible, a binomial of this form must factor as x mx n , where m and n are integers. Note that if this is true: x mx n x 2 bx c x 2 nx mx mn x 2 bx c x 2 m nx mn x 2 bx c We can see in the last statement that the numbers m and n that we seek must add up to b and multiply to c. Example 5: Factor completely x 2 7x 12 Solution: We are looking for two numbers which will add up to 7 and multiply to 12. The numbers (obtained by trial and error) are 3 and 4. Therefore, x 2 7x 12 x 3x 4 Example 6: © Factor completely x 2 − 4x − 21 : Pre-Calculus © : Pre-Calculus - Chapter 1C Solution: We are looking for two numbers which will add up to −3 and multiply to −21. Note that since the product is negative, the numbers must be different signs (one negative and one positive). The numbers are −7 and 3 (Note that if we had chosen 7 and −3, the numbers add up to 4. In general, if the signs are different, the larger number will have the same sign as b, the middle term). Therefore, x 2 − 3x − 21 x − 7x 3 Example 7: Factor completely x 2 − 6x 9 Solution: We are looking for two numbers which will add up to −6 and multiply to 9. Note that since the product is positive, the numbers must be the same sign (both positive or both negative). Since the middle term is negative, the numbers must both be negative. The numbers are −3 and −3. Therefore x 2 − 6x 9 x − 3x − 3 x − 3 2 Note that we could have factored this using the special product a − b 2 a 2 − 2ab b 2 V. Factoring Trinomials (ax 2 bx c ) If possible, a trinomial of this form (where the leading coefficient is not 1) must factor in the form mx snx t , where m, n, s, and t are all integers. It is possible to find these numbers by trial and error; however, such a method may at times be haphazard or tedious. The following is a more straightforward approach utilizing the technique of factoring by grouping. First, we need to split the middle term bx into two terms which will allow us to factor by grouping. To do this, we use a similar technique as before, only now we look for two numbers whose sum is b and whose product is ac . If you are curious as to why, see below: As before, multiply the desired form out: mx snx t ax 2 bx c mnx 2 mtx nsx st ax 2 bx c mnx 2 mt nsx st ax 2 bx c Now note that b mt ns and ac mnst mtns . The numbers we are looking for are mt and ns. How they split up into m, n, s, and t are determined when we factor by grouping. Example 8: Factor completely 9x 2 − 3x − 2 Solution: We first must find two numbers whose sum is −3 and whose product is 9−2 −18. The numbers are −6 and 3. Now we rewrite −3x as −6x 3x and factor by grouping: 9x 2 − 3x − 2 9x 2 − 6x 3x − 2 3x3x − 2 13x − 2 3x 13x − 2 (remember that you can check your answer by multiplying) © : Pre-Calculus © : Pre-Calculus - Chapter 1C Example 9: Factor completely 6x 2 − x − 15 Solution: We must first find two numbers whose sum is −1 and whose product is 6−15 −90. The numbers are 9 and −10. Now we rewrite −x as 9x − 10x and factor by grouping: 6x 2 − x − 15 6x 2 9x −10x − 15 3x2x 3 − 52x 3 (Notice that both signs change in the last term) 3x − 52x 3 If you are afraid of getting the middle terms in the wrong order, not to worry... 6x 2 − x − 15 6x 2 − 10x 9x − 15 2x3x − 5 33x − 5 2x 33x − 5 the same as above. © : Pre-Calculus © : Pre-Calculus - Chapter 1C Exercises for Chapter 1C - Polynomials 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. © Determine which of the following are polynomials: 4x 3 − 8x 2 17x − 12 , −6x 6x −1 , −24 , 3 − x 2 , 16 − x 2 For the polynomials in #1, give the degree, leading coefficient, and constant term For problems 3-6, simplify the expression. −5x 7 2x 4 3x 2 − 2 −9x 2 4 2x 3 − 5x 2 1 −3x 3 7x − 6 − 3x 3 6x 2 − 7x y 3 1 − y 3 − 1 y 2 − 2y 1 y 3 1 − y 3 − 1 − y 2 − 2y 1 Find a polynomial which, when added to x 3 3x 2 − 3x 2, yields 2x 2 4x 7 For problems 8-14, simplify the expression. 