Instructor’s Solutions Manual, Section 2.4
Exercise 1
Solutions to Exercises, Section 2.4
Suppose
p(x) = x 2 + 5x + 2,
q(x) = 2x 3 − 3x + 1,
s(x) = 4x 3 − 2.
In Exercises 1–18, write the indicated expression as a sum of terms, each
of which is a constant times a power of x.
1. (p + q)(x)
solution
(p + q)(x) = (x 2 + 5x + 2) + (2x 3 − 3x + 1)
= 2x 3 + x 2 + 2x + 3
Instructor’s Solutions Manual, Section 2.4
2. (p − q)(x)
solution
(p − q)(x) = (x 2 + 5x + 2) − (2x 3 − 3x + 1)
= −2x 3 + x 2 + 8x + 1
Exercise 2
Instructor’s Solutions Manual, Section 2.4
3. (3p − 2q)(x)
solution
(3p − 2q)(x) = 3(x 2 + 5x + 2) − 2(2x 3 − 3x + 1)
= 3x 2 + 15x + 6 − 4x 3 + 6x − 2
= −4x 3 + 3x 2 + 21x + 4
Exercise 3
Instructor’s Solutions Manual, Section 2.4
4. (4p + 5q)(x)
solution
(4p + 5q)(x) = 4(x 2 + 5x + 2) + 5(2x 3 − 3x + 1)
= 4x 2 + 20x + 8 + 10x 3 − 15x + 5
= 10x 3 + 4x 2 + 5x + 13
Exercise 4
Instructor’s Solutions Manual, Section 2.4
5. (pq)(x)
solution
(pq)(x) = (x 2 + 5x + 2)(2x 3 − 3x + 1)
= x 2 (2x 3 − 3x + 1)
+ 5x(2x 3 − 3x + 1) + 2(2x 3 − 3x + 1)
= 2x 5 − 3x 3 + x 2 + 10x 4 − 15x 2
+ 5x + 4x 3 − 6x + 2
= 2x 5 + 10x 4 + x 3 − 14x 2 − x + 2
Exercise 5
Instructor’s Solutions Manual, Section 2.4
6. (ps)(x)
solution
(ps)(x) = (x 2 + 5x + 2)(4x 3 − 2)
= x 2 (4x 3 − 2) + 5x(4x 3 − 2) + 2(4x 3 − 2)
= 4x 5 − 2x 2 + 20x 4 − 10x + 8x 3 − 4
= 4x 5 + 20x 4 + 8x 3 − 2x 2 − 10x − 4
Exercise 6
Instructor’s Solutions Manual, Section 2.4
7.
p(x)
2
solution
p(x)
2
= (x 2 + 5x + 2)(x 2 + 5x + 2)
= x 2 (x 2 + 5x + 2) + 5x(x 2 + 5x + 2)
+ 2(x 2 + 5x + 2)
= x 4 + 5x 3 + 2x 2 + 5x 3 + 25x 2
+ 10x + 2x 2 + 10x + 4
= x 4 + 10x 3 + 29x 2 + 20x + 4
Exercise 7
Instructor’s Solutions Manual, Section 2.4
8.
q(x)
2
solution
q(x)
2
= (2x 3 − 3x + 1)2
= (2x 3 − 3x + 1)(2x 3 − 3x + 1)
= 2x 3 (2x 3 − 3x + 1) − 3x(2x 3 − 3x + 1)
+ (2x 3 − 3x + 1)
= 4x 6 − 6x 4 + 2x 3 − 6x 4 + 9x 2
− 3x + 2x 3 − 3x + 1
= 4x 6 − 12x 4 + 4x 3 + 9x 2 − 6x + 1
Exercise 8
Instructor’s Solutions Manual, Section 2.4
9.
Exercise 9
2
p(x) s(x)
2
solution Using the expression that we computed for p(x) in the
solution to Exercise 7, we have
2
p(x) s(x)
= (x 4 + 10x 3 + 29x 2 + 20x + 4)(4x 3 − 2)
= 4x 3 (x 4 + 10x 3 + 29x 2 + 20x + 4)
− 2(x 4 + 10x 3 + 29x 2 + 20x + 4)
= 4x 7 + 40x 6 + 116x 5 + 80x 4 + 16x 3
− 2x 4 − 20x 3 − 58x 2 − 40x − 8
= 4x 7 + 40x 6 + 116x 5 + 78x 4
− 4x 3 − 58x 2 − 40x − 8.
Instructor’s Solutions Manual, Section 2.4
10.
