Solutions to Exercises, Section 2.4

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1 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) (p + q)(x) = (x 2 + 5x + 2) + (2x 3 3x + 1) = 2x 3 + x 2 + 2x + 3

2 Instructor s Solutions Manual, Section 2.4 Exercise 2 2. (p q)(x) (p q)(x) = (x 2 + 5x + 2) (2x 3 3x + 1) = 2x 3 + x 2 + 8x + 1

3 Instructor s Solutions Manual, Section 2.4 Exercise 3 3. (3p 2q)(x) (3p 2q)(x) = 3(x 2 + 5x + 2) 2(2x 3 3x + 1) = 3x x + 6 4x 3 + 6x 2 = 4x 3 + 3x x + 4

4 Instructor s Solutions Manual, Section 2.4 Exercise 4 4. (4p + 5q)(x) (4p + 5q)(x) = 4(x 2 + 5x + 2) + 5(2x 3 3x + 1) = 4x x x 3 15x + 5 = 10x 3 + 4x 2 + 5x + 13

5 Instructor s Solutions Manual, Section 2.4 Exercise 5 5. (pq)(x) (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 x 4 15x 2 + 5x + 4x 3 6x + 2 = 2x x 4 + x 3 14x 2 x + 2

6 Instructor s Solutions Manual, Section 2.4 Exercise 6 6. (ps)(x) (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 x 4 10x + 8x 3 4 = 4x x 4 + 8x 3 2x 2 10x 4

7 Instructor s Solutions Manual, Section 2.4 Exercise 7 7. ( p(x) ) 2 ( 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 x x + 2x x + 4 = x x x x + 4

8 Instructor s Solutions Manual, Section 2.4 Exercise 8 8. ( q(x) ) 2 ( 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

9 Instructor s Solutions Manual, Section 2.4 Exercise 9 9. ( p(x) ) 2s(x) Using the expression that we computed for ( p(x) ) 2 in the to Exercise 7, we have ( ) 2s(x) p(x) = (x x x x + 4)(4x 3 2) = 4x 3 (x x x x + 4) 2(x x x x + 4) = 4x x x x x 3 2x 4 20x 3 58x 2 40x 8 = 4x x x x 4 4x 3 58x 2 40x 8.

10 Instructor s Solutions Manual, Section 2.4 Exercise ( q(x) ) 2s(x) Using the expression that we computed for ( q(x) ) 2 in the to Exercise 8, we have ( ) 2s(x) q(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 x x 5 24x 4 + 4x 3 8x x 4 8x 3 18x x 2 = 16x 9 48x 7 + 8x x 5 4x 3 18x x 2.

11 Instructor s Solutions Manual, Section 2.4 Exercise (p q)(x) (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 x 3 + 9x 2 21x + 8

12 Instructor s Solutions Manual, Section 2.4 Exercise (q p)(x) (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 x x x + 4)(x 2 + 5x + 2) 3x 2 15x 5 = 2x x x x x x + 11

13 Instructor s Solutions Manual, Section 2.4 Exercise (p s)(x) (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

14 Instructor s Solutions Manual, Section 2.4 Exercise (s p)(x) (s p)(x) = s ( p(x) ) = s(x 2 + 5x + 2) = 4(x 2 + 5x + 2) 3 2 = 4x x x x x x + 30

15 Instructor s Solutions Manual, Section 2.4 Exercise ( q (p + s) ) (x) ( 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 x x x 2 ) (4x 3 + x 2 + 5x) 12x 3 3x 2 15x + 1 = 128x x x x x x x 3 3x 2 15x + 1

16 Instructor s Solutions Manual, Section 2.4 Exercise ( (q + p) s ) (x) ( (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) (4x 3 2) 2 + 5(4x 3 2) + 2 = 128x 9 176x x 3 13

17 Instructor s Solutions Manual, Section 2.4 Exercise q(2 + x) q(2) x q(2 + x) q(2) x = 2(2 + x)3 3(2 + x) + 1 ( ) x = 2x3 + 12x x x = 2x x + 21

18 Instructor s Solutions Manual, Section 2.4 Exercise s(1 + x) s(1) x s(1 + x) s(1) x = 4(1 + x)3 2 ( ) x = 4x3 + 12x x x = 4x x + 12

19 Instructor s Solutions Manual, Section 2.4 Exercise Find all real numbers x such that x 6 8x = 0. 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 = 3ory = 5 (the same result could have been obtained by using the quadratic formula). Substituting x 3 for y now shows that x 3 = 3orx 3 = 5. Thus x = 3 1/3 or x = 5 1/3.