3x − 25x 7 −9x 2 4x − 5 3x − 2x 2 − 2x 2 4x 5 2 3x 7 2 3x − 7 2x 1x − 3 5x − 3 2x − 1 3 An open box made from a 9-inch square piece of material by cutting equal squares from all corners and folding the sides. The volume of the box is given by x9 − 2x 2 . Give the degree, leading coefficient, and constant term of this polynomial. For problems 16-23, divide by long division. x 2 − 6x 10 2x 3 x 3 − 1 x − 1 x4 − 1 x−1 x5 − 1 x−1 n If n is a positive integer, what do you think x − 1 equals ? x−1 x2 1 x−1 x 3 − 6x 2 12x − 8 x−2 x 4 − 2x 3 4x 2 − 2x − 7 x 2 − 2x 3 For problems 24-38, factor each expression as completely as possible. 6x 3 − 4x x − 22x − 3x − 2 y 3 − 27z 3 2x 3 16 x 4 − 2x 3 − 8x 16 a 3 − 27a : Pre-Calculus © 30. 31. 32. 33. 34. 35. 36. 37. 38. © : Pre-Calculus - Chapter 1C 4x 3 4x 2 − 9x − 9 x 2 11x 10 y 2 − 5y 6 x2 x − 2 x 2 8x − 20 y 3 − 2y 2 − 3y 2x 2 x − 15 3x 2 10x 8 15x 2 − 11x 2 : Pre-Calculus © : Pre-Calculus - Chapter 1C Answers to Exercises for Chapter 1C - Polynomials 1. 2. 3. 4. 5. 6. 7. The polynomials are 4x 3 − 8x 2 17x − 12 , −24, and 3 − x 2 − 6x 6x −1 is not a polynomial because of the negative exponent 16 − x 2 is not a polynomial because of the radical. 4x 3 − 8x 2 17x − 12 : degree 3, leading coefficient 4, and constant term −12 −24 : degree 0, leading coefficient −24, and constant term −24 3 − x 2 −x 2 3 : degree 2, leading coefficient −1, and constant term 3 −5x 7 2x 4 3x 2 − 2 −9x 2 4 −5x 7 2x 4 3x 2 −2−9x 2 4 −5x 7 2x 4 − 6x 2 2 2x 3 − 5x 2 1 −3x 3 7x − 6 − 3x 3 6x 2 − 7x 2x 3 − 5x 2 1 − 3x 3 7x − 6 − 3x 3 − 6x 2 7x 2 − 3 − 3x 3 −5 − 6x 2 7 7x 1 − 6 −4x 3 − 11x 2 14x − 5 y 3 1 − y 3 − 1 y 2 − 2y 1 y 3 1 − y 3 − 1 − y 2 − 2y 1 y 3 1 − y 3 1 − y 2 2y − 1 1 − 1y 3 − y 2 2y 1 1 − 1 −y 2 2y 1 y 3 1 − y 3 − 1 − y 2 − 2y 1 y 3 1 − y 3 − 1 y 2 − 2y 1 y 3 1 − y 3 1 y 2 2y 1 1 − 1y 3 y 2 2y 1 1 1 y 2 2y 3 Recall that if a b c, then a c − b, so the polynomial we seek is 2x 2 4x 7 − x 3 3x 2 − 3x 2 2x 2 4x 7 − x 3 − 3x 2 3x − 2 −x 3 2 − 3x 2 4 3x 7 − 2 −x 3 − x 2 7x 5 3x − 25x 7 8. −2 3x 5x 15x −10x 2 7 21x −14 15x 2 21x − 10x − 14 15x 2 11x − 14 −9x 2 4x − 5 9. −5 x −9x 4 2 −9x 4x 3 −45x 2 2 −20 −9x 3 − 45x 2 4x − 20 3x − 2x 2 − 2x 2 10. © : Pre-Calculus © : Pre-Calculus - Chapter 1C −2x x2 3x 3x −6x 3 2 2 −2 −2x 2 4x 6x −4 3x 3 − 6x 2 − 2x 2 6x 4x − 4 3x 3 − 8x 2 10x − 4 4x 5 2 4x 54x 5 11. 5 4x 4x 16x 2 5 20x 20x 25 16x 2 20x 20x 25 16x 2 40x 25 12. We could square the first term using a b 2 a 2 2ab b 2 , but notice the following: 3x 7 2 3x − 7 3x 73x 73x − 7 Now use difference of squares on the term in brackets: 3x 79x 2 − 49 9x 2 −49 3x 27x 3 −147x 7 63x 2 −343 27x 3 63x 2 − 147x − 343 13. We will use FOIL here because of the multiple operations: 2x 1x − 3 5x − 3 2xx 2x−3 1x 1−3 5x − 15 2x 2 − 6x x − 3 5x − 15 2x 2 − 18 14. 2x − 1 3 2x − 1 2 2x − 1 4x 2 − 4x 12x − 1 2x 4x 2 8x 3 −1 −4x 2 −4x −8x 2 4x 1 2x −1 8x 3 − 12x 2 6x − 1 15. Multiply out the expression: x9 − 2x 2 x18 − 36x 4x 2 4x 3 − 36x 2 18x © : Pre-Calculus © : Pre-Calculus - Chapter 1C Therefore, the degree of the polynomial is 3, the leading coefficient is 4, and the constant term is 0. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. x 2 − 6x 10 1 x − 15 85 4 2x 3 2 42x 3 x3 − 1 x2 x 1 x−1 x4 − 1 x3 x2 x 1 x−1 x5 − 1 x4 x3 x2 x 1 x−1 x n − 1 x n−1 x n−2 x 1 x−1 x2 1 1 x 2 x−1 x−1 x 3 − 6x 2 12x − 8 x 2 − 4x 4 x−2 x 4 − 2x 3 4x 2 − 2x − 7 x 2 1 − 10 x 2 − 2x 3 x 2 − 2x 3 6x 3 − 4x 2x3x 2 − 2 x − 22x − 3x − 2 x − 22x − 3 NOTE: you could multiply out the expression first and then factor the resulting trinomial. Use difference of cubes: y 3 − 27z 3 y 3 − 3z 3 y − 3zy 2 3yz 9z 2 27. Factor the common factor, then use sum of cubes: 2x 3 16 2x 3 8 2x 2x 2 − 2x 4 28. Factor by grouping, then use difference of cubes: x 4 − 2x 3 − 8x 16 x 3 x − 2 − 8x − 2 x 3 − 8x − 2 x − 2x 2 2x 4x − 2 x − 2 2 x 2 2x 4 29. a 3 − 27a aa 2 − 27 We cannot factor this any further (using only integers). 