Exercise 10
2
q(x) s(x)
2
solution Using the expression that we computed for q(x) in the
solution to Exercise 8, we have
2
q(x) s(x)
= (4x 6 − 12x 4 + 4x 3 + 9x 2 − 6x + 1)(4x 3 − 2)
= 4x 3 (4x 6 − 12x 4 + 4x 3 + 9x 2 − 6x + 1)
− 2(4x 6 − 12x 4 + 4x 3 + 9x 2 − 6x + 1)
= 16x 9 − 48x 7 + 16x 6 + 36x 5 − 24x 4 + 4x 3
− 8x 6 + 24x 4 − 8x 3 − 18x 2 + 12x − 2
= 16x 9 − 48x 7 + 8x 6 + 36x 5 − 4x 3
− 18x 2 + 12x − 2.
Instructor’s Solutions Manual, Section 2.4
11. (p ◦ q)(x)
solution
(p ◦ q)(x) = p q(x)
= p(2x 3 − 3x + 1)
= (2x 3 − 3x + 1)2 + 5(2x 3 − 3x + 1) + 2
= (4x 6 − 12x 4 + 4x 3 + 9x 2 − 6x + 1)
+ (10x 3 − 15x + 5) + 2
= 4x 6 − 12x 4 + 14x 3 + 9x 2 − 21x + 8
Exercise 11
Instructor’s Solutions Manual, Section 2.4
12. (q ◦ p)(x)
solution
(q ◦ p)(x) = q p(x)
= q(x 2 + 5x + 2)
= 2(x 2 + 5x + 2)3 − 3(x 2 + 5x + 2) + 1
= 2(x 2 + 5x + 2)2 (x 2 + 5x + 2)
− 3x 2 − 15x − 5
= 2(x 4 + 10x 3 + 29x 2 + 20x + 4)(x 2 + 5x + 2)
− 3x 2 − 15x − 5
= 2x 6 + 30x 5 + 162x 4 + 370x 3
+ 321x 2 + 105x + 11
Exercise 12
Instructor’s Solutions Manual, Section 2.4
13. (p ◦ s)(x)
solution
(p ◦ s)(x) = p s(x)
= p(4x 3 − 2)
= (4x 3 − 2)2 + 5(4x 3 − 2) + 2
= (16x 6 − 16x 3 + 4) + (20x 3 − 10) + 2
= 16x 6 + 4x 3 − 4
Exercise 13
Instructor’s Solutions Manual, Section 2.4
14. (s ◦ p)(x)
solution
(s ◦ p)(x) = s p(x)
= s(x 2 + 5x + 2)
= 4(x 2 + 5x + 2)3 − 2
= 4x 6 + 60x 5 + 324x 4 + 740x 3
+ 648x 2 + 240x + 30
Exercise 14
Instructor’s Solutions Manual, Section 2.4
15.
q ◦ (p + s) (x)
solution
q ◦ (p + s) (x) = q (p + s)(x)
= q p(x) + s(x)
= q(4x 3 + x 2 + 5x)
= 2(4x 3 + x 2 + 5x)3 − 3(4x 3 + x 2 + 5x) + 1
= 2(4x 3 + x 2 + 5x)2 (4x 3 + x 2 + 5x)
− 12x 3 − 3x 2 − 15x + 1
= 2(16x 6 + 8x 5 + 41x 4 + 10x 3 + 25x 2 )
× (4x 3 + x 2 + 5x) − 12x 3 − 3x 2 − 15x + 1
= 128x 9 + 96x 8 + 504x 7 + 242x 6 + 630x 5
+ 150x 4 + 238x 3 − 3x 2 − 15x + 1
Exercise 15
Instructor’s Solutions Manual, Section 2.4
16.
(q + p) ◦ s (x)
solution
(q + p) ◦ s (x) = (q + p) s(x)
= q s(x) + p s(x)
= q(4x 3 − 2) + p(4x 3 − 2)
= 2(4x 3 − 2)3 − 3(4x 3 − 2) + 1
+ (4x 3 − 2)2 + 5(4x 3 − 2) + 2
= 128x 9 − 176x 6 + 88x 3 − 13
Exercise 16
Instructor’s Solutions Manual, Section 2.4
17.
q(2 + x) − q(2)
x
solution
q(2 + x) − q(2)
x
=
2(2 + x)3 − 3(2 + x) + 1 − (2 · 23 − 3 · 2 + 1)
x
=
2x 3 + 12x 2 + 21x
x
= 2x 2 + 12x + 21
Exercise 17
Instructor’s Solutions Manual, Section 2.4
18.