20 Instructor s Solutions Manual, Section 2.4 Exercise Find all real numbers x such that x 6 3x 3 10 = 0. 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 = 5ory = 2 (the same result could have been obtained by using the quadratic formula). Substituting x 3 for y now shows that x 3 = 5orx 3 = 2. Thus x = 5 1/3 or x = 2 1/3.

21 Instructor s Solutions Manual, Section 2.4 Exercise Find all real numbers x such that x 4 2x 2 15 = 0. 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 2. Thus y = 5ory = 3 (the same result could have been obtained by factoring). Substituting x 2 for y now shows that x 2 = 5orx 2 = 3. The equation x 2 = 5 implies that x = 5orx = 5. The equation x 2 = 3 has no s in the real numbers. Thus the only s to our original equation x 4 2x 2 15 = 0 are x = 5orx = 5.

22 Instructor s Solutions Manual, Section 2.4 Exercise Find all real numbers x such that x 4 + 5x 2 14 = 0. 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. 2 Thus y = 2ory = 7 (the same result could have been obtained by factoring). Substituting x 2 for y now shows that x 2 = 2orx 2 = 7. The equation x 2 = 2 implies that x = 2orx = 2. The equation x 2 = 7 has no s in the real numbers. Thus the only s to our original equation x 4 + 5x 2 14 = 0 are x = 2orx = 2.

23 Instructor s Solutions Manual, Section 2.4 Exercise Factor x 8 y 8 as nicely as possible. 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 )

24 Instructor s Solutions Manual, Section 2.4 Exercise Factor x 16 y 8 as nicely as possible. 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 )

25 Instructor s Solutions Manual, Section 2.4 Exercise Find a number b such that 3 is a zero of the polynomial p defined by p(x) = 1 4x + bx 2 + 2x 3. Note that p(3) = b = b. We want p(3) to equal 0. Thus we solve the equation 0 = b, getting b = 43 9.

26 Instructor s Solutions Manual, Section 2.4 Exercise Find a number c such that 2 is a zero of the polynomial p defined by p(x) = 5 3x + 4x 2 + cx 3. 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 Instructor s Solutions Manual, Section 2.4 Exercise Find a polynomial p of degree 3 such that 1, 2, and 3 are zeros of p and p(0) = 1. of p, then If p is a polynomial of degree 3 and 1, 2, and 3 are zeros 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 to 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 be written in the form p(x) = 1 + x 6 2x2 3 + x3 6.

28 Instructor s Solutions Manual, Section 2.4 Exercise Find a polynomial p of degree 3 such that 2, 1, and 4 are zeros of p and p(1) = 2. of p, then If p is a polynomial of degree 3 and 2, 1, and 4 are zeros 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 be written in the form p(x) = x 9 + x2 9 x3 9.

29 Instructor s Solutions Manual, Section 2.4 Exercise 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. that Because 1 and 4 are zeros of p, there is a polynomial q such 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 1orq(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 4or p(x) = x 3 9x x 16. Thus b = 6, c = 9, d = 4orb = 9, c = 24, c = 16.

30 Instructor s Solutions Manual, Section 2.4 Exercise 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. such that Because 3 and 2 are zeros of p, there is a polynomial q 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 + 3orq(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 Thus b = 4, c = 3, d = 18 or b = 1, c = 8, c = 12.

31 Instructor s Solutions Manual, Section 2.4 Problem 31 Solutions to Problems, Section Show that if p and q are nonzero polynomials with deg p<deg q, then deg(p + q) = deg q. 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.

32 Instructor s Solutions Manual, Section 2.4 Problem Give an example of polynomials p and q such that deg(pq) = 8 and deg(p + q) = 5. Define polynomials p and q by the formulas p(x) = x 5 and q(x) = x 3. Then and (pq)(x) = p(x) q(x) = x 5 x 3 = x 8 (p + q)(x) = p(x) + q(x) = x 5 + x 3. Thus deg(pq) = 8 and deg(p + q) = 5.

33 Instructor s Solutions Manual, Section 2.4 Problem Give an example of polynomials p and q such that deg(pq) = 8 and deg(p + q) = 2. 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.

