Instructor’s Solutions Manual, Section 2.4 Exercise 1
Solutions to Exercises, Section 2.4
Suppose
p(x) = x2 + 5x + 2,
q(x) = 2x3 − 3x + 1,
s(x) = 4x3 − 2.
In Exercises 1–18, write the indicated expression as a sum of terms, eachof which is a constant times a power of x.
1. (p + q)(x)solution
(p + q)(x) = (x2 + 5x + 2)+ (2x3 − 3x + 1)
= 2x3 + x2 + 2x + 3
Instructor’s Solutions Manual, Section 2.4 Exercise 2
2. (p − q)(x)solution
(p − q)(x) = (x2 + 5x + 2)− (2x3 − 3x + 1)
= −2x3 + x2 + 8x + 1
Instructor’s Solutions Manual, Section 2.4 Exercise 3
3. (3p − 2q)(x)
solution
(3p − 2q)(x) = 3(x2 + 5x + 2)− 2(2x3 − 3x + 1)
= 3x2 + 15x + 6− 4x3 + 6x − 2
= −4x3 + 3x2 + 21x + 4
Instructor’s Solutions Manual, Section 2.4 Exercise 4
4. (4p + 5q)(x)
solution
(4p + 5q)(x) = 4(x2 + 5x + 2)+ 5(2x3 − 3x + 1)
= 4x2 + 20x + 8+ 10x3 − 15x + 5
= 10x3 + 4x2 + 5x + 13
Instructor’s Solutions Manual, Section 2.4 Exercise 5
5. (pq)(x)
solution
(pq)(x) = (x2 + 5x + 2)(2x3 − 3x + 1)
= x2(2x3 − 3x + 1)
+ 5x(2x3 − 3x + 1)+ 2(2x3 − 3x + 1)
= 2x5 − 3x3 + x2 + 10x4 − 15x2
+ 5x + 4x3 − 6x + 2
= 2x5 + 10x4 + x3 − 14x2 − x + 2
Instructor’s Solutions Manual, Section 2.4 Exercise 6
6. (ps)(x)
solution
(ps)(x) = (x2 + 5x + 2)(4x3 − 2)
= x2(4x3 − 2)+ 5x(4x3 − 2)+ 2(4x3 − 2)
= 4x5 − 2x2 + 20x4 − 10x + 8x3 − 4
= 4x5 + 20x4 + 8x3 − 2x2 − 10x − 4
Instructor’s Solutions Manual, Section 2.4 Exercise 7
7.(p(x)
)2
solution
(p(x)
)2 = (x2 + 5x + 2)(x2 + 5x + 2)
= x2(x2 + 5x + 2)+ 5x(x2 + 5x + 2)
+ 2(x2 + 5x + 2)
= x4 + 5x3 + 2x2 + 5x3 + 25x2
+ 10x + 2x2 + 10x + 4
= x4 + 10x3 + 29x2 + 20x + 4
Instructor’s Solutions Manual, Section 2.4 Exercise 8
8.(q(x)
)2
solution
(q(x)
)2 = (2x3 − 3x + 1)2
= (2x3 − 3x + 1)(2x3 − 3x + 1)
= 2x3(2x3 − 3x + 1)− 3x(2x3 − 3x + 1)
+ (2x3 − 3x + 1)
= 4x6 − 6x4 + 2x3 − 6x4 + 9x2
− 3x + 2x3 − 3x + 1
= 4x6 − 12x4 + 4x3 + 9x2 − 6x + 1
Instructor’s Solutions Manual, Section 2.4 Exercise 9
9.(p(x)
)2s(x)
solution Using the expression that we computed for(p(x)
)2in the
solution to Exercise 7, we have
(p(x)
)2s(x)
= (x4 + 10x3 + 29x2 + 20x + 4)(4x3 − 2)
= 4x3(x4 + 10x3 + 29x2 + 20x + 4)
− 2(x4 + 10x3 + 29x2 + 20x + 4)
= 4x7 + 40x6 + 116x5 + 80x4 + 16x3
− 2x4 − 20x3 − 58x2 − 40x − 8
= 4x7 + 40x6 + 116x5 + 78x4
− 4x3 − 58x2 − 40x − 8.
