Factoring Patterns 5-7 Factoring Pattern for x 2 + bx + c, c positive Objective To factor quadratic trinomials whose quadratic coefficient is 1 and whose constant term is positive. In this lesson you will study trinomials that can be factored as a product of the form (x + r)(x + s), where r and s are both positive or both negative. The diagram shows that the product (x + r)(x + s) and the trinomial x2 + (r + s)x + rs represent the same total area. Notice that the coefficient of the x-term is the sum of r and s, and the constant term is the product of r and s. Example 1 (x + 3)(x + 5) = x2 + 8x + 15 sum of 3 and 5 I T product of 3 and 5 Example 2 (x — 6)(x – 4) = x2 – 10x+ 24 sum of –6 and –4-1 product of –6 and –4 The examples above suggest the following method for factoring trinomials whose quadratic coefficient is 1 and whose constant term is positive. 1. List the pairs of factors whose products equal the constant term. 2. Find the pair of factors in the list whose sum equals the coefficient of the linear term. Examples 1 and 2 suggest that in Step 1 you need to consider only the factors with the same sign as the linear term. Example 3 Factor y2 + 14y + 40. Solution 1. Since the coefficient of the linear term is positive, list the pairs of positive factors of 40. 2. Find the factors whose sum is 14: 4 and 10. 3. y2 + 14y + 40 = (y + 4)(y + 10) Answer Factors of 40 1 2 4 5 Sum of the factors 40 20 10 8 Factoring Polynomials 41 22 14 13 213 Example 4 Factor y2 — 11y + 18. Solution 1. Since —11 is negative, think of the negative factors of 18. 2. Select the factors of 18 with sum —11: —2 and —9 3. .•. y2 — 1 ly + 18 = (y — 2)(y — 9) Answer A polynomial that cannot be expressed as a product of polynomials of lower degree is said to be irreducible. An irreducible polynomial with integral coefficients whose greatest monomial factor is 1 is a prime polynomial. Example 5 Factor x2 — 10x + 14. Solution 1. The pairs of negative factors of 14 are: —1, —14; —2, —7. 2. Neither of these pairs has the sum —10. 3. x2 — 10x + 14 cannot be factored. It is a prime polynomial. Answer Oral Exercises The area of each rectangle is represented by the trinomial below it. Use the diagram to factor the trinomial. You may wish to make models from grid paper, using a 10-by-10 square for x2, a 10-by-1 rectangle for x and a 1-by-1 square for 1. x Sample 6 Solution x2 x 12 12 x2 + 8x + 12 = (x + ?)(x + ?) 9 1. .v x2 + 8x + 12 = (x + 2)(x + 6) 2. 3. ?x 9.A- 4 x2 + 5x + 4 = (x + ?)(x + ?) 8 Chapter 5 x2 ?x- 12 x 2 + 6x + 8 = (x + ?)(x + ?) 214 x x2 + 7x + 12 = (x + ?)(x + ?) Find two integers with the given sum and product. Example 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. Sum 5=4+1 5 7 -6 -6 8 10 -9 -11 -10 12 Product 4=4-1 6 6 8 9 15 16 18 24 24 32 For each trinomial tell which two factors of the constant term have a sum equal to the coefficient of the linear term. Sample x2 - 13x + 22 Solution (-2)(-11) = 22 and -2 + (-11) = -13 -2 and -11 are the correct factors. Answer 14. x2 + 8x + 7 17. y2 + 7y + 12 20. r2 + 9r + 20 23. y2 + 25y + 24 15. 18. 21. 24. z2 - 6z + 5 c2 - 15c + 14 s2 - 12s + 20 x2 + 1 lx + 28 16. 19. 22. 25. p2 - Sp + 6 u2 + llu + 18 x2 - 14x + 24 n2 - 17n + 30 Written Exercises Factor. Check by multiplying the factors. If the polynomial is not factorable, write prime. A 1. x2 + 5x + 4 4. c2 - 10c + 16 7. q2 16q + 15 10. s2 - 12s + 40 13. u2 + 12u + 28 16. 64 - 20s + s2 3. r2 - 6r + 8 6. p2 - 14p + 13 9. a2 - 15a + 26 12. z2 + 16z + 39 15. 42 - 23k + k2 18. 75 + 27u + u2 2. z2 + 9z + 8 5. y2 - 9y + 14 8. n2 + 10n + 21 11. x2 + 20x + 36 14. x2 - 22x + 72 17. 75 + 20r + r2 Sample x2 - 10xy Solution x2 - 10xy + 21y2 = (x - ?)(x - ?) = (x - 3y)(x - 7y) 21y2 Check: (x - 3y)(x - 7y) = x 2 - 3xy - 7.xy + 21y2 = x2 - 10xy + 21y2 19. p2 + 19pq + 34q2 22. x2 - 15xy + 72y2 20. a2 + 10ab + 24b2 23. u2 - 50uv + 49v2 25. x2 - 16xy + 45y2 28. p2 + 20pq + 50q2 26. m2 + 20mn + 51n2 21. c2 - 16cd + 48d2 24. h2 - 14hk + 49k2 27. a2 + 17ab + 52b2 29. r2 - 15rs + 54s2 30. a2 - 12ab + 27b2 Factoring Polynomials 215 Factor. Check by multiplying the factors. B 31. y2 + 20yz + 91z2 34. 108 - 24y + y2 32. w2 + 20wm + 96m2 35. 112a2 - 22ab + b2 33. 124 - 35y + y2 36. 