30. Factor by grouping, then use difference of squares: 4x 3 4x 2 − 9x − 9 4x 2 x 1 − 9x 1 4x 2 − 9x 1 2x 32x − 3x 1 31. We need two numbers whose product is 10 and whose sum is 11; the numbers are 10 and 1: x 10x 1 32. We need two numbers whose product is 6 and whose sum is −5; the numbers are −2 and −3 y − 2y − 3 33. We need two numbers whose product is −2 and whose sum is 1; the numbers are −1 and 2: x 2x − 1 34. We need two numbers whose product is −20 and whose sum is 8; the numbers are 10 and −2: © : Pre-Calculus © : Pre-Calculus - Chapter 1C x 10x − 2 35. Common factor first: y 3 − 2y 2 − 3y yy 2 − 2y − 3 We need two numbers whose product is −3 and whose sum is −2; the numbers are −3 and 1: yy − 3y 1 36. We need two numbers whose product is 2−15 −30 and whose sum is 1; the numbers are 6 and -5: 2x 2 6x − 5 − 15 2xx 3 − 5x 3 2x − 5x 3 37. We need two numbers whose product is 38 24 and whose sum is 10; the numbers are 6 and 4: 3x 2 6x 4x 8 3xx 2 4x 2 3x 4x 2 38. We need two numbers whose product is 152 30 and whose sum is −11; the numbers are −5 and −6: 15x 2 − 5x − 6x 2 5x3x − 1 − 23x − 1 5x − 23x − 1 © : Pre-Calculus © : Pre-Calculus - Chapter 1D Chapter 1D - Rational Expressions Definition of a Rational Expression A rational expression is the quotient of two polynomials. (Recall: A function px is a polynomial in x of degree n, if there are constants a 0 , a 1 , , a n , with a n ≠ 0 such that px a 0 a 1 x a n x n .) p More formally, a rational expression is an expression of the form q , where p and q are polynomials, and qx cannot be the zero polynomial. The denominator of a rational expression can be a constant polynomial though. For example, rx x is a polynomial of degree 1, and it is also a rational expression, for rx x x . 1 3 4x − 7 Example 1: , 2 , x − 28 , and x2 5 are examples of rational expressions. 6x − 16 x − 9 2 x 25 Example 2: x 2 12 is not a rational expression. x7 A rational expression involves a quotient, and since division by 0 is not defined for real numbers, there are sometimes restrictions on the variable x. In particular, the restrictions occur wherever the denominator is zero. Example 3: For the rational expression 4x − 7 , the values of x which make the denominator zero 6x − 16 can be found by solving the equation 6x − 16 0 6x 16 x 16 8 6 3 8 Therefore, the variable x cannot equal in the rational expression 4x − 7 . We also say that the 3 6x − 16 domain of this expression is all x not equal to 8 . 3 Example 4: For the rational expression 2 , the values of x which make the denominator zero x−9 can be found by solving the equation x−9 0 x9 Therefore, the variable x cannot equal 9 in the rational expression 2 . The domain of this x−9 expression is all x not equal to 9. For the rational expression x2 5 , there are no values of x which make the x 25 denominator zero since x 2 25 is always positive. Therefore, there are no restrictions on the variable x. The domain of this expression is all x. Example 5: © : Pre-Calculus © : Pre-Calculus - Chapter 1D 2 For the rational expression x 2 − 9 , the values of x which make the denominator zero x −1 can be found by solving the equation x2 − 1 0 Example 6: x2 1 x 1 2 Therefore, the variable x cannot equal 1 in the rational expression x 2 − 9 . The domain of this x −1 expression is all x not equal to 1. © : Pre-Calculus © : Pre-Calculus - Chapter 1D Simplifying Rational Expressions To simplify a rational expression means to reduce it to lowest terms. From working with fractions, you may recall that simplifying is done by cancelling common factors. Therefore, the key to simplifying rational expressions (and to most problems