Instructor’s Solutions Manual, Section 2.4 Exercise 10
10.(q(x)
)2s(x)
solution Using the expression that we computed for(q(x)
)2in the
solution to Exercise 8, we have
(q(x)
)2s(x)
= (4x6 − 12x4 + 4x3 + 9x2 − 6x + 1)(4x3 − 2)
= 4x3(4x6 − 12x4 + 4x3 + 9x2 − 6x + 1)
− 2(4x6 − 12x4 + 4x3 + 9x2 − 6x + 1)
= 16x9 − 48x7 + 16x6 + 36x5 − 24x4 + 4x3
− 8x6 + 24x4 − 8x3 − 18x2 + 12x − 2
= 16x9 − 48x7 + 8x6 + 36x5 − 4x3
− 18x2 + 12x − 2.
Instructor’s Solutions Manual, Section 2.4 Exercise 11
11. (p ◦ q)(x)solution
(p ◦ q)(x) = p(q(x))= p(2x3 − 3x + 1)
= (2x3 − 3x + 1)2 + 5(2x3 − 3x + 1)+ 2
= (4x6 − 12x4 + 4x3 + 9x2 − 6x + 1)
+ (10x3 − 15x + 5)+ 2
= 4x6 − 12x4 + 14x3 + 9x2 − 21x + 8
Instructor’s Solutions Manual, Section 2.4 Exercise 12
12. (q ◦ p)(x)solution
(q ◦ p)(x) = q(p(x))= q(x2 + 5x + 2)
= 2(x2 + 5x + 2)3 − 3(x2 + 5x + 2)+ 1
= 2(x2 + 5x + 2)2(x2 + 5x + 2)
− 3x2 − 15x − 5
= 2(x4 + 10x3 + 29x2 + 20x + 4)(x2 + 5x + 2)
− 3x2 − 15x − 5
= 2x6 + 30x5 + 162x4 + 370x3
+ 321x2 + 105x + 11
Instructor’s Solutions Manual, Section 2.4 Exercise 13
13. (p ◦ s)(x)solution
(p ◦ s)(x) = p(s(x))= p(4x3 − 2)
= (4x3 − 2)2 + 5(4x3 − 2)+ 2
= (16x6 − 16x3 + 4)+ (20x3 − 10)+ 2
= 16x6 + 4x3 − 4
Instructor’s Solutions Manual, Section 2.4 Exercise 14
14. (s ◦ p)(x)solution
(s ◦ p)(x) = s(p(x))= s(x2 + 5x + 2)
= 4(x2 + 5x + 2)3 − 2
= 4x6 + 60x5 + 324x4 + 740x3
+ 648x2 + 240x + 30
Instructor’s Solutions Manual, Section 2.4 Exercise 15
15.(q ◦ (p + s))(x)
solution
(q ◦ (p + s))(x) = q((p + s)(x))= q(p(x)+ s(x))= q(4x3 + x2 + 5x)
= 2(4x3 + x2 + 5x)3 − 3(4x3 + x2 + 5x)+ 1
= 2(4x3 + x2 + 5x)2(4x3 + x2 + 5x)
− 12x3 − 3x2 − 15x + 1
= 2(16x6 + 8x5 + 41x4 + 10x3 + 25x2)
× (4x3 + x2 + 5x)− 12x3 − 3x2 − 15x + 1
= 128x9 + 96x8 + 504x7 + 242x6 + 630x5
+ 150x4 + 238x3 − 3x2 − 15x + 1
Instructor’s Solutions Manual, Section 2.4 Exercise 16
16.((q + p) ◦ s)(x)
solution
((q + p) ◦ s)(x) = (q + p)(s(x))= q(s(x))+ p(s(x))= q(4x3 − 2)+ p(4x3 − 2)
= 2(4x3 − 2)3 − 3(4x3 − 2)+ 1
+ (4x3 − 2)2 + 5(4x3 − 2)+ 2
= 128x9 − 176x6 + 88x3 − 13
Instructor’s Solutions Manual, Section 2.4 Exercise 17
17.q(2+ x)− q(2)
xsolution
q(2+ x)− q(2)x
= 2(2+ x)3 − 3(2+ x)+ 1− (2 · 23 − 3 · 2+ 1)x
= 2x3 + 12x2 + 21xx
= 2x2 + 12x + 21
Instructor’s Solutions Manual, Section 2.4 Exercise 18
18.s(1+ x)− s(1)
xsolution
s(1+ x)− s(1)x
= 4(1+ x)3 − 2− (4 · 13 − 2)x
= 4x3 + 12x2 + 12xx
= 4x2 + 12x + 12
Instructor’s Solutions Manual, Section 2.4 Exercise 19
19. Find all real numbers x such that
x6 − 8x3 + 15 = 0.
solution This equation involves x3 and x6; thus we make thesubstitution x3 = y . Squaring both sides of the equation x3 = y givesx6 = y2. With these substitutions, the equation above becomes
y2 − 8y + 15 = 0.