117x2 - 22xy + y2 Find all the integral values of k for which the trinomial can be factored. Sample x2 + kx + 28 Solution 28 can be factored as a product of two integers in these ways: (1)(28) (-1)(-28) (2)(14) (-2)(-14) (4)(7) (-4)(-7) The corresponding values of k are 29, 16, 11, -29, -16, and -11. Answer 37. y2 + ky + 14 40. p2 + kp + 18 38. x2 + kx + 10 41. n2 + kn + 9 39. z2 + kz + 12 42. r2 + kr + 20 Find all positive integral values of k for which the trinomial can be factored. 43. n2 + 6n + k 44. z2 + 7z + k 46. x2 + 9x + k 45. y2 + 8y + k Factor completely. C 47. (y + 2) 2 - 6(y + 2) + 5 49. (y + 3) 2 + 6(y + 3) + 9 51. x4 - 5x2 + 4 48. (t + 3)2 + 8(t + 3) + 15 50. z6 - 17z4 + 16z2 52. r4 - 29r2 + 100 54. (a - 4)2 + 5(a - 4)(a + 2) + 6(a + 2)2 53. t5 - 20t3 + 64t 55. Factor a2n 30anb2n 209b4n, where n is a positive integer. 56. Factor p4" - 30p2nqn + 221q2n, where n is a positive integer. Mixed Review Exercises Solve. 1. -13 + x = -9 4. n + 2 =13 - 61 2. d + (-5) = -6 5. 19m = 76 3. -15 + b = 8 6. 3p + 18 = -72 7. - 1x= 12 8. -r3- - 4 = 5 9. -21x = 252 Simplify. 10. (5y + '7)(5y - 7) 216 Chapter 5 11. (2xy 3)3 12. (2x2)6 5-8 Factoring Pattern for x 2 + bx + c, c negative Objective To factor quadratic trinomials whose quadratic coefficient is 1 and whose constant term is negative. The factoring that you did in the last lesson had this pattern: x2 + bx + c = (x + r)(x + T T c positive r and s are both positive or both negative. The factoring that you will do in this lesson has the following pattern: x2 + bx + c = (x + r)(x + s) T T r and s have opposite signs. c negative T When you find the product (x + r)(x + s), you obtain x2 + bx + c = x2 + (r + s)x + rs Therefore, the method used in this lesson is the same as before. You find two numbers, r and s, whose product is c and whose sum is b. Since c is negative, one of r and s must be negative and the other must be positive. Example 1 Factor x2 — x Solution — 20. 1. List the factors of —20 by writing them down or reviewing them mentally. 2. Find the pair of factors with sum —1: 4 and —5. 3. ... x2 — x — 20 = (x + 4)(x — 5) Answer You can check the result by multiplying (x + 4) and (x — 5). Factors of —20 1 —1 2 —2 4 —4 Sum of the factors —20 20 —10 10 —5 5 —19 19 —8 8 —1 1 Example 2 Factor a2 + 29a — 30. Solution 1. The factoring pattern is (a + ?)(a — ?). 2. Find the pair of factors of —30 with sum 29: 30 and —1. 3. .•. a2 + 29a — 30 = (a + 30)(a — 1) Answer Example 3 Factor x2 4kx — 12k2 . Solution 1. The factoring pattern is (x + ?)(x — ?). 2. Find the pair of factors of —12k 2 with a sum of —4k: 2k and —6k. 3. x2 — 4kx — 12k2 = (x + 2k)(x — 6k) Answer Factoring Polynomials 217 Oral Exercises Find two integers with the given sum and product. Example Sum 1 = 3 + (-2) Product -6 = 3(-2) 5. 4. 8. 7. 6. 9. 10. 2. 3. 2 -3 -2 15 3 -7 1 0 2 -10 -3 -10 -15 -16 -18 -18 -30 -25 -24 -24 1. For each trinomial tell which two factors of the constant term have a sum equal to the coefficient of the linear term. Sample X2 - Solution (-7)(4) = -28 and -7 + 4 = -3 -7 and 4 are the correct factors. 3x - 28 11. z2 + 3z - 4 14. p2 + p - 12 17. x2 + 2x - 15 Answer 12. z2 - 4z - 5 15. y2 - 5y - 14 18. u2 - u 2 13. c2 - c - 6 16. r2 - 2r - 8 19. k2 + 8k - 9 - ■ 111•11 11111111111W Written Exercises Factor. Check by multiplying the factors. If the polynomial is not factorable, write prime. A 1. y2 + 5y - 6 4. x2 + 2x - 8 7. n2 + 2n - 6 10. p2 + 7p 18 13. x2 - 25x - 54 16. z2 + z - 72 19. u2 + 9uv - 70v2 22. m2 + mn - 56n2 - 3. x2 - 6x - 16 6. u2 - 10u - 9 9. b2 - 13b - 30 12. y2 - 4y - 32 15. - 21y - 72 18. r2 - 2Ors - 44s 2 14. t2 - 16t - 40 17. a2 - ab - 42b2 21. h2 - 25hk 54k2 24. a2 - 13ab - 48b2 20. x2 - 2xy - 63y2 23. p2 - 16pq - 36q2 . - Sample 1 - iox - 24x2 Solution Find two factors of -24x 2 whose sum is -10x: 2x and -12x. 1 - 10x - 24x2 = (1 + 2x)(1 - 12x) Answer B 25. 1 - 2n - 48n2 27. x2 - 10xy - 75y 2 218 2. v2 - 3v - 4 5. c2 - 4c - 12 8. a2 - 5a - 24 11. y2 + 12y - 36 Chapter 5 26. 1 + 15c - 34c2 28. a2 + 5ab 84b2 - 30. 1 - 15mn - 100m 2n2 32. n2 + 9n - 400 29. 1 + llpq - 80p2q2 31. p2 + 2p - 360 33. -380 + x + x2 34. -800 - 20a + a 2 Find all the integral values of k for which the given polynomial can be factored. 35. y2 + ky - 28 36. c2 + kc - 20 38. x2 + kx - 36 37. p= + kp - 35 Find two negative values of k for which the given polynomial can be factored. (There may be many possible values.) 