involving rational expressions) is to factor the polynomials whenever possible. Example 1: Solution: cancel the x’s: 4 Write 8x 6 in reduced form. 12x Factor the numbers and cancel common factors. Use properties of exponents to help 8x 4 4 2 x 4 12x 6 4 3 x4 x2 22 3x Example 2: Write y 3 5y 2 6y in reduced form. y2 − 4 Solution: yy 2 5y 6 y 3 5y 2 6y 2 y −4 y2 − 4 yy 2y 3 y 2y − 2 yy 3 y 2 3y ; y ≠ 2 y−2 y−2 Question: Why is y ≠ 2? Answer: In order for our answer to be equivalent to the original fraction, the variable must have the same restrictions. Since y cannot equal 2 in the original expression (the denominator would then be zero), we must restrict the domain of our answer in order for these fractions to be equivalent. Example 3: Write 4x − x 3 in reduced form. x −x−2 2 Solution: 2 4x − x 3 x4 − x 2 x −x−2 x −x−2 x2 x2 − x x − 2x 1 2 At this point, it doesn’t look like anything will cancel. However, if we factor −1 from the last term in the numerator, we obtain the following: −x2 xx − 2 x − 2x 1 2 −xx 2 −x − 2x ; x ≠ 2 or − 1 x1 x1 Question: Answer: © What property of real numbers tells us that −x2 x −xx 2 ? The commutative property of addition. x 2 2 x. : Pre-Calculus © : Pre-Calculus - Chapter 1D Operations with Rational Expressions The operations of addition, subtraction, multiplication, and division with rational expressions follow the same rules that are used with common fractions. The key, as with simplifying rational expressions, is to factor the polynomials. I. Multiplication of Rational Expressions Recall with fractions a and c that a c ac . Also recall when mulitplying fractions that you b d b d bd may cross-cancel; that is, reduce common factors of any numerator with any denominator. Example 1: 2 3 9 10 2 3 33 25 1 1 3 5 1 15 The same is true of multiplying rational expressions. As before, the key is to factor the polynomials first. 2 Example 2: Simplify 5 x − 1 x − 1 25x − 50 Solution: First factor the polynomials. Don’t forget to cancel common factors! 5 x 2 − 1 5 x 1x − 1 x−1 x − 1 25x − 50 25x − 2 1 x1 1 5x − 2 x 1 ; x ≠ 1 or 2 5x − 10 Before looking at the next example, recall our strategies for factoring: 1. Factor out any common factors 2. If more than 3 terms, try factoring by grouping 3. Recognize special products 4. Factor trinomials using product/sum strategies: a) x 2 bx c: factors into x mx n where mn c and m n b b) ax 2 bx c: split bx into mx nx, where mn ac and m n b, then factor by grouping. 2 3 2 2x x 5 Example 3: Simplify 2x2 x − 6 x − 3x 2 x3 − 8 x 4x − 5 4x − 6x Solution: 2x 2 x − 6 x 3 − 3x 2 2x x 5 x3 − 8 x 2 4x − 5 4x 2 − 6x 2x − 3x 2 xx − 2x − 1 x5 2x2x − 3 x 5x − 1 x − 2x 2 2x 4 x2 x2 ; x ≠ −5, 1, 0, 3 , 2 2 2x 2 2x 4 2x 2 4x 8 Example 4: © Simplify 2x2 − 4 x −4 : Pre-Calculus © Solution: : Pre-Calculus - Chapter 1D 2x − 2 2x − 4 2 , x ≠ 2. x2 x − 2x 2 x2 − 4 II. Division of Rational Expressions a/b Recall with fractions a and c that a c a dc . In other words, dividing by a fraction b d b d b c/d is the same as multiplying by its reciprocal. The same is true for rational expressions. As before, the key to making the work easier is to factor the polynomials first. x 2 2x − 15 x 2 7x 10 Example 5: x2 2 Solution: Begin by writing the second term as x 7x 10 : 1 x 2 2x − 15 x 2 7x 10 x 2 2x − 15 1 x2 1 x2 x 2 7x 10 x 5x − 3 1 x2 x 5x 2 x − 3 2 2 x − 3 ; x ≠ −5, − 2 x 4x 4 x 2 Notice that the x 2 terms cannot be cancelled since both are in the denominator. Multiplication and Division can also appear together. Only take the reciprocal of the fractions you are dividing. y 2 3y − 28 3y 21 y 2 − 7y 12 Example 6: 2 2 4y 24 y 3y − 18 y 12y 36 Solution: We only take the reciprocal of the second fraction: y 2 3y − 28 3y 21 y 2 − 7y 12 2 2 4y 24 y 3y − 18 y 12y 36 y 2 − 7y 12 y 2 12y 36 3y 21 2 2 4y 24 y 3y − 18 y 3y − 28 3y 7 y − 3y − 4 y 6 2 y 6y − 3 y 7y − 4 4y 6 3; y ≠ −7, − 6, 3, 4 Question: Explain why the values −7, −6, 3, and 4 are excluded in the previous example. Answer: The easiest way to understand why these values have been excluded is to write y 2 3y − 28 3y 21 y 2 − 7y 12 2 as a fraction. 