This new equation can now be solved either by factoring the left side orby using the quadratic formula. Let’s factor the left side, getting
(y − 3)(y − 5) = 0.
Thus y = 3 or y = 5 (the same result could have been obtained byusing the quadratic formula).
Substituting x3 for y now shows that x3 = 3 or x3 = 5. Thus x = 31/3
or x = 51/3.
Instructor’s Solutions Manual, Section 2.4 Exercise 20
20. Find all real numbers x such that
x6 − 3x3 − 10 = 0.
solution This equation involves x3 and x6; thus we make thesubstitution x3 = y . Squaring both sides of the equation x3 = y givesx6 = y2. With these substitutions, the equation above becomes
y2 − 3y − 10 = 0.
This new equation can now be solved either by factoring the left side orby using the quadratic formula. Let’s factor the left side, getting
(y − 5)(y + 2) = 0.
Thus y = 5 or y = −2 (the same result could have been obtained byusing the quadratic formula).
Substituting x3 for y now shows that x3 = 5 or x3 = −2. Thus x = 51/3
or x = −21/3.
Instructor’s Solutions Manual, Section 2.4 Exercise 21
21. Find all real numbers x such that
x4 − 2x2 − 15 = 0.
solution This equation involves x2 and x4; thus we make thesubstitution x2 = y . Squaring both sides of the equation x2 = y givesx4 = y2. With these substitutions, the equation above becomes
y2 − 2y − 15 = 0.
This new equation can now be solved either by factoring the left side orby using the quadratic formula. Let’s use the quadratic formula, getting
y = 2±√4+ 602
= 2± 82
.
Thus y = 5 or y = −3 (the same result could have been obtained byfactoring).
Substituting x2 for y now shows that x2 = 5 or x2 = −3. The equationx2 = 5 implies that x = √5 or x = −√5. The equation x2 = −3 has nosolutions in the real numbers. Thus the only solutions to our originalequation x4 − 2x2 − 15 = 0 are x = √5 or x = −√5.
Instructor’s Solutions Manual, Section 2.4 Exercise 22
22. Find all real numbers x such that
x4 + 5x2 − 14 = 0.
solution This equation involves x2 and x4; thus we make thesubstitution x2 = y . Squaring both sides of the equation x2 = y givesx4 = y2. With these substitutions, the equation above becomes
y2 + 5y − 14 = 0.
This new equation can now be solved either by factoring the left side orby using the quadratic formula. Let’s use the quadratic formula, getting
y = −5±√25+ 562
= −5± 92
.
Thus y = 2 or y = −7 (the same result could have been obtained byfactoring).
Substituting x2 for y now shows that x2 = 2 or x2 = −7. The equationx2 = 2 implies that x = √2 or x = −√2. The equation x2 = −7 has nosolutions in the real numbers. Thus the only solutions to our originalequation x4 + 5x2 − 14 = 0 are x = √2 or x = −√2.
Instructor’s Solutions Manual, Section 2.4 Exercise 23
23. Factor x8 −y8 as nicely as possible.
solution
x8 −y8 = (x4 −y4)(x4 +y4)
= (x2 −y2)(x2 +y2)(x4 +y4)
= (x −y)(x +y)(x2 +y2)(x4 +y4)
Instructor’s Solutions Manual, Section 2.4 Exercise 24
24. Factor x16 −y8 as nicely as possible.
solution
x16 −y8 = (x8 −y4)(x8 +y4)
= (x4 −y2)(x4 +y2)(x8 +y4)
= (x2 −y)(x2 +y)(x4 +y2)(x8 +y4)
Instructor’s Solutions Manual, Section 2.4 Exercise 25
25. Find a number b such that 3 is a zero of the polynomial p defined by
p(x) = 1− 4x + bx2 + 2x3.
solution Note that
p(3) = 1− 4 · 3+ b · 32 + 2 · 33
= 43+ 9b.
