In Problems 55 through 58, evaluate the given sum. 55. (3j 1) 56.

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1 Appendix A2 Factoring Polynomials and Solving Systems of Equations 605 SUMMATION NOTATION In Problems 55 through 58, evaluate the given sum. 4 j 1 10 j (3j 1) j 2 j 1 5 j ( 1) j j In Problems 59 through 64, use summation notation to express the given sum x 1 2x 2 2x 3 2x 4 2x 5 2x x x 2 x 3 x 4 x 5 ECOLOGY 65. Every square centimeter of the earth s surface is loaded with 1 kg of atmosphere. (a) Assuming the earth is a sphere of radius R 6,440 km, use the formula S 4 R 2 to calculate the surface area of the earth and then find the total mass of the atmosphere. (b) Oxygen occupies approximately 22% of the total mass of the atmosphere, and it is estimated that plant life produces approximately kg of oxygen per year. If none of this oxygen were used up by plants or animals (or combustion), how long would it take to build up the total mass of oxygen in the atmosphere (part a)?* A2 Factoring Polynomials and Solving Systems of Equations A polynomial is an expression of the form a 0 a 1 x a 2 x 2... a n x n where n is a nonnegative integer and a 0, a 1, a 2,..., a n are real numbers, known as the coefficients of the polynomial. If a n 0, n is said to be the degree of the polynomial. For example, 3x 5 7x 2 12 is a polynomial of degree 5. Polynomial expressions appear throughout calculus, as examples and in practical applications. In this section, we shall study techniques for factoring polynomials and shall also see how to solve systems of equations involving polynomials. * Adapted from a problem in E. Batschelet, Introduction to Mathematics for Life Scientists, 2nd ed., New York: Springer-Verlag, 1976, page 31.

2 606 Appendix Algebra Review FACTORING POLYNOMIALS WITH INTEGER COEFFICIENTS Many of the polynomials that arise in practice have integer coefficients (or are closely related to polynomials that do). Techniques for factoring polynomials with integer coefficients are illustrated in the following examples. In each, the goal is to rewrite the given polynomial as a product of polynomials of lower degree that also have integer coefficients. EXAMPLE A2.1 Factor the polynomial x 2 2x 3 using integer coefficients. The goal is to write the polynomial as a product of the form x 2 2x 3 (x a)(x b) where a and b are integers. The distributive law implies that (x a)(x b) x 2 (a b)x ab Hence, the goal is to find integers a and b such that or, equivalently, such that From the list x 2 2x 3 x 2 (a b)x ab a b 2 and ab 3 1, 3 and 1, 3 of pairs of integers whose product is 3, choose a 3 and b 1 as the only pair whose sum is 2. It follows that x 2 2x 3 (x 3)(x 1) which you should check by multiplying out the right-hand side. EXAMPLE A2.2 Factor the polynomial 2x 2 x 10 using integer coefficients. We wish to find integers a, b, c, and d so that 2x 2 x 10 (ax b)(cx d) (distributive law) (ac)x 2 (bc ad)x (bd)

3 Appendix A2 Factoring Polynomials and Solving Systems of Equations 607 so we must have ac 2 bc ad 1 bd 10 Because a, b, c, and d must all be integers, there are only a limited number of possibilities for the choices of a, c and b, d; namely, 4 possible choices for the pair a, c: 2, 1; 1, 2; 2, 1; and 1, 2 8 possible choices for the pair b, d: 2, 5; 5, 2; 1, 10; 10, 1; 2, 5; 5, 2; 1, 10; and 10, 1 There are (4)(8) 32 possible ways of forming the expression bc ad. We find that the condition bc ad 1 is satisfied when a 2, b 5, c 1, d 2, so that 2x 2 x 10 (2x 5)(x 2) (Note that bc ad 1 is also satisfied by a 2, b 5, c 1, and d 2. Why is it not necessary to list this as a second factorization of the given polynomial? Factor the polynomial x 3 8 using integer coefficients. The fact that tells you that x 2 must be a factor of this expression. That is, there are integers a and b for which Since EXAMPLE A2.3 x 3 8 (x 2)(x 2 ax b) (x 2)(x 2 ax b) x 3 (a 2)x 2 (b 2a)x 2b (which you should check for yourself), the goal is to find integers a and b for which a 2 0 b 2a 0 and 2b 8 Clearly, the only such integers are a 2 and b 4. Hence, x 3 8 (x 2)(x 2 2x 4) Convince yourself by examining pairs of integers whose product is 4 that the polynomial x 2 2x 4 cannot be factored further with integer coefficients.