2 4y 24 y 3y − 18 y 12y 36 y 2 − 7y 12 y 2 3y − 18 3y 21 y 2 3y − 28 4y 24 y 2 12y 36 y ≠ −6 follows since 4y 24 cannot equal 0. We also get y ≠ 3 because the term y 2 − 7y 12 y 2 − 7y 12 y − 6y 3 y 2 3y − 18 © : Pre-Calculus © : Pre-Calculus - Chapter 1D The next observation is that the denominator of y 2 − 7y 12 y 2 3y − 18 y 2 3y − 28 y 2 12y 36 y 2 3y − 28 that is, 2 cannot equal zero. Factoring the numerator and denominator of this rational y 12y 36 expression we have y 2 3y − 28 y 7y − 4 The numerator is zero if y −7 or if y 4. This accounts for excluding the numbers −7 and 4. The denominator of y 2 3y − 28 y 2 12y 36 factors into y 6 2 which means we have to exclude −6, but we have already done that. What values of x are not allowed in the rational expression x − 1 x − 3 ? x−4 x−6 x − 3 is zero at x 3 and we cannot Answer: 3, 4, 6 (The value 3 is not allowed because x−4 divide by 0. x 4 is not allowed because we have the term x − 4 in the denominator. Similarly x 6 is not allowed because the term x − 6 is the denominator of x − 1 .) x−6 Question: III. Addition and Subtraction of Rational Expressions Recall that addition and subtraction is done by first finding a common denominator. The least common denominator (LCD) of several fractions is the product of all prime factors of the denominators. A factor only occurs more than once in the LCD if it occurs more than once in any one fraction. Once you have the LCD, convert each fraction to an equivalent fraction with the LCD, then add or subtract the numerators. 1 − 1 Example 7: x x−1 Solution: The LCD here is xx − 1. We convert each fraction to an equivalent one with this x denominator: That is, 1x x − 1 and 1 x−1 xx − 1 xx − 1 1x 1 − 1 1x − 1 − x x−1 xx − 1 x − 1x x x−1 − xx − 1 xx − 1 x − 1 − x −1 xx − 1 xx − 1 Example 8: Solution: © y−4 y 2 3y − 28 − 2 y6 y 12y 36 We must factor the denominators first to find the LCD: : Pre-Calculus © : Pre-Calculus - Chapter 1D y−4 y 2 3y − 28 − y6 y 6 2 The LCD is y 6 2 . We now convert each fraction to an equivalent fraction using this denominator: y − 4y 6 y−4 y6 y 6 2 y 2 3y − 28 y 2 3y − 28 2 y 6 y 6 2 Next subtract the two expressions y − 4y 6 y 2 3y − 28 y 2 3y − 28 y−4 − 2 − 2 y6 y 12y 36 y 6 2 y 6 y 2 2y − 24 y 2 3y − 28 − 2 y 6 y 6 2 y 2 2y − 24 − y 2 3y − 28 y 6 2 y 2 2y − 24 − y 2 − 3y 28 y 6 2 −y 4 y 6 2 Note that the minus sign in the numerator was distributed across the expression −y 2 3y − 28 −y 2 − 3y 28. Question: Answer: © What is the common denominator in x − 7x 1x 2 − 4 : Pre-Calculus x−5 − 21 ? x − 7x 1 x −4 © : Pre-Calculus - Chapter 1D Compound Fractions A compound fraction (also called a complex fraction) is an expression which contains nested fractions. In general, there is one ”main fraction” which will have one or more fractions in the numerator and/or denominator. There are two strategies which may be used to simplify compound fractions. First, simplify both the numerator and the denominator individually, then divide the numerator by the denominator by multiplying with the reciprocal of the denominator. 5 6 − y 2y 1 Example 1: 6 4 y Solution: First simplify the numerator using the common denominator y2y 1: Example 5y 62y 1 5 6 − − y y2y 1 2y 1y 2y 1 6 4 6 4 y y 12y 6 − 5y y2y 1 6 4 y 7y 6 y2y 1 6 4 y Now simplify the denominator using the common denominator y (remember that 4 4 : 1 7y 6 5 6 − y y2y 1 2y 1 6 4 6 4y y y 1y 7y 6 y2y 1 6 4y y 7y 6 6 4y We now divide the two fractions and y : y2y 1 5 6 − y 2y 1 7y 6 6 4y y 6 4 y2y 1 y 7y 6 y y2y 1 6 4y 7y 6 ; y ≠ −3 , −1 , 0 . 