We want p(3) to equal 0. Thus we solve the equation 0 = 43+ 9b,getting b = −43
9 .
Instructor’s Solutions Manual, Section 2.4 Exercise 26
26. Find a number c such that −2 is a zero of the polynomial p defined by
p(x) = 5− 3x + 4x2 + cx3.
solution Note that
p(−2) = 5− 3(−2)+ 4(−2)2 + c(−2)3
= 27− 8c.
We want p(−2) to equal 0. Thus we solve the equation 0 = 27− 8c,getting c = 27
8 .
Instructor’s Solutions Manual, Section 2.4 Exercise 27
27. Find a polynomial p of degree 3 such that −1, 2, and 3 are zeros of pand p(0) = 1.
solution If p is a polynomial of degree 3 and −1, 2, and 3 are zerosof p, then
p(x) = c(x + 1)(x − 2)(x − 3)
for some constant c. We have p(0) = c(0+ 1)(0− 2)(0− 3) = 6c. Thusto make p(0) = 1 we must choose c = 1
6 . Thus
p(x) = (x + 1)(x − 2)(x − 3)6
,
which by multiplying together the terms in the numerator can also bewritten in the form
p(x) = 1+ x6− 2x2
3+ x
3
6.
Instructor’s Solutions Manual, Section 2.4 Exercise 28
28. Find a polynomial p of degree 3 such that −2, −1, and 4 are zeros of pand p(1) = 2.
solution If p is a polynomial of degree 3 and −2, −1, and 4 are zerosof p, then
p(x) = c(x + 2)(x + 1)(x − 4)
for some constant c. We have p(1) = c(1+ 2)(1+ 1)(1− 4) = −18c.Thus to make p(1) = 2 we must choose c = −1
9 . Thus
p(x) = −(x + 2)(x + 1)(x − 4)9
,
which by multiplying together the terms in the numerator can also bewritten in the form
p(x) = 89+ 10x
9+ x
2
9− x
3
9.
Instructor’s Solutions Manual, Section 2.4 Exercise 29
29. Find all choices of b, c, and d such that 1 and 4 are the only zeros ofthe polynomial p defined by
p(x) = x3 + bx2 + cx + d.
solution Because 1 and 4 are zeros of p, there is a polynomial q suchthat
p(x) = (x − 1)(x − 4)q(x).
Because p has degree 3, the polynomial q must have degree 1. Thus qhas a zero, which must equal 1 or 4 because those are the only zeros ofp. Furthermore, the coefficient of x in the polynomial q must equal 1because the coefficient of x3 in the polynomial p equals 1.
Thus q(x) = x − 1 or q(x) = x − 4. In other words,p(x) = (x − 1)2(x − 4) or p(x) = (x − 1)(x − 4)2. Multiplying outthese expressions, we see that p(x) = x3 − 6x2 + 9x − 4 orp(x) = x3 − 9x2 + 24x − 16.
Thus b = −6, c = 9, d = −4 or b = −9, c = 24, c = −16.
Instructor’s Solutions Manual, Section 2.4 Exercise 30
30. Find all choices of b, c, and d such that −3 and 2 are the only zeros ofthe polynomial p defined by
p(x) = x3 + bx2 + cx + d.
solution Because −3 and 2 are zeros of p, there is a polynomial qsuch that
p(x) = (x + 3)(x − 2)q(x).
Because p has degree 3, the polynomial q must have degree 1. Thus qhas a zero, which must equal −3 or 2 because those are the only zerosof p. Furthermore, the coefficient of x in the polynomial q must equal1 because the coefficient of x3 in the polynomial p equals 1.
Thus q(x) = x + 3 or q(x) = x − 2. In other words,p(x) = (x + 3)2(x − 2) or p(x) = (x + 3)(x − 2)2. Multiplying outthese expressions, we see that p(x) = x3 + 4x2 − 3x − 18 orp(x) = x3 − x2 − 8x + 12.
Thus b = 4, c = −3, d = −18 or b = −1, c = −8, c = 12.