39. r2 - 2r + k 42. k - 7r + r2 41. k + 5x + x2 44. k - 6z + z 2 40. y2 + 4y + k 43. k + 4t + t2 Factor completely. C 45. 46. t4 - 7t2 - 18 48. (x + 2) 2 - 4(x + 2) - 21 50. (p + q)2 - 2(p + q) - 15 52. (p + q)2 - 2r(p + q) - 15r2 54. (a + b)4 - (a - b)4 x4 - 3x2 - 4 47. x4 - 15x2y2 - 16y4 49. (y + 3) 2 + 5(y + 3) - 24 51. (a + b)2 - (a + b) - 2 53. (a + b)2 - c(a + b) - 2c2 55. Factor x2n - 4Xny2n 221y4n, where n is a positive integer. 4x2nyn - 252y2n, where n is a positive integer. 56. Factor x4' Mixed Review Exercises Simplify. 1. (9x2y)(3xy2)(2x2) 4. (3x - 4)2 5 00 0 6 10(xy)3 10. (n + 3p)2 7. 2. (3x - 4)(2x + 3) 5 . (7x.sy2)3 3. -7x(3x2 - 2x + 4) 6. 5y(2y2 + 3y + 5) 8. 0. (-n)6 -4ab -12ab3 11. (a - 6)(5a + 2) _ nio 12. (2y + 7)2 Factor. 13. 15m - 21n + 9 16. a2 - 13ab + 42b2 19. u2 - 10u + 25 22. 49w4 - 16x2 25. 9x2 - 24xy + 16y2 28. a2 + 13a - 68 14. 17. 20. 23. 26. 121k2 - 81 16x2 + 24x 44 + 15y + y2 4m2 + 20m + 24 56 - 15z + z2 29. 25w6 - 144x6 15. 18. 21. 24. 27. a2 + 18a + 81 64 - n2 7a2b3 - 14ab c2 - 11c - 26 x2 - 1 30. 25a2 + 20ab + 4b2 Factoring Polynomials 219 5-9 Factoring Pattern for ax e + bx + c Objective To factor general quadratic trinomials with integral coefficients. If axe + bx + c (a> 1) can be factored, the factorization will have the pattern (px + r)(qx + s). Example 1 Factor 2x2 + 7x — 9. Solution Clue 1 Because the trinomial has a negative constant term, one of r and s will be negative and the other will be positive. Clue 2 You can list the possible factors of the quadratic term, 2x2, and the possible factors of the constant term, —9. Test the possibilities to see which produces the correct linear term, 7x. Making a chart will help you do this. Since (2x + 9)(x — 1) gives the correct linear term, 2x2 + 7x — 9 = (2x + 9)(x — 1). Answer Factors of 2x2 2x, x Factors of —9 1, —9 —1, 9 3, —3 —3, 3 9, —1 —9, 1 Possible factors Linear term (2x + 1)(x — 9) (-18 + 1)x = —17x (2x + 3)(x — 3) (-6 + 3)x = —3x (2x + 9)(x — 1) (-2 + 9)x = 7x (2x — 1)(x + 9) (18 — 1)x = 17x (2x — 3)(x + 3) (6 — 3)x = 3x (2x — 9)(x + 1) (2 — 9)x = —7x Example 2 Factor 14x2 — 17x + 5. Solution Clue 1 Because the trinomial has a positive c onstant term and a negative linear term, both r and s will be negative. Clue 2 List the factors of the quadratic term, 14x2 , and the negative factors of the constant term, 5. Test the possibilities to see which produces the correct linear term, —17x. Since (2x — 1)(7x — 5) gives the correct linear term, 14x2 — 17x + 5 = (2x — 1)(7x — 5). Answer 220 Chapter 5 Factors of 14x2 x, 14x 2x, 7x Possible factors (x — 1)(14x — 5) (x — 5)(14x — 1) (2x — 1)(7x — 5) (2x — 5)(7x — 1) Factors of 5 —1, —5 —5, —1 Linear term (-5 — 14)x = —19x (-1 — 70)x = —71x (-10 — 7)x = —17x (-2 — 35)x = —37x Remember to check each factorization by multiplying the factors. After some practice you will be able to select the correct factors without writing down all the possibilities. When the coefficient of the quadratic term is negative, it may be helpful to begin by factoring —1 from each term. Example 3 Factor 10 + llx — 6x2 . I Arrange the terms by Idecreasing degree. 10 + 1 lx — 6x2 = —6x2 + 1 lx + 10 { Factor —1 from each term. = ( —1)(6x2 — 1 lx — 10) { Factor the resulting trinomial. = (-1)(2x — 5)(3x + 2) = —(2x — 5)(3x + 2) Answer Note: If you factor 10 + l lx — 6x2 directly, you will get (5 — 2x)(2 + 3x). Since (5 — 2x) = —(2x — 5), the two answers are equivalent. Solution Example 4 Factor 5a 2 — ab — 22b2 . Solution {Write the factors of 5a 2 . ) )(5a { Test possibilities. = (a + ?)(5a — ?) = (a + 2b)(5a — 11b) Answer If you write (a — ?)(5a + ?) as the second step, you will not find a combination of factors that produces the desired linear term. 5a2 — ab — 22b2 = (a Note: Oral Exercises The area of the rectangle is represented by the trinomial below. Use the diagram to factor the trinomial. Sample Solution 4x' 6.v 2x 3 4x2 + 8x + 3 4.v2 1 2.v 4x2 + 8x + 3 =(2x+3)(2x+ 1) 1. 3 v= 3x 3x2 + 5x + 2 2x 6x= 4x 3x 6x2 + 7x + 2 Factoring Polynomials 221 For each quadratic trinomial tell whether its factorization will have the form (px + r)(qx + s), (px + r)(qx - s), or (px - r)(qx - s), where p, q, r, and s represent positive integers. 3. 2x2 + x - 6 6. 