2 2 2y 16 4y As you can see, this is often a long, tedious process. A second technique is to multiply the numerator and the denominator of the ”main fraction” by the LCD of all the smaller fractions and simplify the result. © : Pre-Calculus © : Pre-Calculus - Chapter 1D 5 6 − y 2y 1 Example 2: 6 4 y Solution: The LCD of all the smaller fractions is y2y 1. We multiply the numerator and denominator by this LCD: 5 5 6 − 6 − y2y 1 y y 2y 1 2y 1 6 4 6 4 y2y 1 y y 6 y2y 1 − 5 y2y 1 y 2y 1 6 y2y 1 4y2y 1 y 62y 1 − 5y 62y 1 4y2y 1 12y 6 − 5y 2y 16 4y 7y 6 2y 16 4y The following example is a common one in many calculus classes: 2x h 1 −1 − 2x 1 −1 Example 3: Simplify the expression . h First rewrite the expression without negative exponents. (Recall x −1 1x ) 2 − 2 2x h 1 −1 − 2x 1 −1 x1 x h 1 h h Method I Solution: Now simplify the numerator using the common denominator x h 1x 1: 2x h 1 2x 1 − 2x h 1 −1 − 2x 1 −1 x 1x h 1 x h 1x 1 h h 2x 2 − 2x 2h 2 x h 1x 1 h 2x 2 − 2x − 2h − 2 x h 1x 1 h −2h x h 1x 1 h h The denominator is h which is already simplified. We now divide the two fractions 1 −2h h and : 1 x h 1x 1 2x h 1 −1 − 2x 1 −1 −2h 1 h x h 1x 1 h −2 ; h≠0 x h 1x 1 Solution: © : Pre-Calculus © : Pre-Calculus - Chapter 1D Method II Solution: The LCD of all the smaller fractions is x h 1x 1. We mulitply the numerator and denominator by this LCD: 2 − 2 x h 1x 1 x1 xh1 hx h 1x 1 © 2 2 x h 1x 1 − x h 1x 1 x1 xh1 hx h 1x 1 2x 1 − 2x h 1 hx h 1x 1 2x 2 − 2x − 2h − 2 hx h 1x 1 −2h −2 ; h≠0 hx h 1x 1 x h 1x 1 : Pre-Calculus © : Pre-Calculus - Chapter 1D Exercises for Chapter 1D - Rational Expressions 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. For problems 1-5, determine if the given expression is a rational expression. If it is a rational expression, state any restrictions on the variable. 3x 2 − 6x 7 5x − 10 5x 10 x8 x 2 4x 6 5x 7 4 2 x−4 For problems 6-8, reduce the rational expression if possible. x3y x2y2 x 2 4x − 12 5x − 10 x 3 − 4x x2 x − 2 For problems 9-20, perform the indicated operations and simplify. x3 8 x3 − 8 2 2 x − 4 x 2x 4 z 2 − 5z − 24 9z 2 − 12z 4 z 2 − z − 6 9z − 12 z2 − 9 z 2 − 10z 16 3 6x 9 x−1 x 2 − 9x 8 y 2 8y − 20 y3 1 y2 − y 1 y 2 11y 10 y 3 − 8 y 2 2y 4 3 y − 27 4y 2y − 6 2 2 3 y − 5y 6 y − 2y 2y x 2 x−3 3x 4 1 − 3 x3 x−2 1 1 − 10 x3 x−3 x2 − 9 7 8x − 4 2x 1 2x − 1 3 − 2 x3 x x−1 x2 − 1 2 y − 7y 12 y 2 − 25 (HINT: Simplify first) y 2 3y − 18 y 2 − 8y 15 y−7 − 5 y1 y 2 2y 1 For problems 21-25, simplify the expression completely. 3x y 4 21. xy 2 © : Pre-Calculus © 22. 23. 24. 25. © : Pre-Calculus - Chapter 1D 5 x x−1 x1 2 1 x2 − 1 1 1 1 a b 1 − 1 x 2 x−2 x h −2 x −2 h : Pre-Calculus © : Pre-Calculus - Chapter 1D Answers to Exercises for Chapter 1D - Rational Expressions 1. 2. This is a rational function and there are no restrictions on the variable x. Is a rational function Restriction: 5x 10 ≠ 0 5x ≠ −10 x ≠ −2 3. 4. 5. Not a rational function because of the radical in the numerator. This is a rational function and there are no restrictions on the variable x. Is a rational function. Restriction: x−4 ≠ 0 x≠4 6. x2 x y x3y 2 xy ; x ≠ 0 and y ≠ 0 2 2 x y x yy 7. x 2 4x − 12 x 6x − 2 x 6 ; x ≠ 2 5 5x − 10 5x − 2 8. x 3 − 4x xx 2x − 2 xx − 2 ; x ≠ −2 and x ≠ 1 x−1 x 2x − 1 x x−2 2 9. x − 2x 2 2x 4 x 2x 2 − 2x 4 x3 − 8 x3 8 x 2 − 2x 4; x ≠ 2, −2 2 2 x 2x − 2 x − 4 x 2x 4 x 2 2x 4 10. z 2 − 5z − 24 9z 2 − 12z 4 z 2 − z − 6 9z − 12 z2 − 9 z 2 − 10z 16 2 z − 8z 3 3z − 2 z − 3z 2 33z − 4 z 3z − 3 z − 2z − 8 2 z2 1 3z − 2 z ≠ −3, 3, 8, −2, 4 3 3 z − 23z − 4 11. 2 x − 8x − 1 3 6x 9 3 x − 9x 8 3 x − 8 ; x ≠ 1, 8 2 x−1 x−1 2x 3 6x 9 x−1 32x 3 x − 9x 8 12. © : Pre-Calculus © : Pre-Calculus - Chapter 1D y 2 8y − 20 y3 1 y2 − y 1 3 2 2 y 11y 10 y − 8 y 2y 4 y 1y 2 − y 1 y 10y − 2 y 2 2y 4 y 10y 1 y − 2y 2 2y 4 y2 − y 1 1; y ≠ −10, −1, 2 13. 