4 608 Appendix Algebra Review THE DIFFERENCE OF TWO SQUARES Here is a formula (which you can verify by multiplication) for the factorization of the difference between two perfect squares. It is particularly useful, and you should memorize it. The Difference of Two Squares For any real numbers a and b, a 2 b 2 (a b)(a b) The use of this formula is illustrated in the next example. EXAMPLE A2.4 Factor the polynomial x 5 4x 3 using integer coefficients. First factor out x 3 to get x 5 4x 3 x 3 (x 2 4) and then factor x 2 4 (which is the difference of two squares) to conclude that x 5 4x 3 x 3 (x 2)(x 2) THE SOLUTION OF EQUATIONS BY FACTORING The solutions of an equation are the values of the variable that make the equation true. For example, x 2 is a solution of the equation because substitution of 2 for x gives x 3 6x 2 12x (2 2 ) 12(2) In the following two examples, you will see how factoring can be used to solve certain equations. The technique is based on the fact that if the product of two (or more) terms is equal to zero, then at least one of the terms must be equal to zero. For example, if ab 0, then either a 0 or b 0 (or both). EXAMPLE A2.5 Solve the equation x 2 3x 10.

5 Appendix A2 Factoring Polynomials and Solving Systems of Equations 609 First subtract 10 from both sides to get x 2 3x 10 0 and then factor the resulting polynomial on the left-hand side to get (x 5)(x 2) 0 Since the product (x 5)(x 2) can be zero only if one (or both) of its factors is zero, it follows that the solutions are x 5 (which makes the first factor zero) and x 2 (which makes the second factor zero). 1 2 Solve the equation 1 0. x x 2 Put the fractions on the left-hand side over the common denominator x 2 and add to get or EXAMPLE A2.6 x 2 x 2 0 x 2 x 2 x 2 x 2 x 2 x 2 0 Now factor the polynomial in the numerator to get (x 1)(x 2) x 2 A quotient is zero only if its numerator is zero and its denominator is not zero, so it follows that x 1 and x 2 are the required solutions. 0 THE SOLUTION OF QUADRATIC EQUATIONS An equation of the form ax 2 bx c 0 (for a 0) is said to be a quadratic equation. A quadratic equation can have at most two solutions. As you have seen, one way to find the solutions is to factor the equation. When

6 610 Appendix Algebra Review the factors are not obvious or when the equation cannot be factored at all, you can use the following special formula to solve quadratic equations. The Quadratic Formula The solutions of the quadratic equation are given by the formula ax 2 bx c 0 (for a 0) x b b2 4ac 2a The term b 2 4ac in the quadratic formula is called the discriminant of the quadratic equation. If the discriminant is positive, the equation has two solutions, one coming from the formula with the sign replaced by and the other with replaced by. If the discriminant is zero, the equation has only one solution since the formula reduces to x b. If the discriminant is negative, the equation has no real 2a solutions since negative numbers do not have real square roots. The use of the quadratic formula is illustrated in the following examples. EXAMPLE A2.7 Solve the equation x 2 3x 1 0. This is a quadratic equation with a 1, b 3, and c 1. Using the quadratic formula, you get x and x EXAMPLE A2.8 Solve the equation x 2 18x 81 0.