4x2 - 4x - 3 9. 3x2 + 4x - 4 12. 9x2 + 6x - 8 4. 5x2 - 13x + 6 7. 2x2 - x - 10 5. 4x2 + 8x + 3 8. 6x2 + 5x + 1 11. 8x2 - 25x + 3 14. 10x2 - 10x - 9 10. 5x2 - 1 lx + 2 13. 14x2 + 13x + 3 Written Exercises Factor. Check by multiplying the factors. If the polynomial is not factorable, write prime. A 1. 3x2 + 7x + 2 3. 3c2 - 8c + 5 5. 5y2 + 4y - 1 7. 5u2 - 6u - 2 9. 7x2 + 8x + 1 5x2 - 17x + 6 3. 3p2 + 7p - 6 15. 4y2 - y - 3 4. 2x2 - 15x + 7 6. 3a2 + 4a - 4 8. 3r2 - 2r - 5 10. 2p2 + 7p + 3 12. 7m2 - 9m + 2 14. 4c2 + 4c - 3 16. 6a2 - 5a - 2 5 + 7x - 6x2 9. 1 - 5b - 8b2 21. 3m2 + 1 lmn + 6n2 18. 9 + 6k - 8k2 20. 7 - 12s - 4s2 22. 2p2 - 7pq + 6q2 23. 2x2 + xy - 3y2 24. 5a2 - 2ab - 7b2 B 25. 9m2 - 25mn - 6n2 27. 6r2 - 1 lrp + 5p2 29. 21c2 + 4c - 12 31. 6 + 7a - 20a2 26. 28. 30. 32. 33. 32n2 - 4n - 15 35. 21c2 + 22c - 24 34. 33u2 - u - 14 36. 35y2 + 2y - 24 C 37. 2(a + 2)2 + 5(a + 2) - 3 39. 2(a + 2b) 2 + 5(a + 2b)c - 3c2 41. 4x4 - 17x2 + 4 43. (y2 + 3y - 1)2 - 9 222 2. 2x2 + 5x + 3 Chapter 5 6h2 + 17hk + 10k2 4x2 + 16xy - 9y2 18z2 + 19z - 12 8 + 45r - 18r2 38. 2(x - 1)2 - 9(x - 1) - 5 40. 2(x - y)2 - 9(x - y)z - 5z2 42. 2x4 - 15x2 - 27 44. (a2 - 4a - 1)2 - 16 45. Show that (15x2 — 14x + 3)(6x2 + 19x — 7)(10x2 + 29x — 21) is a perfect square by showing that it is the square of a polynomial. 46. Factor 90a8'1+1b2 25,24n+t b2n+2 _ 240ab 4n+2, where n is a positive integer. Mixed Review Exercises Factor. 1. x2 — 225 4. c2 — 6c + 9 7. y2 + 15y + 56 10. m2 — m — 72 2. x2 — 9x + 20 5. 9y2 — 289x2 8. p2 + 12p + 36 11. n2 + 15n + 36 3. r2 — 5r — 14 6. 4a4 — 49 9. 16y2 + 24y + 9 12. b2 — 2b — 24 Self-Test 3 Vocabulary irreducible polynomial (p. 214) prime polynomial (p. 214) Factor. 1. a2 + 12a + 35 3. n2 — 3n — 28 5. 2r2 — 7r + 6 Obj. 5-7, p. 213 2. x2 — 10x + 16 4. c2 + 3cd — 40d2 Obj. 5-8, p. 217 6. 3x2 + 10xy — 8y2 Obj. 5-9, p. 220 ■ MIA I I VA I I I /Mil/AMP/AM I I/AM I VA INIAI Challenge According to the legend, the inventor of the game of chess asked to be rewarded by having one grain of wheat put on the first square of a chessboard, two grains on the second, four grains on the third, eight grains on the fourth, and so on. The total number of grains would be 264 — 1, which is several thousand times the world's annual wheat yield. 1. To find how large 264 is approximately, you could enter the number 2 on the calculator and press the squaring button a number of times. How many times must you press this button? 2. Factor 264 — 1 as a difference of squares to show that it is divisible by 3, 5, and 17. Factoring Polynomials 223 General Factoring and Its Application 5-10 Factoring by Grouping Objective To factor a polynomial by grouping terms. A key tool in factoring polynomials is the distributive property: ba + ca = (b +c)a This property is valid not only when a represents a monomial, but also when a represents any polynomial. For example: If a = x + 2, you have b(x + 2) + c(x + 2) = (b + c)(x + 2) If a = 3r — s + 7, you have b(3r — s + 7) + c(3r — s + 7) = (b + c)(3r — s + 7) Example 1 Factor: a. 5(x + y) + w(x + y) b. m(m + 4n) — (m + 4n) Solution a. (5 + w)(x + b. (m — 1)(m + 4n) Another helpful tool is recognizing factors that are opposites of each other. Factor x—y 4 — a2 2n — 3k — 1 Opposite —(x — y) —(4 — a2) —(2n — 3k — 1) or or or —x + y —4 + a2 —2n + 3k + 1 or y — x or a2 — 4 or 3k — 2n + 1 Example 2 Factor 5(a — 3) — 2a(3 — a). Solution Notice that a — 3 and 3 — a are opposites. 5(a — 3) — 2a(3 — a) = 5(a — 3) — 2a[—(a — 3)] = 5(a — 3) + 2a(a — 3) = (5 + 2a)(a — 3) Answer In Example 3 you first group terms with common factors, and then factor. Example 3 Factor 2ab — 6ac + 3b — 9c. Solution 1 2ab — 6ac + 3b — 9c = (2ab — 6ac) + (3b — 9c) = 2a(b — 3c) + 3(b — 3c) = (2a + 3)(b — 3c) Answer 224 Chapter 5 Solution 2 2ab — 6ac + 3b — 9c = (2ab + 3b) — (6ac + 9c) = b(2a + 3) — 3c(2a + 3) = (b — 3c)(2a + 3) Answer Example 4 uses what you know about factoring perfect square trinomials and differences of squares. Example 4 Factor 4p2 — 4q2 + 4qr — r2 . Solution A trinomial square 4p2 — 4q2 + 4qr — r2 = 4p2 — (4q2 — 4qr + r2) The difference of two squares. < = (2p) 2 — (2q = [2p + (2q — r)][2p — (2q — r)] = (2p + 2q — r)(2p — 2q + r) Answer In Example 4 you could have tried the grouping (4p 2 — 4q2) + (4qr — r2) and factored the groups to obtain 4(p + q)(p — q) + r(4q — r). But this doesn't lead anywhere. There are different approaches to factoring a polynomial. You may need to try several before arriving at one that works. Oral Exercises Factor. 