Instructor’s Solutions Manual, Section 2.4 Problem 31
Solutions to Problems, Section 2.4
31. Show that if p and q are nonzero polynomials with degp < degq, then
deg(p + q) = degq.
solution Let n = degq. Thus q(x) includes a term of the form cxn
with c �= 0, and q(x) contains no nonzero terms with higher degree.Because degp < n, the term cxn cannot be canceled by any of theterms of p(x) in the sum p(x)+ q(x). Thus deg(p + q) = n = degq.
Instructor’s Solutions Manual, Section 2.4 Problem 32
32. Give an example of polynomials p and q such that deg(pq) = 8 anddeg(p + q) = 5.
solution Define polynomials p and q by the formulas
p(x) = x5 and q(x) = x3.
Then(pq)(x) = p(x) · q(x) = x5x3 = x8
and(p + q)(x) = p(x)+ q(x) = x5 + x3.
Thus deg(pq) = 8 and deg(p + q) = 5.
Instructor’s Solutions Manual, Section 2.4 Problem 33
33. Give an example of polynomials p and q such that deg(pq) = 8 anddeg(p + q) = 2.
solution Define polynomials p and q by the formulas
p(x) = x2 + x4 and q(x) = x2 − x4.
Then
(pq)(x) = p(x) · q(x) = (x2 + x4)(x2 − x4) = x4 − x8
and
(p + q)(x) = p(x)+ q(x) = (x2 + x4)+ (x2 − x4) = 2x2.
Thus deg(pq) = 8 and deg(p + q) = 2.
Instructor’s Solutions Manual, Section 2.4 Problem 34
34. Suppose q(x) = 2x3 − 3x + 1.
(a) Show that the point (2,11) is on the graph of q.
(b) Show that the slope of a line containing (2,11) and a point on thegraph of q very close to (2,11) is approximately 21.
[Hint: Use the result of Exercise 17.]
solution
(a) Note thatq(2) = 2 · 23 − 3 · 2+ 1 = 11.
Thus the point (2,11) is on the graph of q.
(b) Suppose x is a very small nonzero number. Thus((2+ x,q(2+ x)) is a
point on the graph of q that is very close to (2,11). The slope of theline containing (2,11) and
((2+ x,q(2+ x)) is
q(2+ x)− 11(2+ x)− 2
= q(2+ x)− q(2)x
= 2x2 + 12x + 21,
where the last equality comes from Exercise 17. Because x is verysmall, 2x2 + 12x is also very small, and thus the last equation showsthat the slope of this line is approximately 21.
Instructor’s Solutions Manual, Section 2.4 Problem 35
35. Suppose s(x) = 4x3 − 2.
(a) Show that the point (1,2) is on the graph of s.
(b) Give an estimate for the slope of a line containing (1,2) and apoint on the graph of s very close to (1,2).
[Hint: Use the result of Exercise 18.]
solution
(a) Note thats(1) = 4 · 13 − 2 = 2.
Thus the point (1,2) is on the graph of q.
(b) Suppose x is a very small nonzero number. Thus((1+ x, s(1+ x)) is a
point on the graph of s that is very close to (1,2). The slope of the linecontaining (1,2) and
((1+ x, s(1+ x)) is
s(1+ x)− 2(1+ x)− 1
= s(1+ x)− s(1)x
= 4x2 + 12x + 12,
where the last equality comes from Exercise 18. Because x is verysmall, 4x2 + 12x is also very small, and thus the last equation showsthat the slope of this line is approximately 12.
Instructor’s Solutions Manual, Section 2.4 Problem 36
36. Give an example of polynomials p and q of degree 3 such thatp(1) = q(1), p(2) = q(2), and p(3) = q(3), but p(4) �= q(4).solution One example is to take
p(x) = (x − 1)(x − 2)(x − 3) and q(x) = 2(x − 1)(x − 2)(x − 3).
Then p(1) = q(1) = p(2) = q(2) = p(3) = q(3) = 0. However, p(4) = 6and q(4) = 12, and thus p(4) �= q(4).Of course there are also many other correct examples.
Instructor’s Solutions Manual, Section 2.4 Problem 37
37. Suppose p and q are polynomials of degree 3 such that p(1) = q(1),p(2) = q(2), p(3) = q(3), and p(4) = q(4). Explain why p = q.
solution Define a polynomial r by
r(x) = p(x)− q(x).
Because p and q are polynomials of degree 3, the polynomial r has noterms with degree higher than 3. Thus either r is the zero polynomialor r is a polynomial with degree at most 3.
