Solutions to Exercises, Section 2 - UW-Madison …park/Fall2013/precalculus/2.4sol.pdfInstructor’s...

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


2. (p − q)(x)solution

(p − q)(x) = (x2 + 5x + 2)− (2x3 − 3x + 1)

= −2x3 + x2 + 8x + 1


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


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


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


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


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


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


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.


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.


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


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


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


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


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


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


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


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


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.



x6 − 3x3 − 10 = 0.


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.



x4 − 2x2 − 15 = 0.


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.



x4 + 5x2 − 14 = 0.


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.


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)


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)


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 .


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 .


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.


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.


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.


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.


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.


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.


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.


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.


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.


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.


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


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


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


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


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


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).


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.


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.


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.


Thus none of the four possibilities are zeros of p. Hence p has nozeros that are integers.


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.


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.


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.


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.


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.



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.



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.


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.


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.


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


(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).


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.

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