1. a(a — 2) + 3(a — 2) 4. x(x + 2y) — (x + 2y) 7. x(x — 4) — (4 — x) 2. p(q + 1) — 4(q + 1) 5. u(u + v) — v(u + v) 8. m(n — m) — n(m — n) 3. 2r(r — 3) — 5(r — 3) 6. h(h — 2) + 2(2 — h) 9. 2r(r — s) + s(s — r) Written Exercises Factor. Check by multiplying the factors. A 1. 3(x + y) + z(x + y) 3. e(f — g) — 4(f — g) 5. 7(r — s) + t(s — r) 7. 2a(a + 3) — (3 + a) 9. 2x(x — y) + y(y — x) 11. 2u(u — 2v) + v(u — 2v) + (u — 2v) 2. 7(r — s) + t(r — s) 4. w(x — y) — 8(x — y) 6. 7(m — n) + p(n — m) 8. u(v — 2) + 2(2 — v) 10. 3p(2q — p) — 2q(p — 2q) 12. a(a — b) + 4b(a — b) — a(a — b) 13. x(2w — 3v + u) — (2w — 3v + u) 15. (s2 — 2ps + 2s) — (2s — 4p + 4) 14. r(r — s — 2t) + s(r — s — 2t) 16. (x2 — xy + x) — (y — x — 1) 17. (3t — 3st) + (rs — r) 18. (9p — 3pq) + (2nq — 6n) 19. (12x2 — 8xy) — 5(3xz — 2yz) 20. (p2 — 2pq) — 2(2qr — pr) Factoring Polynomials 225 Factor. Check by multiplying the factors. 21. 3a + ab + 3c + bc 23. x2 - 2x + xy - 2y 25. h2 - hk + hr - kr 22. rs + 5r + st + 5t 24. u2 - 2u + uv - 2v 27. p3 - 2p2 + 4p - 8 29. p2 - 2pq + pr - 2qr 31. 3hk - 2k - 12h + 8 33. 4z3 - 6z2 - 6z + 9 35. (h2k2 + 4k2) + (h2k + 4k) 37. x3 - 3x2 - x + 3 28. 3a3 + a2 + 6a + 2 30. u2 - 3uv - 6uw + 18vw 32. 3ab - b - 4 + 12a 34. 3u3 - u2 - 9u + 3 36. (a2b2 + 2a2) - (2ab2 + 4a) 38. n 3 + 2n2 - 4n - 8 26. x2 - 2xy + 4xz - 8yz Factor each expression as a difference of squares. 40. (a + 2b)2 - 9c2 39. x2 - (y - z) 2 41. (u - 2v)2 - 4w2 42. 4p2 - (q - 2r)2 B 43. 44. 4(x + y) 2 - (2y - z) 2 46. x2 - 2xy + y2 - 4 (a + 2b)2 - (2b + c)2 45. a2 + 4a + 4 - b2 47. u2 - v2 + 2v - 1 49. h2 - 4k2 - 4h + 4 51. p 2 a2 r.2 _ 2pr 48. m2 - n2 - 2m + 1 50. a2 - b2 - 2a + 1 52. 4s2 - 4t2 + 4s + 1 Factor. 53. x2 - 4y2 + 4z2 - 4xz 55. a2 + b2 + 2ab + 2a + 2b 54. m2 - 9n2 + 9 - 6m 56. p2 - q2 - 2p + 2q 57. x4 - y4 - 4x2 + 4 59. p2 + q2 - r2 - 2pq + 2r - 1 58. a4 + b4 - c4 + 2a2b2 60. h2 - 4k2 + 4h - 8k C 61. Factor x4 + 4 by writing it as (x4 + 4x2 + 4) - 4x2 , a difference of two squares. 62. Use the method of Exercise 61 to factor (a) 64x4 + 1 and (b) x4 + 4a4. 63. Factor a2n± 1 b2"+1 a2nb2n + a to. where n is a positive integer. Mixed Review Exercises Solve. 1. -12 + x = -29 4. 16 = 1 + 3z 7. 14x -= 700 10. 10n = 2n - 24 226 Chapter 5 2. -n + 10 = 2 5. 10m - 6m = 36 8. -13n = 156 11. 19m = 55 + 14m 3. 18 + x = 32 6. 5n - 2n + 8 = 9 9. 9b = 108 12. lOy + 6 = 4(19 - y) 5-11 Using Several Methods of Factoring Objective To factor polynomials completely. A polynomial is factored completely when it is expressed as the product of a monomial and one or more prime polynomials 1. Factor out the greatest monomial factor first. 2. Look for a difference of squares. 3. Look for a perfect square trinomial. 4. If a trinomial is not a square, look for a pair of binomial factors. 5. If a polynomial has four or more terms, look for a way to group the terms in pairs or in a group of three terms that is a perfect square trinomial. 6. Make sure that each binomial or trinomial factor is prime. 7. Check your work by multiplying the factors. Example 1 Factor —4n4 + 40n3 — 100n2 completely. Solution —4n4 + 40n3 — 100n2 = —4n2(n2 — 10n + 25) perfect square trinomial greatest monomial factor = —4n2(n — 5) 2 Answer Example 2 Factor 5a3b2 3a4b — 2a2b3 completely. Solution First rewrite the polynomial in order of decreasing degree in a. 5a3b2 3a4b — 2a2b3 = 3a4b 5a3b2 — 2a2b3 = a2b(3a2 + 5ab — 2b2) trinomial greatest monomial factor --I = a2b(3a — b)(a + 2b) Answer Example 3 Factor a2bc — 4bc + a2b — 4b completely. Solution a2bc — 4bc + a2b — 4b = b(a2c — 4c + a2 — 4) = b[c(a2 — 4) + (a2 — 4)] {Factor by grouping. = b(c + 1)(a 2 4) < Difference of squares = b(c + 1)(a + 2)(a — 2) Answer Factoring Polynomials 227 Oral Exercises State the greatest monomial factor of each polynomial. 1. 6a2 - 9ab - 15b2 2b - 15a3b4.6ab+9 3. 15r3 + 20r2s - 20rs2 6. 12x2y - 36xy2 + 27y3 2. 18x - 8x3 5. 4(z - 4)2 - 16 Factor completely. 7. 10a2 - 15ab2 4 + 6y3 + 9y2 10.y 8. -4x + 6x2 11. p3 - 2p2q + pq2 9. t3 - 9f 12. u3v - uv3 Written Exercises Factor completely. A 1-6. The polynomials in Oral Exercises 1-6. 