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.

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