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Appendix A2
SUMMATION NOTATION
Factoring Polynomials and Solving Systems of Equations
605
In Problems 55 through 58, evaluate the given sum.
4
55.
5
(3j 1)
56.
j1
10
57.
(1)
j
2
j1
5
j
58.
j1
2
j
j1
In Problems 59 through 64, use summation notation to express the given sum.
59. 1 1
1
1
1
1
2
3
4
5
6
60. 3 6 9 12 15 18 21 24 27 30
61. 2x1 2x2 2x3 2x4 2x5 2x6
62. 1 1 1 1 1 1
63. 1 2 3 4 5 6 7 8
64. x x2 x3 x4 x5
ECOLOGY
A2
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 4R2 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 0.9 1013 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)?*
A polynomial is an expression of the form
a0 a1x a2x2 . . . anxn
Factoring
Polynomials
and Solving
Systems of
Equations
where n is a nonnegative integer and a0, a1, a2, . . . , an are real numbers, known as
the coefficients of the polynomial. If an 0, n is said to be the degree of the polynomial. For example, 3x5 7x2 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.
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FACTORING POLYNOMIALS WITH
INTEGER COEFFICIENTS
Appendix
Algebra Review
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 x2 2x 3 using integer coefficients.
Solution
The goal is to write the polynomial as a product of the form
x2 2x 3 (x a)(x b)
where a and b are integers. The distributive law implies that
(x a)(x b) x2 (a b)x ab
Hence, the goal is to find integers a and b such that
x2 2x 3 x2 (a b)x ab
or, equivalently, such that
a b 2
and
ab 3
From the list
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
x2 2x 3 (x 3)(x 1)
which you should check by multiplying out the right-hand side.
EXAMPLE A2.2
Factor the polynomial 2x2 x 10 using integer coefficients.
Solution
We wish to find integers a, b, c, and d so that
2x2 x 10 (ax b)(cx d)
(distributive law)
2
(ac)x (bc ad)x (bd)
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
2x2 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?
EXAMPLE A2.3
Factor the polynomial x3 8 using integer coefficients.
Solution
The fact that 23 8 tells you that x 2 must be a factor of this expression. That is,
there are integers a and b for which
x3 8 (x 2)(x2 ax b)
Since
(x 2)(x2 ax b) x3 (a 2)x2 (b 2a)x 2b
(which you should check for yourself), the goal is to find integers a and b for which
a20
b 2a 0
and
2b 8
Clearly, the only such integers are a 2 and b 4. Hence,
x3 8 (x 2)(x2 2x 4)
Convince yourself by examining pairs of integers whose product is 4 that the polynomial x2 2x 4 cannot be factored further with integer coefficients.
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THE DIFFERENCE
OF TWO SQUARES
Appendix
Algebra Review
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,
a2 b2 (a b)(a b)
The use of this formula is illustrated in the next example.
EXAMPLE A2.4
Factor the polynomial x5 4x3 using integer coefficients.
Solution
First factor out x3 to get
x5 4x3 x3(x2 4)
and then factor x2 4 (which is the difference of two squares) to conclude that
x5 4x3 x3(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
x3 6x2 12x 8 0
because substitution of 2 for x gives
23 6(22) 12(2) 8 8 24 24 8 0
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 x2 3x 10.
Appendix A2
Factoring Polynomials and Solving Systems of Equations
609
Solution
First subtract 10 from both sides to get
x2 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).
EXAMPLE A2.6
Solve the equation 1 1
2
2 0.
x
x
Solution
Put the fractions on the left-hand side over the common denominator x2 and add to
get
x2
x
2
2 2 2 0
x
x
x
or
x2 x 2
0
x2
Now factor the polynomial in the numerator to get
(x 1)(x 2)
0
x2
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.
THE SOLUTION OF QUADRATIC
EQUATIONS
An equation of the form
ax2 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
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
ax bx c 0
2
(for a 0)
are given by the formula
x
b b2 4ac
2a
The term b2 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
b
formula reduces to x . 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 x2 3x 1 0.
Solution
This is a quadratic equation with a 1, b 3, and c 1. Using the quadratic formula, you get
x
3 5
0.38
2
EXAMPLE A2.8
Solve the equation x2 18x 81 0.
and
x
3 5
2.62
2
Appendix A2
Factoring Polynomials and Solving Systems of Equations
611
Solution
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
18
18 0
9
2
2
EXAMPLE A2.9
Solve the equation x2 x 1 0.
Solution
This is a quadratic equation with a 1, b 1, and c 1. Using the quadratic formula, you get
x
1 3
2
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.
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Appendix
Algebra Review
EXAMPLE A2.10
ELIMINATION BY
MULTIPLICATION AND ADDITION
Solve the system
4x 3y 13
3x 2y 7
Solution
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
8x 6y 26
9x 6y 21
Then add the equations to get
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
4(5) 3(11) 20 33 13
and from the second equation you get
3(5) 2(11) 15 22 7
as required.
EXAMPLE A2.11
ELIMINATION BY SUBSTITUTION
Solve the system
2y2 x2 14
xy1
Solution
Solve the second equation for x to get
xy
1
Appendix A2
Factoring Polynomials and Solving Systems of Equations
613
and substitute this into the first equation to eliminate x. This gives
2y2 (y 1)2 14
2y2 (y2 2y 1) 14
2y2 y2 2y 1 14
y2 2y 15 0
or
(y 3)(y 5) 0
from which it follows that
y 3
y5
or
If y 3, the second equation gives
x (3) 1
or
x 2
and if y 5, the second equation gives
x51
or
x6
Hence the system has two solutions,
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
2(52) 62 50 36 14
and if x 2 and y 3, you get
2(3)2 (2)2 18 4 14
as required.
P . R . O . B . L . E . M . S
FACTORING POLYNOMIALS WITH
INTEGER COEFFICIENTS
A2
In Problems 1 through 14 factor the given polynomial using integer coefficients.
1. x2 x 2
2. x2 3x 10
3. x2 7x 12
4. x2 8x 12
5. x2 2x 1
6. x2 6x 9
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Appendix
Algebra Review
7. 16x2 25
9. x3 1
SOLUTION OF EQUATIONS
BY FACTORING
QUADRATIC FORMULA
SYSTEMS OF EQUATIONS
8. 3x2 x 14
10. x3 27
11. x7 x5
12. x3 2x2 x
13. 2x3 8x2 10x
14. x4 5x3 14x2
In Problems 15 through 30 solve the given equation by factoring.
15. x2 2x 8 0
16. x2 4x 3 0
17. x2 10x 25 0
18. x2 8x 16 0
19. x2 16 0
20. x2 25 0
21. 2x2 3x 1 0
22. x2 2x 1 0
23. 4x2 12x 9 0
24. 6x2 7x 3 0
25. 1 4
5
20
x
x
26.
9
6
10
x2
x
27. 2 2
4
20
x
x
28.
3
5
20
2 x
x
29.
x
4
10
0
x 2 x 3 x2 x 6
30.
3
11x 10
x
2
0
x 1 2x 3 2x 5x 3
In Problems 31 through 36 use the quadratic formula to solve the given equation.
31. 2x2 3x 1 0
32. x2 3x 1 0
33. x2 2x 3 0
34. x2 2x 1 0
35. 4x2 12x 9 0
36. x2 12 0
In Problems 37 through 42 solve the given system of equations.
37. x 5y 13
3x 10y 11
38. 2x 3y 4
3x 5y 2
39. 5x 4y 12
2x 3y 2
40. 3x2 9y 0
3y2 9x 0
41. 2y2 x2 1
x 2y 3
42. 2x2 y2 7
2x y 1