7. 5a2 + 10ab + 5b2 9. 4m3 - m 11. y4 - 2y2 - y3 10. 3xy 2 - 27x3 12. -n4 - 3n2 - 2n3 13. x2 - xy - x + y - 14. 15. -41a + 10 + 21a2 17. a3 - 2a2b + 3a2 - 6ab 19. 6U2V 11U2V2 10U2v3 21. k(k + 1)(k + 2) - 3k(k + 1) 23. 2u5 - 7u3 - 4u 16. 80 - 120p + 45p2 -m2 + mn + 2m - 2n 25. r2 - 6r - 9s2 + 9 27. u2 - 4v2 + 3u - 6v 8p3q - 18pq3 180x2y - 108xy2 - 75x3 n(n2 - 1) + n(n - 1) 81a + 18a3 + a5 26. x2 - 4y2 - 4x + 4 28. a2 - b2 + ac - bc 29. p2 - 1 - 4q2 - 4q 30. x2 - 2x - 4y2 - 4y B 31. 100 + 4x2 - 16y2 - 40x 33. a4 - b4 35. 2pq + 2pr + q2 - r2 37. (a + b)2 - (a - c)2 39. x3 x2y xy2 + y3 41. a(a + 2)(a - 3) - 8(a - 3) 43. 16c 16 - 16 45. a(a2 - 9) - 2(a + 3)2 47. 9u2 - 9v2 - 36w2 + 36vw 228 8. 6c2 + 18cd + 12d2 Chapter 5 18. 20. 22. 24. 32. 16x2 + 16y - y2 - 64 34. m8 - n8 36. 8a3 + 4a2b - 2ab2 - b3 38. 3x5 + 15x3 - 108x 40. 4 - 4x2 - 4y2 + 8xy 42. x(x + 1)(x - 4) + 4(x + 1) 44. (u - v) 3 + v - u 46. (x - 2)(x2 - 1) - 6x - 6 48. x4 - x2 + 4x - 4 C 49. x2(x + 2) — x(x + 2) — 12(x + 2) 51. et — 10t2 + 9 50. (a + b)3 — 6(a + b)2 — 7(a + b) 52. 16/4 — 8t2 + 1 53. a2 + b2 — c 2 — d2 — 2ab + 2cd 54. (u2 — v2)2 — w2(u + v)2 55. Factor x4 + x2 + 1 by writing it as (x4 + 2x2 + 1) — x2 , a difference of squares. 56. Factor a4 a2b2 + b4 . (Hint: See Exercise 55.) 57. Factor a3 + b3 by writing it as a 3 alb — a2b — ab2 + ab2 + b3 and grouping the terms by pairs. 58. Factor a3 — b3 . (Hint: See Exercise 57.) . Mixed Review Exercises Simplify. 1. (-1)4)(40) 2. 1(56) 8 4. 140b 5. 52 + (*) 3. 91 (72) ( ) 6. 625 + (-5) Factor. 7. x2 — 12x + 35 8. x2 + 3x — 28 10. 2n2 + 19n + 9 11. 3x2 + llx + 10 9. x2 — x — 2 12. (3x -- 12) — 2n(4 — x) Career Note / Draftsperson Manufacturers and construction workers rely on detailed plans of buildings and manufactured products as a guide for production. The plans are prepared by a draftsperson using many different tools. For example, he or she may use a compass, a protractor, a triangle, and a calculator. A draftsperson also makes use of math skills, such as working with fractions, making measurements, and making drawings to different scales. Today draftspersons use computeraided design (CAD) systems to allow them to see many variations of a design. They often specialize in a particular field of work, such as mechanical, electrical, aeronautical, or architectural drafting. A draftsperson needs coursework in mathematics, mechanical drawing, and drafting. Factoring Polynomials 229 5-12 Solving Equations by Factoring Objective To use factoring in solving polynomial equations. The multiplicative property of zero can be stated as follows: If a = 0 or b = 0, then ab = 0. statement The statement above is given in "if-then" form. The converse of a statement in "if-then" form is obtained by interchanging the "if" and "then" parts of the statement as shown below. If ab = 0, then a = 0 or b = 0. <— converse The converse of a true statement is not necessarily true. You can show that the particular converse displayed above is true (Exercise 55, page 233). The words "if and only if" are used to combine a statement and its converse when both are true. The zero-product property stated below combines the multiplicative property of zero and its converse. Zero-Product Property For all real numbers a and 12: ab = 0 if and only if a = 0 or b = 0. A product of factors is zero if and only if one or more of the factors is zero. The zero-product property is true for any number of factors. You can use this property to solve certain equations. Example 1 Solve (x + 2)(x — 5) = 0. Solution One of the factors on the left side must equal zero. Therefore, x + 2 = 0 or x — 5 = 0 x = —2 x=5 Just by looking at the original equation, you might have seen that when x = —2 or x = 5 one of the factors will be zero. Either method gives the solution set {-2, 5}. Example 2 Solve 5n(n — 3)(n — 4) = 0. Solution 5n = 0 or n — 3 = 0 or n=3 n=0 the solution set is {0, 3, 4}. 