34 Instructor s Solutions Manual, Section 2.4 Problem 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.] (a) Note that q(2) = = 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) 11 (2 + x) 2 = q(2 + x) q(2) x = 2x x + 21, where the last equality comes from Exercise 17. Because x is very small, 2x x is also very small, and thus the last equation shows that the slope of this line is approximately 21.

35 Instructor s Solutions Manual, Section 2.4 Problem 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.] (a) Note that s(1) = = 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) 2 (1 + x) 1 = s(1 + x) s(1) x = 4x x + 12, where the last equality comes from Exercise 18. Because x is very small, 4x x is also very small, and thus the last equation shows that the slope of this line is approximately 12.

36 Instructor s Solutions Manual, Section 2.4 Problem 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). 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.

37 Instructor s Solutions Manual, Section 2.4 Problem 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. 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.

38 Instructor s Solutions Manual, Section 2.4 Problem Verify that (x + y) 3 = x 3 + 3x 2 y + 3xy 2 + y 3. (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

39 Instructor s Solutions Manual, Section 2.4 Problem Verify that x 3 y 3 = (x y)(x 2 + xy + y 2 ). (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 = x 3 y 3

40 Instructor s Solutions Manual, Section 2.4 Problem Verify that x 3 + y 3 = (x + y)(x 2 xy + y 2 ). (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 = x 3 + y 3

41 Instructor s Solutions Manual, Section 2.4 Problem Verify that x 5 y 5 = (x y)(x 4 + x 3 y + x 2 y 2 + xy 3 + y 4 ). (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 = x 5 y 5

42 Instructor s Solutions Manual, Section 2.4 Problem Verify that x = (x 2 + 2x + 1)(x 2 2x + 1). (x 2 + 2x + 1)(x ) = ( (x 2 + 1) + 2x )( (x 2 + 1) 2x ) = (x 2 + 1) 2 ( 2x) 2 = x 4 + 2x x 2 = x 4 + 1

43 Instructor s Solutions Manual, Section 2.4 Problem Write the polynomial x as the product of two polynomials of degree 2. [Hint: Use the result from the previous problem with x replaced by x 2.] Replacing x by x 2 on both sides of the result from the previous problem, we have ( x 2 which can be rewritten as ) = ( ( x 2 ) x x 4 ( x = 4 + )( ( x 2 ) 2 ) x , 2 )( x 2 2 ) 2 x x + 1. 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 = (x x + 1)(x 2 2 2x + 1).

44 Instructor s Solutions Manual, Section 2.4 Problem Show that (a + b) 3 = a 3 + b 3 if and only if a = 0orb = 0ora = b. First we expand (a + b) 3 : (a+b) 3 = (a+b)(a+b) 2 = (a+b)(a 2 +2ab+b 2 ) = a 3 +3a 2 b+3ab 2 +b 3. Thus (a + b) 3 = a 3 + b 3 if and only if 0 = 3a 2 b + 3ab 2 = 3ab(a + b), which happens if and only if and only if a = 0orb = 0ora = b.

45 Instructor s Solutions Manual, Section 2.4 Problem Suppose d is a real number. Show that if and only if d = 0. (d + 1) 4 = d First we expand (d + 1) 4 : (d + 1) 4 = ( (d + 1) 2) 2 = (d 2 + 2d + 1) 2. = d 4 + 4d 3 + 6d 2 + 4d + 1. Thus (d + 1) 4 = d if and only if 0 = 4d 3 + 6d 2 + 4d = 2d(2d 2 + 3d + 2), which happens if and only if d = 0or2d 2 + 3d + 2 = 0. However, the quadratic formula shows that there are no real numbers d such that 2d 2 + 3d + 2 = 0. Hence we conclude that (d + 1) 4 = d if and only if d = 0.

46 Instructor s Solutions Manual, Section 2.4 Problem Suppose p(x) = 3x 7 5x 3 + 7x 2. (a) Show that if m is a zero of p, then 2 m = 3m6 5m (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. (a) Suppose m is a zero of p. Then 0 = p(m) = 3m 7 5m 3 + 7m 2. Adding 2 to both sides and then dividing by m shows that 2 m = 3m6 5m (b) Suppose m is an integer and is a zero of p. Because m is an integer, 3m 6 5m is also an integer. Thus part (a) implies that 2 m is an integer, which implies that m = 2orm = 1orm = 1orm = 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.

47 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.