4y 2y − 6 y 3 − 27 2 2 y − 5y 6 y − 2y 2y 3 y − 3y 2 3y 9 yy − 2 4y 3 y − 2y − 3 2y − 3 2y 2 y 3y 9 ; y≠2 yy − 3 14. LCD: x − 33x 4 2 3x 2 4x 2x − 6 x3x 4 2x − 3 x 2 3x 6x − 6 x−3 3x 4 x − 33x 4 x − 33x 4 3x 4x − 3 x − 33x 4 15. LCD: x − 2x 3 1x 3 3x − 2 x 3 − 3x − 6 −2x 9 1 − 3 − x3 x−2 x − 2x 3 x 3x − 2 x − 2x 3 x − 2x 3 16. LCD: x 3x − 3 1x − 3 1x 3 10 − x 3x − 3 x − 3x 3 x 3x − 3 x − 3 x 3 − 10 x 3x − 3 2x − 10 x 3x − 3 17. 7 − 8x 4 2x 1 2x − 1 1 LCD: 2x 12x − 1 72x − 1 8x2x 1 42x − 12x 1 − 2x 12x − 1 2x − 12x 1 2x − 12x 1 14x − 7 − 16x 2 8x 16x 2 − 4 2x 12x − 1 6x − 11 2x 12x − 1 18. 3 − 2 x3 x x−1 x2 − 1 x3 3 − 2x x−1 x 1x − 1 LCD: xx 1x − 1 © : Pre-Calculus © : Pre-Calculus - Chapter 1D 3xx 1 2x 1x − 1 x 3x − xx 1x − 1 x − 1xx 1 x 1x − 1x 2 2 2 3x 3x − 2x − 2 x 3x xx − 1x 1 2 2x 6x 2 xx − 1x 1 19. y 2 − 25 y 2 − 7y 12 2 2 y 3y − 18 y − 8y 15 y − 4y − 3 y 5y − 5 y 6y − 3 y − 5y − 3 y−4 y5 ; y≠5 y6 y−3 So LCD is y 6y − 3 y − 4y − 3 y 5y 6 y 6y − 3 y − 3y 6 y 2 − 7y 12 y 2 11y 30 y 6y − 3 2 2y 4y 42 ; y≠5 y 6y − 3 20. y−7 − 5 y1 y 2y 1 y−7 − 5 y1 y 1 2 2 So LCD y 1 2 5y 1 y−7 − 2 1 y y 1 2 y − 7 − 5y 5 y 1 2 −4y − 12 y 1 2 21. 3x y 4 3x y 4 3 xy 2 2x y 4 2 22. © : Pre-Calculus © : Pre-Calculus - Chapter 1D 5 x x1 x−1 2 1 x2 − 1 5 x x 1x − 1 x1 x−1 2 1 x 1x − 1 x 1x − 1 5x − 1 xx 1 2 x 1x − 1 5x − 5 x 2 x 2 x2 − 1 2 x 6x − 5 ; x ≠ 1, −1 x2 1 23. 1ab ab ; a ≠ 0; b ≠ 0 ba 1 1 ab a b 24. 1 − 1 x 2 x−2 1 − 1 2x x 2 x − 2 2x 1 2−x 2xx − 2 −x − 2 2xx − 2 −1 2x 25. x h −2 − x −2 h © : Pre-Calculus 1 − 12 x h 2 x 2 2 x x h hx h 2 x 2 x 2 − x h 2 hx 2 x h 2 x 2 − x 2 2xh h 2 hx 2 x h 2 x 2 − x 2 − 2xh − h 2 hx 2 x h 2 −2xh − h 2 hx 2 x h 2 h−2x − h hx 2 x h 2 −2x − h ; h ≠ 0 2 x x h 2 © : Pre-Calculus - Chapter 1E Chapter 1E - Complex Numbers Definition of a Complex Number Not everything can be done using real numbers. For example, there is no real number x such that x 2 −1. To handle this, mathematicians (such as William R. Hamilton 1805-1865) expanded the real number system by introducing the imaginary unit i −1 . Putting real and imaginary numbers together creates the Complex Number System. Definition A complex number is a number that can be written in the form a bi, where a and b are real numbers. a is referred to as the real part and b is referred to as the imaginary part. This is called the standard form of the complex number. Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal. Example 1: Solution: Write the number 24 − −64 in its standard form a bi. We begin by simplifying the radicals, noting that −64 64 −1 : 24 − −64 4 6 − 64 −1 2 6 − 8 −1 2 6 − 8i The real part is 2 6 and the imaginary part is −8. Example 2: Solution: Find real numbers a and b such that 2a − 1 − 4i −7 Find a by setting the real parts equal: 2a − 1 −7 b i 3 2a −6 a −3 Then find b by setting the imaginary parts equal: −4 b 3 − 12 b We now define the absolute value of a complex number. If z a bi is a complex number in standard form, the absolute value of z, |z|, is defined as |z| |a bi| a 2 b 2 . If z 1 and z 2 are two complex numbers, then the distance between these two numbers is defined to be (We’ll see how to subtract complex numbers soon.) |z 1 − z 2 | If z is a real number, then the absolute value of z is the same as its real number absolute value. © : Pre-Calculus © : Pre-Calculus - Chapter 1E Example 3: Compute the absolute values of the following complex numbers. 2 2 −3 2 |2 − 3i| |5| |5 0i| |−7| |−7 0i| |7 11i| |16 − 5i| |−5| 49 52 02 5 2 5. −7 2 0 2 7 2 11 2 16 2 5 2 13 . 49 7. 49 121 281 . 170 . −5 2 0 2 5. Some properties of the absolute value of a complex number: 1. 2. For z a bi a complex number, the absolute value of z represents the distance from the origin 0, 0 to the point with coordinates a, b. If you have not learned about the Cartesian coordinate system for the Euclidean plane yet, then ignore this for now. Let z 1 and z 2 be any two complex numbers, then |z 1 z 2 | |z 1 ||z 2 | . 