230 Chapter 5 n— 4 = 0 n=4 Answer Answer A polynomial equation is an equation whose sides are both polynomials. Polynomial equations usually are named by the term of highest degree. If a 0 0: ax + b = 0 is a linear equation. axe + bx + c = 0 is a quadratic equation. ax3 + bx2 + cx + d = 0 is a cubic equation. Many polynomial equations can be solved by factoring and then using the zero-product property. Often the first step is to transform the equation into standard form in which one side is zero. The other side should be a simplified polynomial arranged in order of decreasing degree of the variable. Example 3 Solve the quadratic equation 2x 2 + 5x = 12. Solution 1. Transform the equation into standard form. 2. Factor the left side. 2x2 + 5x — 12 = 0 (2x — 3)(x + 4) = 0 3. Set each factor equal to 0 and solve. 2x — 3 = 0 or x + 4 = 0 2x = 3 x = —4 x= 3 4. Check the solutions in the original equation. 4- 2() 2 + 54) 1 12 2(-4)2 + 5(-4) 1 12 1 12 2(16) — 20 1 12 2(4) + 2++ — 24 = 12 ,/ .*. the solution set is {-}, —4}. 32 — 20 = 12 Answer Example 4 Solve the cubic equation 18y 3 + 8y + 24y2 = 0. Solution 1. Transform the equation into standard form. 2. Factor completely. 3. Solve by inspection or by equating each factor to 0. 4. The check is left to you. .*. the solution set is {0, —1}. 18y3 + 24y2 + 8y = 0 2y(9y2 + 12y + 4) = 0 2y(3y + 2)2 = 0 y = 0 or y = 2 Or y = 2 Answer The factorization in Example 4 produced two identical factors. Since the factor 3y + 2 occurs twice in the factored form of the equation, is a double or multiple root. Notice that we list it only once in the solution set. Factoring Polynomials 231 Caution: Never transform an equation by dividing by an expression containing a variable. Notice that in Example 4, the solution 0 would have been lost if both sides of 2y(9y2 + 12y + 4) = 0 had been divided by 2y. Oral Exercises Solve. 1. x(x - 6) = 0 4. (y -- 2)(y + 3) = 0 2. 2a(a + 1) = 0 5. 0 = (3t - 2)(t - 3) 3. 0 = 3p(2p - 1) 6. x(2x - 5)(2x + 1) = 0 Explain how you could solve the given equation. Then solve. 8. a3 = 4a 11. 9x2 = x3 7. 4x2 - x3 = 0 10. m3 - 2m = m2 9. k2 + 4 = 4k 12. 0 = - n3 + n 13. Give an example of a true "if-then" statement with a false converse. Written Exercises Solve. A 1. (y + 5)(y - 7) = 0 4. 2x(x - 20) = 0 7. 3x(2x + 1)(2x + 5) = 0 10. p2 - p - 6 = 0 13. m2 - 36 = 16m 16. x2 = 20x - 100 19. 4x2 - 9 = 0 22. 3x2 + x = 2 25. 7x2 = 18x - 11 28. 10u3 - 5u2 = 0 31. 8y2 - 9y + 1 = 0 34. 25x2 - 90x = -81 B 37. 4x3 - 12x2 + 8x = 0 40. 9x3 + 25x = 30x2 Sample 1 Solution 232 2. (n + 1)(n + 9) = 0 5. (2t - 3)(3t - 2) = 0 8. n(5n - 2)(2n + 5) = 0 11. 0 = x2 + 14x + 48 14. r2 + 9 = lOr 17. y2 = 16y 20. 25m2 - 16 = 0 23. 4s - 4s2 = 1 26. 2y2 = 25y + 13 29. 0 = 4y3 - 2y2 32. 6h2 + 17h + 12 = 0 35. 4p2 + 121 = 44p 38. 2n3 - 30n2 + 100n = 0 41. y4 - 10y2 + 9 = 0 3. 15n(n + 15) = 0 6. (2u + 7)(3u - 1) = 0 9. y2 - 3y + 2 = 0 12. 0 = k2 - 12k + 35 15. s2 = 4s + 32 18. 9k2 = 4k 21. 6n2 + n = 2 24. r - 6r2 = -1 27. 8u3 - 2u2 = 0 30. 0 = 10x3 - 15x2 33. 15u2 - 14u = 49 36. 6c2 - 72 = llc 39. 9x3 + 9x = 30x2 42. u5 - 13u3 + 36u = 0 (x - 1)(x + 3) = 12 " x2 1- - Chapter 5 3 - 12 = 0 (x - 3)(x + 5) = 0 the solution set is {3, -5} Solve. See Sample 1 on page 232. = 16 43. (z + 1)(z — 5) 44. (2t — 5)(t — 1) = 2 45. (x — 2)(x + 3) = 6 46. (a — 5)(a — 2) = 28 47. x(x — 6) = 4(x — 4) 48. 3(m + 2) = m(m — 2) Find an equation in standard form with integral coefficients that has the given solution set. Sample 2 Solution -4} - 1)(x + 4) = 0 3(x — 2 Multiply by 3 for integral coefficients. + 4) = 0 (3x — 2)(x + 4) = 0 3x2 + 10x — 8 = 0 Answer 51. 2' —21 49. {2, —3} 52. { C 55. — 3' — 54. { --1 2 1} 2'5 Supply the missing reasons in the proof of: If ab = 0, then a = 0 or b = 0. Case 1: If a = 0, then the theorem is true; there is nothing to prove. Case 2: Suppose that a 0 and show that then b = 0. a. Given ab = 0 a. b b. 1 exists a 1 1 c. 71 (ab) = —a-(0) d. a (ab) = 0 d a)b = 0 e 1• b = 0 b=0 f. g e. a f. g. ? Mixed Review Exercises Evaluate if x = 2 and y = 4. 1. (x — y) 4. (5x) 3 7. 4(x + y) 2 2. x4 • y2 5. 4x + y 2 8. (yx)2 3. 5x3 6. 4x2 + y 9. y 2x2 Simplify. 