Note that
r(1) = p(1)− q(1) = 0;
r(2) = p(2)− q(2) = 0;
r(3) = p(3)− q(3) = 0;
r(4) = p(4)− q(4) = 0.
Thus the polynomial r has at least four zeros. However, a nonconstantpolynomial of degree at most 3 can have at most 3 zeros. Thus r mustbe the zero polynomial, which implies that p = q.
Instructor’s Solutions Manual, Section 2.4 Problem 38
38. Verify that(x +y)3 = x3 + 3x2y + 3xy2 +y3.
solution
(x +y)3 = (x +y)(x +y)2
= (x +y)(x2 + 2xy +y2)
= x(x2 + 2xy +y2)+y(x2 + 2xy +y2)
= x3 + 2x2y + xy2 + x2y + 2xy2 +y3
= x3 + 3x2y + 3xy2 +y3
Instructor’s Solutions Manual, Section 2.4 Problem 39
39. Verify thatx3 −y3 = (x −y)(x2 + xy +y2).
solution
(x −y)(x2 + xy +y2) = x(x2 + xy +y2)−y(x2 + xy +y2)
= x3 + x2y + xy2 − x2y − xy2 −y3
= x3 −y3
Instructor’s Solutions Manual, Section 2.4 Problem 40
40. Verify thatx3 +y3 = (x +y)(x2 − xy +y2).
solution
(x +y)(x2 − xy +y2) = x(x2 − xy +y2)+y(x2 − xy +y2)
= x3 − x2y + xy2 + x2y − xy2 +y3
= x3 +y3
Instructor’s Solutions Manual, Section 2.4 Problem 41
41. Verify that
x5 −y5 = (x −y)(x4 + x3y + x2y2 + xy3 +y4).
solution
(x −y)(x4 + x3y + x2y2 + xy3 +y4)
= x(x4 + x3y + x2y2 + xy3 +y4)−y(x4 + x3y + x2y2 + xy3 +y4)
= x5 + x4y + x3y2 + x2y3 + xy4 − x4y − x3y2 − x2y3 − xy4 −y5
= x5 −y5
Instructor’s Solutions Manual, Section 2.4 Problem 42
42. Verify thatx4 + 1 = (x2 +√2x + 1)(x2 −√2x + 1).
solution
(x2 +√2x + 1)(x2 −√2+ 1) = ((x2 + 1)+√2x)((x2 + 1)−√2x
)= (x2 + 1)2 − (√2x)2
= x4 + 2x2 + 1− 2x2
= x4 + 1
Instructor’s Solutions Manual, Section 2.4 Problem 43
43. Write the polynomial x4 + 16 as the product of two polynomials ofdegree 2.[Hint: Use the result from the previous problem with x replaced by x
2 .]
solution Replacing x by x2 on both sides of the result from the
previous problem, we have
(x2
)4 + 1 =((x
2
)2 +√2x2+ 1
)((x2
)2 −√2x2+ 1
),
which can be rewritten as
x4
16+ 1 =
(x2
4+√
22x + 1
)(x2
4−√
22x + 1
).
Now multiply both sides of the equation above by 16, but on the rightside do this by multiplying the first factor by 4 and by multiplying thesecond factor by 4, getting
x4 + 16 = (x2 + 2√
2x + 1)(x2 − 2√
2x + 1).
Instructor’s Solutions Manual, Section 2.4 Problem 44
44. Show that(a+ b)3 = a3 + b3
if and only if a = 0 or b = 0 or a = −b.
solution First we expand (a+ b)3:
(a+b)3 = (a+b)(a+b)2 = (a+b)(a2+2ab+b2) = a3+3a2b+3ab2+b3.
Thus (a+ b)3 = a3 + b3 if and only if
0 = 3a2b + 3ab2 = 3ab(a+ b),
which happens if and only if and only if a = 0 or b = 0 or a = −b.
Instructor’s Solutions Manual, Section 2.4 Problem 45
45. Suppose d is a real number. Show that
(d+ 1)4 = d4 + 1
if and only if d = 0.
solution First we expand (d+ 1)4:
(d+ 1)4 = ((d+ 1)2)2 = (d2 + 2d+ 1)2. = d4 + 4d3 + 6d2 + 4d+ 1.