s(1 + x) − s(1)
x
solution
s(1 + x) − s(1)
x
=
4(1 + x)3 − 2 − (4 · 13 − 2)
x
=
4x 3 + 12x 2 + 12x
x
= 4x 2 + 12x + 12
Exercise 18
Instructor’s Solutions Manual, Section 2.4
Exercise 19
19. Find all real numbers x such that
x 6 − 8x 3 + 15 = 0.
solution This equation involves x 3 and x 6 ; thus we make the
substitution x 3 = y. Squaring both sides of the equation x 3 = y gives
x 6 = y 2 . With these substitutions, the equation above becomes
y 2 − 8y + 15 = 0.
This new equation can now be solved either by factoring the left side or
by 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 by
using the quadratic formula).
Substituting x 3 for y now shows that x 3 = 3 or x 3 = 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
x 6 − 3x 3 − 10 = 0.
solution This equation involves x 3 and x 6 ; thus we make the
substitution x 3 = y. Squaring both sides of the equation x 3 = y gives
x 6 = y 2 . With these substitutions, the equation above becomes
y 2 − 3y − 10 = 0.
This new equation can now be solved either by factoring the left side or
by 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 by
using the quadratic formula).
Substituting x 3 for y now shows that x 3 = 5 or x 3 = −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
x 4 − 2x 2 − 15 = 0.
solution This equation involves x 2 and x 4 ; thus we make the
substitution x 2 = y. Squaring both sides of the equation x 2 = y gives
x 4 = y 2 . With these substitutions, the equation above becomes
y 2 − 2y − 15 = 0.
This new equation can now be solved either by factoring the left side or
by using the quadratic formula. Let’s use the quadratic formula, getting
y=
2±
√
2±8
4 + 60
=
.
2
2
Thus y = 5 or y = −3 (the same result could have been obtained by
factoring).
Substituting x 2 for y now shows that x 2 = 5 or x 2 = −3. The equation
√
√
x 2 = 5 implies that x = 5 or x = − 5. The equation x 2 = −3 has no
solutions in the real numbers. Thus the only solutions to our original
√
√
equation x 4 − 2x 2 − 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
x 4 + 5x 2 − 14 = 0.
solution This equation involves x 2 and x 4 ; thus we make the
substitution x 2 = y. Squaring both sides of the equation x 2 = y gives
x 4 = y 2 . With these substitutions, the equation above becomes
y 2 + 5y − 14 = 0.
This new equation can now be solved either by factoring the left side or
by using the quadratic formula. Let’s use the quadratic formula, getting
y=
−5 ±
√
−5 ± 9
25 + 56
=
.
2
2
Thus y = 2 or y = −7 (the same result could have been obtained by
factoring).
Substituting x 2 for y now shows that x 2 = 2 or x 2 = −7. The equation
√
√
x 2 = 2 implies that x = 2 or x = − 2. The equation x 2 = −7 has no
solutions in the real numbers. Thus the only solutions to our original
√
√
equation x 4 + 5x 2 − 14 = 0 are x = 2 or x = − 2.
Instructor’s Solutions Manual, Section 2.4
23. Factor x 8 − y 8 as nicely as possible.
solution
x 8 − y 8 = (x 4 − y 4 )(x 4 + y 4 )
= (x 2 − y 2 )(x 2 + y 2 )(x 4 + y 4 )
= (x − y)(x + y)(x 2 + y 2 )(x 4 + y 4 )
Exercise 23
Instructor’s Solutions Manual, Section 2.4
24. Factor x 16 − y 8 as nicely as possible.
solution
x 16 − y 8 = (x 8 − y 4 )(x 8 + y 4 )
= (x 4 − y 2 )(x 4 + y 2 )(x 8 + y 4 )
= (x 2 − y)(x 2 + y)(x 4 + y 2 )(x 8 + y 4 )
Exercise 24
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 + bx 2 + 2x 3 .
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,
43
getting b = − 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 + 4x 2 + cx 3 .
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,
27
getting c = 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 p
and p(0) = 1.
solution If p is a polynomial of degree 3 and −1, 2, and 3 are zeros
of 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. Thus
1
to make p(0) = 1 we must choose c = 6 . Thus
p(x) =
(x + 1)(x − 2)(x − 3)
,
6
which by multiplying together the terms in the numerator can also be
written in the form
p(x) = 1 +
2x 2
x3
x
−
+
.