48 Instructor s Solutions Manual, Section 2.4 Problem 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}. Suppose m is an integer that is a zero of p. Then 0 = p(m) = am 3 + bm 2 + cm + 9. Subtracting 9 from both sides and then dividing by m shows that 9 m = am2 bm c. Because a, b, c, and m are all integers, am 2 bm c is also an integer. Thus the equation above shows that 9 m is an integer, which implies that m equals 9, 3, 1, 1, 3, or 9.

49 Instructor s Solutions Manual, Section 2.4 Problem Suppose p(x) = a 0 + a 1 x + +a n x n, where a 1,a 2,...,a n are integers. Suppose m is a nonzero integer that is a zero of p. Show that a 0 /m is an integer. Because m is a zero of p, we have 0 = p(m) = a 0 + a 1 m + +a n m n. Subtracting a 0 from both sides and then dividing both sides by m shows that a 0 m = a 1 a 2 m a n m n 1. Because a 1,a 2,...,a n and m are all integers, a 1 a 2 m a n m n 1 is also an integer. Thus the equation above shows that a 0 /m is an integer.

50 Instructor s Solutions Manual, Section 2.4 Problem Give an example of a polynomial of degree 5 that has exactly two zeros. 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.

51 Instructor s Solutions Manual, Section 2.4 Problem Give an example of a polynomial of degree 8 that has exactly three zeros. 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.

52 Instructor s Solutions Manual, Section 2.4 Problem Give an example of a polynomial p of degree 4 such that p(7) = 0 and p(x) 0 for all real numbers x. 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 x x , which explicitly shows that p is a polynomial of degree 4.

53 Instructor s Solutions Manual, Section 2.4 Problem Give an example of a polynomial p of degree 6 such that p(0) = 5 and p(x) 5 for all real numbers x. Define p by p(x) = x Then clearly p is a polynomial of degree 6 and p(0) = 5 and p(x) 5 for all real numbers x.

54 Instructor s Solutions Manual, Section 2.4 Problem Give an example of a polynomial p of degree 8 such that p(2) = 3 and p(x) 3 for all real numbers x. Define p by p(x) = x 6 (x 2) 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.

55 Instructor s Solutions Manual, Section 2.4 Problem Explain why there does not exist a polynomial p of degree 7 such that p(x) 100 for every real number x. Suppose p is a polynomial of the form p(x) = a 0 + a 1 x + a 2 x 2 + a 3 x 3 + a 4 x 4 + a 5 x 5 + a 6 x 6 + a 7 x 7, where a 7 0. Then p behaves approximately the same as a 7 x 7 near ±. Ifa 7 > 0, this means that p(x) is a negative number with very large absolute value for x near. Ifa 7 < 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.

56 Instructor s Solutions Manual, Section 2.4 Problem Explain why the composition of two polynomials is a polynomial. Suppose q is a polynomial and k is a positive integer. Define a function r k by r k (x) = ( q(x) ) k = q(x) q(x) q(x). }{{} k times Then r k is a polynomial because the product of polynomials is a polynomial. Suppose now that p is a polynomial defined by Thus p(x) = a 0 + a 1 x + a 2 x 2 + +a m x m. (p q)(x) = p ( q(x) ) = a 0 + a 1 q(x) + a 2 ( q(x) ) 2 + +am ( q(x) ) m. The equation above shows that p q = a 0 + a 1 r 1 + a 2 r 2 + +a m r m. Each term a k r k is a polynomial because each r k 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.

57 Instructor s Solutions Manual, Section 2.4 Problem Show that if p and q are nonzero polynomials, then deg(p q) = (deg p)(deg q). Suppose q is a polynomial with degree n and k is a positive integer. Define a function r k by Thus r k (x) = ( q(x) ) k = q(x) q(x) q(x). }{{} k times deg r k = 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 where a m 0. Thus p(x) = a 0 + a 1 x + a 2 x 2 + +a m x m,

58 Instructor s Solutions Manual, Section 2.4 Problem 56 (p q)(x) = p ( q(x) ) = a 0 + a 1 q(x) + a 2 ( q(x) ) 2 + +am ( q(x) ) m. The equation above shows that p q = a 0 + a 1 r 1 + a 2 r 2 + +a m r m. Each term a k r k with a k 0 is a polynomial with degree kn. In particular, the term a m r m 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 a m r m (x). Thus deg(p q) = mn = (deg p)(deg q).

59 Instructor s Solutions Manual, Section 2.4 Problem In the first figure in the 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, the graph of p seems to be on or above the x-axis for all values of p in the interval [0, ]. Explain. There is actually no contradiction between the two graphs in the 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 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.

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