7 Appendix A2 Factoring Polynomials and Solving Systems of Equations 611 This is a quadratic equation with a 1, b 18, and c 81. Using the quadratic formula, you find that the discriminant is zero and that the formula for x gives x EXAMPLE A2.9 Solve the equation x 2 x 1 0. This is a quadratic equation with a 1, b 1, and c 1. Using the quadratic formula, you get x Since there is no real square root of 3, it follows that the equation has no real solution. SYSTEMS OF EQUATIONS A collection of equations that are to be solved simultaneously is called a system of equations. Some of the calculus problems in Chapter 7 involve the solution of systems of two (or more) equations in two (or more) unknowns. A typical example is to find the real numbers x and y that satisfy the system 2x 3y 5 x 2y 4 The procedure for solving a system of two equations in two unknowns is to (temporarily) eliminate one of the variables, thereby reducing the problem to a single equation in one variable, which you then solve for its variable. Once you have found the value of one of the variables, you can substitute it into either of the original equations and solve to get the value of the other variable. The most common techniques for the elimination of variables are illustrated in the next two examples.

8 612 Appendix Algebra Review EXAMPLE A2.10 ELIMINATION BY MULTIPLICATION AND ADDITION Solve the system 4x 3y 13 3x 2y 7 To eliminate y, multiply both sides of the first equation by 2 and both sides of the second equation by 3 so that the system becomes Then add the equations to get 8x 6y 26 9x 6y 21 x 0 5 or x 5 To find y, you can substitute x 5 into either of the original equations. If you choose the second equation, you get 3( 5) 2y 7 2y 22 or y 11 That is, the solution of the system is x 5 and y 11. To check this answer, substitute x 5 and y 11 into each of the original equations. From the first equation you get and from the second equation you get as required. 4( 5) 3(11) ( 5) 2(11) EXAMPLE A2.11 ELIMINATION BY SUBSTITUTION Solve the system Solve the second equation for x to get 2y 2 x 2 14 x y 1 x y 1

9 Appendix A2 Factoring Polynomials and Solving Systems of Equations 613 and substitute this into the first equation to eliminate x. This gives or from which it follows that If y 3, the second equation gives and if y 5, the second equation gives Hence the system has two solutions, 2y 2 (y 1) y 2 (y 2 2y 1) 14 2y 2 y 2 2y 1 14 y 2 2y 15 0 (y 3)(y 5) 0 y 3 or y 5 x ( 3) 1 or x 2 x 5 1 or x 6 x 6, y 5 and x 2, y 3 To check these answers, substitute each pair x, y into the first equation. If x 6 and y 5, you get and if x 2 and y 3, you get as required. 2(5 2 ) ( 3) 2 ( 2) P. R. O. B. L. E. M. S A2 P. R. O. B. L. E. M. S A2 FACTORING POLYNOMIALS WITH INTEGER COEFFICIENTS In Problems 1 through 14 factor the given polynomial using integer coefficients. 1. x 2 x 2 2. x 2 3x x 2 7x x 2 8x x 2 2x 1 6. x 2 6x 9

10 614 Appendix Algebra Review SOLUTION OF EQUATIONS BY FACTORING QUADRATIC FORMULA SYSTEMS OF EQUATIONS 7. 16x x 2 x x x x 7 x x 3 2x 2 x 13. 2x 3 8x 2 10x 14. x 4 5x 3 14x 2 In Problems 15 through 30 solve the given equation by factoring. 15. x 2 2x x 2 4x x 2 10x x 2 8x x x x 2 3x x 2 2x x 2 12x x 2 7x x x 2 x 2 x x x 2 x 2 x x 29. x 2 4 x 3 10 x 2 x 6 0 x 30. x x 10 2x 3 2x 2 5x 3 0 In Problems 31 through 36 use the quadratic formula to solve the given equation x 2 3x x 2 3x x 2 2x x 2 2x x 2 12x x In Problems 37 through 42 solve the given system of equations. 37. x 5y x 3y 4 3x 10y 11 3x 5y x 4y x 2 9y 0 2x 3y 2 3y 2 9x y 2 x x 2 y 2 7 x 2y 3 2x y 1