Thus (d+ 1)4 = d4 + 1 if and only if
0 = 4d3 + 6d2 + 4d = 2d(2d2 + 3d+ 2),
which happens if and only if d = 0 or 2d2 + 3d+ 2 = 0. However, thequadratic formula shows that there are no real numbers d such that2d2 + 3d+ 2 = 0. Hence we conclude that (d+ 1)4 = d4 + 1 if and onlyif d = 0.
Instructor’s Solutions Manual, Section 2.4 Problem 46
46. Suppose p(x) = 3x7 − 5x3 + 7x − 2.
(a) Show that if m is a zero of p, then
2m= 3m6 − 5m2 + 7.
(b) Show that the only possible integer zeros of p are −2, −1, 1, and 2.
(c) Show that no zero of p is an integer.
solution
(a) Suppose m is a zero of p. Then
0 = p(m) = 3m7 − 5m3 + 7m− 2.
Adding 2 to both sides and then dividing by m shows that
2m= 3m6 − 5m2 + 7.
(b) Suppose m is an integer and is a zero of p. Because m is an integer,3m6 − 5m2 + 7 is also an integer. Thus part (a) implies that 2
m is aninteger, which implies that m = −2 or m = −1 or m = 1 or m = 2.
(c) We know from part (b) that no integer other than possibly −2, −1, 1,and 2 can be a zero of p. Thus we need to check only those fourpossibilities. Doing some arithmetic, we see that
p(−2) = −360, p(−1) = −7, p(1) = 3, p(2) = 356.
Instructor’s Solutions Manual, Section 2.4 Problem 46
Thus none of the four possibilities are zeros of p. Hence p has nozeros that are integers.
Instructor’s Solutions Manual, Section 2.4 Problem 47
47. Suppose a, b, and c are integers and that
p(x) = ax3 + bx2 + cx + 9.
Explain why every zero of p that is an integer is contained in the set{−9,−3,−1,1,3,9}.solution Suppose m is an integer that is a zero of p. Then
0 = p(m) = am3 + bm2 + cm+ 9.
Subtracting 9 from both sides and then dividing by −m shows that
9m= −am2 − bm− c.
Because a, b, c, and m are all integers, −am2 − bm− c is also aninteger. Thus the equation above shows that 9
m is an integer, whichimplies that m equals −9, −3, −1, 1, 3, or 9.
Instructor’s Solutions Manual, Section 2.4 Problem 48
48. Suppose p(x) = a0 + a1x + · · · + anxn, where a1, a2, . . . , an areintegers. Suppose m is a nonzero integer that is a zero of p. Show thata0/m is an integer.
solution Because m is a zero of p, we have
0 = p(m) = a0 + a1m+ · · · + anmn.
Subtracting a0 from both sides and then dividing both sides by −mshows that
a0
m= −a1 − a2m− · · · − anmn−1.
Because a1, a2, . . . , an and m are all integers,−a1 −a2m− · · · −anmn−1 is also an integer. Thus the equation aboveshows that a0/m is an integer.
Instructor’s Solutions Manual, Section 2.4 Problem 49
49. Give an example of a polynomial of degree 5 that has exactly two zeros.
solution One example is the polynomial p defined by
p(x) = x4(x − 1) = x5 − x4.
Then p has exactly two zeros, namely 0 and 1.
Of course there are also many other correct examples.
Instructor’s Solutions Manual, Section 2.4 Problem 50
50. Give an example of a polynomial of degree 8 that has exactly threezeros.
solution One example is the polynomial p defined by
p(x) = x6(x − 1)(x − 2) = x8 − 3x7 + 2x6.
Then p has exactly three zeros, namely 0, 1, and 2.
Of course there are also many other correct examples.
Instructor’s Solutions Manual, Section 2.4 Problem 51
51. Give an example of a polynomial p of degree 4 such that p(7) = 0 andp(x) ≥ 0 for all real numbers x.
solution Define p byp(x) = (x − 7)4.
Then clearly p(7) = 0 and p(x) ≥ 0 for all real numbers x.
Expanding the expression above shows that
p(x) = x4 − 28x3 + 294x2 − 1372x + 2401,
which explicitly shows that p is a polynomial of degree 4.
Instructor’s Solutions Manual, Section 2.4 Problem 52
52. Give an example of a polynomial p of degree 6 such that p(0) = 5 andp(x) ≥ 5 for all real numbers x.
solution Define p byp(x) = x6 + 5.