6
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 p
and p(1) = 2.
solution If p is a polynomial of degree 3 and −2, −1, and 4 are zeros
of 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.
1
Thus to make p(1) = 2 we must choose c = − 9 . Thus
p(x) = −
(x + 2)(x + 1)(x − 4)
,
9
which by multiplying together the terms in the numerator can also be
written in the form
p(x) =
x2
x3
8 10x
+
+
−
.
9
9
9
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 of
the polynomial p defined by
p(x) = x 3 + bx 2 + cx + d.
solution Because 1 and 4 are zeros of p, there is a polynomial q such
that
p(x) = (x − 1)(x − 4)q(x).
Because p has degree 3, the polynomial q must have degree 1. Thus q
has a zero, which must equal 1 or 4 because those are the only zeros of
p. Furthermore, the coefficient of x in the polynomial q must equal 1
because the coefficient of x 3 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 out
these expressions, we see that p(x) = x 3 − 6x 2 + 9x − 4 or
p(x) = x 3 − 9x 2 + 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 of
the polynomial p defined by
p(x) = x 3 + bx 2 + cx + d.
solution Because −3 and 2 are zeros of p, there is a polynomial q
such that
p(x) = (x + 3)(x − 2)q(x).
Because p has degree 3, the polynomial q must have degree 1. Thus q
has a zero, which must equal −3 or 2 because those are the only zeros
of p. Furthermore, the coefficient of x in the polynomial q must equal
1 because the coefficient of x 3 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 out
these expressions, we see that p(x) = x 3 + 4x 2 − 3x − 18 or
p(x) = x 3 − x 2 − 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 deg p < deg q, then
deg(p + q) = deg q.
solution Let n = deg q. Thus q(x) includes a term of the form cx n
with c = 0, and q(x) contains no nonzero terms with higher degree.
Because deg p < n, the term cx n cannot be canceled by any of the
terms of p(x) in the sum p(x) + q(x). Thus deg(p + q) = n = deg q.
Instructor’s Solutions Manual, Section 2.4
Problem 32
32. Give an example of polynomials p and q such that deg(pq) = 8 and
deg(p + q) = 5.
solution Define polynomials p and q by the formulas
p(x) = x 5
and q(x) = x 3 .
Then
(pq)(x) = p(x) · q(x) = x 5 x 3 = x 8
and
(p + q)(x) = p(x) + q(x) = x 5 + x 3 .
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 and
deg(p + q) = 2.
solution Define polynomials p and q by the formulas
p(x) = x 2 + x 4
and
q(x) = x 2 − x 4 .
Then
(pq)(x) = p(x) · q(x) = (x 2 + x 4 )(x 2 − x 4 ) = x 4 − x 8
and
(p + q)(x) = p(x) + q(x) = (x 2 + x 4 ) + (x 2 − x 4 ) = 2x 2 .
Thus deg(pq) = 8 and deg(p + q) = 2.
Instructor’s Solutions Manual, Section 2.4
Problem 34
34. Suppose q(x) = 2x 3 − 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 the
graph of q very close to (2, 11) is approximately 21.
[Hint: Use the result of Exercise 17.]
solution
(a) Note that
q(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 the
line containing (2, 11) and (2 + x, q(2 + x) is
q(2 + x) − q(2)
q(2 + x) − 11
=
= 2x 2 + 12x + 21,
(2 + x) − 2
x
where the last equality comes from Exercise 17. Because x is very
small, 2x 2 + 12x is also very small, and thus the last equation shows
that the slope of this line is approximately 21.
Instructor’s Solutions Manual, Section 2.4
Problem 35
35. Suppose s(x) = 4x 3 − 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 a
point on the graph of s very close to (1, 2).
[Hint: Use the result of Exercise 18.]
solution
(a) Note that
s(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 line
containing (1, 2) and (1 + x, s(1 + x) is
s(1 + x) − s(1)
s(1 + x) − 2
=
= 4x 2 + 12x + 12,
(1 + x) − 1
x
where the last equality comes from Exercise 18. Because x is very
small, 4x 2 + 12x is also very small, and thus the last equation shows
that 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 that
p(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) = 6
and 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 no
terms with degree higher than 3. Thus either r is the zero polynomial
or 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 nonconstant
polynomial of degree at most 3 can have at most 3 zeros. Thus r must
be the zero polynomial, which implies that p = q.