3. Let z 1 and z 2 be any two complex numbers, then z 1 |z 1 | . z2 |z 2 | 4. Let z 1 and z 2 be any two complex numbers, then |z 1 z 2 | ≤ |z 1 | |z 2 | . This inequality is called the triangle inequality. Example 4: Compute the absolute value of the following compex numbers. 2 − 5i, 7 3i, −11 − 4i, and 3i. Solution: |2 − 5i| 4 25 29 ≈ 5. 385 2 |7 3i| 49 9 58 ≈ 7. 615 8 |−11 − 4i| |3i| 121 16 137 ≈ 11. 705 9 3 Example 5: Show that the absolute value of the product of 3 − 2i and −4 i equals the product of their absolute values. Solution: |3 − 2i−4 i| |−10 11i| |3 − 2i||−4 i| © : Pre-Calculus 100 121 9 4 16 1 13 17 221 13 ∘ 17 221 . © : Pre-Calculus - Chapter 1E Example 6: Show that the absolute value of the sum of −8 i and 4 3i is less than or equal to the sum of their absolute values. Solution: |−8 i 4 3i| |−4 4i| |−8 i| |4 3i| We now need to show that 32 ≤ 32 65 . 16 16 64 1 16 9 32 65 25 65 5 . 65 5, but this is obvious since 32 65 and therefore We now define the conjugate of a complex number. The conjugate of the complex number z a bi is defined to be a − bi, where we assume that z is in standard form. The conjugate of a complex number z is commonly denoted by writing a line above z. That is, z̄ is used to denote the conjugate of z. Example 7: Compute the conjugate of the following complex numbers: 1 − i, 3i, −8, and −3 − 6i. Solution: The conjugate of 1 − i 1 − i 1 i. The conjugate of 3i 3i −3i. The conjugate of −8 −8 −8. The conjugate of −3 − 6i −3 − 6i −3 6i. Some Properties of the Conjugate of a Complex Number In the following z, z 1 , and z 2 represent arbitrary complex numbers. 1. z z |z| 2 , a bia − bi a 2 b 2 0i a 2 b 2 |a bi| 2 . 2. If x is a real number, then x x. 3. z̄ z (The conjugate of the conjugate of z equals z.) 4. z 1 z 2 z 1 z 2 (The conjugate of a sum is the sum of the conjugates.) 5. z 1 z 2 z 1 z 2 (The conjugate of a product is the product of the conjugates.) z 1 z 1 (The conjugate of a quotient is the quotient of the conjugates.) 6. z2 z2 © : Pre-Calculus © : Pre-Calculus - Chapter 1E Exercises for Chapter 1E - Complex Numbers 1. Write each of the following in standard form: −2 −16 , 5 − −50 , 2i 2 − 4i, − 9 − 3 4 4 2. Find real numbers a and b such that a bi 14i − 3 3. Find real numbers a and b such that 2a − 5 − 6i 11 − 3bi 4. What is the absolute value and conjugate of z 5? 5. What is the absolute value and conjugate of z −7i? 6. If the conjugate of the complex number z equals −2 7i, what does z equal? 7. What is the absolute value and conjugate of z −3 − 5i? 8. If the real part of z equals −5 and the absolute value of z 5, what must z equal? 9. If the imaginary part of z equals −13, what does the imaginary part of z̄ equal? 10. If the real part of z equals −23, what does the real part of z̄ equal? 11. If the real and imaginary parts of a complex number are equal, and the real part equals −5, what is the complex number? 12. What are the real and imaginary parts of the complex number 2 − 5i? © : Pre-Calculus © : Pre-Calculus - Chapter 1E Answers to Exercises for Chapter 1E - Complex Numbers 1. 2. 3. −2 −16 −2 4i 5 − −50 5 − 5 2 i 2i 2 − 4i −2 − 4i −9 − 3 −3 3i 4 4 4 2 a −3 b 14 2a − 5 − 6i 11 − 3bi This equation of complex numbers implies the following two real equations. 2a − 5 11 − 6 −3b 4. 5. 6. 7. Thus, 2a 16 or a 8, and b −6 2. −3 5̄ 5 |5| 5 −7i 7i |−7i| 7 Since z̄ −2 7i, and z̄ z, we have z z̄ −2 7i −2 − 7i. −3 − 5i −3 5i |−3 − 5i| 9 25 34 Suppose z a bi. We know that a −5. Thus, 5 |z| −5 2 b 2 .Squaring both sides of this equality, we have 25 25 b 2 , which implies that b 0. Thus, z −5. 9. 13 10. −23 11. −5 − 5i 12. Real part of 2 − 5i is 2, and the imaginary part of 2 − 5i is −5. 8. © : Pre-Calculus