Then clearly p is a polynomial of degree 6 and p(0) = 5 and p(x) ≥ 5for all real numbers x.
Instructor’s Solutions Manual, Section 2.4 Problem 53
53. Give an example of a polynomial p of degree 8 such that p(2) = 3 andp(x) ≥ 3 for all real numbers x.
solution Define p by
p(x) = x6(x − 2)2 + 3.
Then clearly p(2) = 3 and p(x) ≥ 3 for all real numbers x.
Expanding the expression above shows that
p(x) = x8 − 4x7 + 4x6 + 3,
which explicitly shows that p is a polynomial of degree 8.
Instructor’s Solutions Manual, Section 2.4 Problem 54
54. Explain why there does not exist a polynomial p of degree 7 such thatp(x) ≥ −100 for every real number x.
solution Suppose p is a polynomial of the form
p(x) = a0 + a1x + a2x2 + a3x3 + a4x4 + a5x5 + a6x6 + a7x7,
where a7 �= 0. Then p behaves approximately the same as a7x7 near±∞. If a7 > 0, this means that p(x) is a negative number with verylarge absolute value for x near −∞. If a7 < 0, this means that p(x) is anegative number with very large absolute value for x near ∞. Eitherway, we cannot have that p(x) ≥ −100 for every real number x.
Instructor’s Solutions Manual, Section 2.4 Problem 55
55. Explain why the composition of two polynomials is a polynomial.
solution Suppose q is a polynomial and k is a positive integer.Define a function rk by
rk(x) =(q(x)
)k = q(x) · q(x) · · · · · q(x)︸ ︷︷ ︸k times
.
Then rk is a polynomial because the product of polynomials is apolynomial.
Suppose now that p is a polynomial defined by
p(x) = a0 + a1x + a2x2 + · · · + amxm.
Thus
(p ◦ q)(x) = p(q(x)) = a0 + a1q(x)+ a2(q(x)
)2 + · · · + am(q(x)
)m.The equation above shows that
p ◦ q = a0 + a1r1 + a2r2 + · · · + amrm.
Each term akrk is a polynomial because each rk is a polynomial and aconstant times a polynomial is a polynomial. The sum of polynomials isa polynomial, thus the equation above implies that p ◦q is a polynomial.
Instructor’s Solutions Manual, Section 2.4 Problem 56
56. Show that if p and q are nonzero polynomials, then
deg(p ◦ q) = (degp)(degq).
solution Suppose q is a polynomial with degree n and k is a positiveinteger. Define a function rk by
rk(x) =(q(x)
)k = q(x) · q(x) · · · · · q(x)︸ ︷︷ ︸k times
.
Thus
deg rk = deg(q · q · · · · · q︸ ︷︷ ︸k times
)
= degq + degq + · · · + degq︸ ︷︷ ︸k times
= n+n+ · · · +n︸ ︷︷ ︸k times
= kn.
Suppose now that p is a polynomial with degree m defined by
p(x) = a0 + a1x + a2x2 + · · · + amxm,
where am �= 0. Thus
Instructor’s Solutions Manual, Section 2.4 Problem 56
(p ◦ q)(x) = p(q(x)) = a0 + a1q(x)+ a2(q(x)
)2 + · · · + am(q(x)
)m.The equation above shows that
p ◦ q = a0 + a1r1 + a2r2 + · · · + amrm.
Each term akrk with ak �= 0 is a polynomial with degree kn. Inparticular, the term amrm is a polynomial with degree mn, and none ofthe other terms has high enough degree to cancel the multiple of xmn
that appears in amrm(x). Thus
deg(p ◦ q) =mn = (degp)(degq).
Instructor’s Solutions Manual, Section 2.4 Problem 57
57. In the first figure in the solution to Example 5, the graph of thepolynomial p clearly lies below the x-axis for x in the interval[5000,10000]. Yet in the second figure in the same solution, the graphof p seems to be on or above the x-axis for all values of p in theinterval [0,1000000]. Explain.
solution There is actually no contradiction between the two graphsin the solution to Example 5. The scale of the two graphs is vastlydifferent. Thus although the graph of p is indeed below the x-axis inthe interval [5000,10000], in the second graph in the solution toExample 5 the scale is so huge that the amount by which the graph of pis below the x-axis is too small for our eyes to see.