Instructor’s Solutions Manual, Section 2.4
Problem 38
38. Verify that
(x + y)3 = x 3 + 3x 2 y + 3xy 2 + y 3 .
solution
(x + y)3 = (x + y)(x + y)2
= (x + y)(x 2 + 2xy + y 2 )
= x(x 2 + 2xy + y 2 ) + y(x 2 + 2xy + y 2 )
= x 3 + 2x 2 y + xy 2 + x 2 y + 2xy 2 + y 3
= x 3 + 3x 2 y + 3xy 2 + y 3
Instructor’s Solutions Manual, Section 2.4
Problem 39
39. Verify that
x 3 − y 3 = (x − y)(x 2 + xy + y 2 ).
solution
(x − y)(x 2 + xy + y 2 ) = x(x 2 + xy + y 2 ) − y(x 2 + xy + y 2 )
= x 3 + x 2 y + xy 2 − x 2 y − xy 2 − y 3
= x3 − y 3
Instructor’s Solutions Manual, Section 2.4
Problem 40
40. Verify that
x 3 + y 3 = (x + y)(x 2 − xy + y 2 ).
solution
(x + y)(x 2 − xy + y 2 ) = x(x 2 − xy + y 2 ) + y(x 2 − xy + y 2 )
= x 3 − x 2 y + xy 2 + x 2 y − xy 2 + y 3
= x3 + y 3
Instructor’s Solutions Manual, Section 2.4
Problem 41
41. Verify that
x 5 − y 5 = (x − y)(x 4 + x 3 y + x 2 y 2 + xy 3 + y 4 ).
solution
(x − y)(x 4 + x 3 y + x 2 y 2 + xy 3 + y 4 )
= x(x 4 + x 3 y + x 2 y 2 + xy 3 + y 4 ) − y(x 4 + x 3 y + x 2 y 2 + xy 3 + y 4 )
= x 5 + x 4 y + x 3 y 2 + x 2 y 3 + xy 4 − x 4 y − x 3 y 2 − x 2 y 3 − xy 4 − y 5
= x5 − y 5
Instructor’s Solutions Manual, Section 2.4
42. Verify that
x 4 + 1 = (x 2 +
√
2x + 1)(x 2 −
Problem 42
√
2x + 1).
solution
(x 2 +
√
2x + 1)(x 2 −
√
√ √ 2 + 1) = (x 2 + 1) + 2x (x 2 + 1) − 2x
√
= (x 2 + 1)2 − ( 2x)2
= x 4 + 2x 2 + 1 − 2x 2
= x4 + 1
Instructor’s Solutions Manual, Section 2.4
Problem 43
43. Write the polynomial x 4 + 16 as the product of two polynomials of
degree 2.
[Hint: Use the result from the previous problem with x replaced by
solution Replacing x by
previous problem, we have
x 4
2
+1=
x
2
x 2
2
x
2 .]
on both sides of the result from the
+
√ x
x 2 √ x
2 +1
− 2 +1 ,
2
2
2
which can be rewritten as
√
√
x2
x 2
2
2
x4
+1=
+
x+1
−
x+1 .
16
4
2
4
2
Now multiply both sides of the equation above by 16, but on the right
side do this by multiplying the first factor by 4 and by multiplying the
second factor by 4, getting
√
√
x 4 + 16 = (x 2 + 2 2x + 1)(x 2 − 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 +3a2 b+3ab2 +b3 .
Thus (a + b)3 = a3 + b3 if and only if
0 = 3a2 b + 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 :
2
(d + 1)4 = (d + 1)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, the
quadratic formula shows that there are no real numbers d such that
2d2 + 3d + 2 = 0. Hence we conclude that (d + 1)4 = d4 + 1 if and only
if d = 0.
Instructor’s Solutions Manual, Section 2.4
Problem 46
46. Suppose p(x) = 3x 7 − 5x 3 + 7x − 2.
(a) Show that if m is a zero of p, then
2
= 3m6 − 5m2 + 7.
m
(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
2
= 3m6 − 5m2 + 7.
m
(b) Suppose m is an integer and is a zero of p. Because m is an integer,
2
3m6 − 5m2 + 7 is also an integer. Thus part (a) implies that m is an
integer, 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 four
possibilities. 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 no
zeros that are integers.
Instructor’s Solutions Manual, Section 2.4
Problem 47
47. Suppose a, b, and c are integers and that
p(x) = ax 3 + bx 2 + 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
9
= −am2 − bm − c.
m
Because a, b, c, and m are all integers, −am2 − bm − c is also an
9
integer. Thus the equation above shows that m is an integer, which
implies that m equals −9, −3, −1, 1, 3, or 9.
Instructor’s Solutions Manual, Section 2.4
Problem 48
48. Suppose p(x) = a0 + a1 x + · · · + an x n , where a1 , a2 , . . . , an are
integers. Suppose m is a nonzero integer that is a zero of p. Show that
a0 /m is an integer.
solution Because m is a zero of p, we have
0 = p(m) = a0 + a1 m + · · · + an mn .
Subtracting a0 from both sides and then dividing both sides by −m
shows that
a0
= −a1 − a2 m − · · · − an mn−1 .
m
Because a1 , a2 , . . . , an and m are all integers,
−a1 − a2 m − · · · − an mn−1 is also an integer. Thus the equation above
shows 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) = x 4 (x − 1) = x 5 − x 4 .
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 three
zeros.
solution One example is the polynomial p defined by
p(x) = x 6 (x − 1)(x − 2) = x 8 − 3x 7 + 2x 6 .
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 and
p(x) ≥ 0 for all real numbers x.
solution Define p by
p(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) = x 4 − 28x 3 + 294x 2 − 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 and
p(x) ≥ 5 for all real numbers x.
solution Define p by
p(x) = x 6 + 5.
Then clearly p is a polynomial of degree 6 and p(0) = 5 and p(x) ≥ 5
for 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 and
p(x) ≥ 3 for all real numbers x.
solution Define p by
p(x) = x 6 (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) = x 8 − 4x 7 + 4x 6 + 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 that
p(x) ≥ −100 for every real number x.
solution Suppose p is a polynomial of the form
p(x) = a0 + a1 x + a2 x 2 + a3 x 3 + a4 x 4 + a5 x 5 + a6 x 6 + a7 x 7 ,
where a7 = 0. Then p behaves approximately the same as a7 x 7 near
±∞. If a7 > 0, this means that p(x) is a negative number with very
large absolute value for x near −∞. If a7 < 0, this means that p(x) is a
negative number with very large absolute value for x near ∞. Either
way, 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
k
rk (x) = q(x) = q(x) · q(x) · · · · · q(x) .
k times
Then rk is a polynomial because the product of polynomials is a
polynomial.
Suppose now that p is a polynomial defined by
p(x) = a0 + a1 x + a2 x 2 + · · · + am x m .
Thus
2
m
(p ◦ q)(x) = p q(x) = a0 + a1 q(x) + a2 q(x) + · · · + am q(x) .
The equation above shows that
p ◦ q = a 0 + a 1 r1 + a 2 r 2 + · · · + a m r m .
Each term ak rk is a polynomial because each rk is a polynomial and a
constant times a polynomial is a polynomial. The sum of polynomials is
a 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) = (deg p)(deg q).
solution Suppose q is a polynomial with degree n and k is a positive
integer. Define a function rk by
k
rk (x) = q(x) = q(x) · q(x) · · · · · q(x) .
k times
Thus
deg rk = deg(q · q · · · · · q)
k times
= deg q + deg q + · · · + deg q
k times
= n + n +· · · + n
k times
= kn.
Suppose now that p is a polynomial with degree m defined by
p(x) = a0 + a1 x + a2 x 2 + · · · + am x m ,
where am = 0. Thus
Instructor’s Solutions Manual, Section 2.4
Problem 56
2
m
(p ◦ q)(x) = p q(x) = a0 + a1 q(x) + a2 q(x) + · · · + am q(x) .
The equation above shows that
p ◦ q = a 0 + a 1 r1 + a 2 r2 + · · · + a m r m .
Each term ak rk with ak = 0 is a polynomial with degree kn. In
particular, the term am rm is a polynomial with degree mn, and none of
the other terms has high enough degree to cancel the multiple of x mn
that appears in am rm (x). Thus
deg(p ◦ q) = mn = (deg p)(deg q).
Instructor’s Solutions Manual, Section 2.4
Problem 57
57. In the first figure in the solution to Example 5, the graph of the
polynomial p clearly lies below the x-axis for x in the interval
[5000, 10000]. Yet in the second figure in the same solution, the graph
of p seems to be on or above the x-axis for all values of p in the
interval [0, 1000000]. Explain.
solution There is actually no contradiction between the two graphs
in the solution to Example 5. The scale of the two graphs is vastly
different. Thus although the graph of p is indeed below the x-axis in
the interval [5000, 10000], in the second graph in the solution to
Example 5 the scale is so huge that the amount by which the graph of p
is below the x-axis is too small for our eyes to see.