10. (3x3y)(-2xy5) 11. (9a) 3 12. — 5(x + 2) Factoring Polynomials 233 5-13 Using Factoring to Solve Problems Objective To solve problems by writing and factoring quadratic equations. The problems in this lesson all lead to polynomial equations that can be solved by factoring. Sometimes a solution of an equation may not satisfy some of the conditions of the problem. For example, a negative number cannot represent a length or an age. You reject solutions of an equation that do not make sense for the problem. Example 1 A decorator plans to place a rug in a 9 m by 12 m room so that a uniform strip of flooring around the rug will remain uncovered. How wide will this strip be if the area of the rug is to be half the area of the room? Solution Step 1 The problem asks for the width of the strip. Step 2 Let x = the width of the strip. Then 12 — 2x = the length of the rug and 9 — 2x = the width of the rug. Step 3 9m Area of the rug = (Area of the room) (12 — 2x)(9 — 2x) = • 9 • 12 Step 4 108 — 42x + 4x2 = 54 4x2 — 42x + 54 = 0 2(2x2 — 21x + 27) = 0 2[(2x — 3)(x — 9,)] = 0 or 2x — 3 = 0 x= —32 ' or 1.5 2m x— 9 = 0 x=9 When x = 1.5, the area of the rug When x = 9, the length, is (12 — 2x)(9 — 2x) = 9 • 6 = 54 12 — 2x, and width, 9 — 2x, are negative. Since a negative = —1 (Area of the room) ✓ 2length or width is meaningless, reject x = 9 as an answer. the strip around the rug will be 1.5 m wide. Answer Step 5 Check: 234 Chapter 5 The equation in Step 3 of Example 1 has a root that does not check because this equation does not meet the "hidden" requirements that the rug have positive length (12 — 2x > 0) and positive width (9 — 2x > 0). Usually it is easier to write only the equation and then check its roots against other conditions stated or implied in the problem. In the next example both solutions of the equation satisfy the conditions of the problem. You can use the formula h = rt — 4.9t2 to obtain a good approximation of the height h (in meters) of an object t seconds after it is projected upward with an initial speed of r meters per second (m/s). Example 2 An arrow is shot upward with an initial speed of 34.3 m/s. When will it be at a height of 49 m? Solution Step 1 The problem asks for the time when the arrow is 49 m high. Step 2 Let t = the number of seconds after being shot that the arrow is 49 m high. Let h = the height of arrow = 49 m. Let r = initial speed = 34.3 m/s. Step 3 Substitute in the formula: h = rt — 4.9t2 49 = 34.3t — 4.9t2 4.9t2 — 34.3t + 49 = 0 4.9(t2 — 7t + 10) = 0 4.9(t — 2)(t — 5) = 0 Step 4 Completing the solution and checking the result are left for you. A calculator may be helpful. the arrow is 49 m high both 2 s and 5 s after being shot. Answer Problems Solve. A 1. If a number is added to its square, the result is 56. Find the number. 2. If a number is subtracted from its square, the result is 72. Find the number. 3. A positive number is 30 less than its square. Find the number. 4. A negative number is 42 less than its square. Find the number. 5. Find two consecutive negative integers whose product is 90. 6. Find two consecutive positive odd integers whose product is 143. 7. The sum of the squares of two consecutive positive even integers is 340. Find the integers. 8. The sum of the squares of two consecutive negative even integers is 100. Find the integers. Factoring Polynomials 235 Solve. 9. The length of a rectangle is 8 cm greater than its width. Find the dimensions of the rectangle if its area is 105 cm 2 . 10. The length of a rectangle is 6 cm less than twice its width. Find the dimensions of the rectangle if its area is 108 cm 2 . 1. Find the dimensions of a rectangle whose perimeter is 46 m and whose area is 126 m2 . (Hint: Let the width be w. Use the perimeter to find the length in terms of w.) 12. Find the dimensions of a rectangle whose perimeter is 42 m and whose area is 104 m2 . 13. The sum of two numbers is 25 and the sum of their squares is 313. Find the numbers. (Hint: Let one of the numbers be x. Express the other number in terms of x.) 0 The difference of two positive numbers is 5 and the sum of their squares is 233. What are the numbers? 15. Originally the dimensions of a rectangle were 20 cm by 23 cm. When both dimensions were decreased by the same amount, the area of the rectangle decreased by 120 cm 2 . Find the dimensions of the new rectangle. 16. Originally a rectangle was twice as long as it was wide. When 4 m were added to its length and 3 m subtracted from its width, the resulting rectangle had an area of 600 m 2 . Find the dimensions of the new rectangle. In Exercises 17-23, use the formula h = rt — 4.9t2 where h is in meters and the formula h = rt — 1612 where h is in feet. A calculator may be helpful. B A ball is thrown upward with an initial speed of 24.5 m/s. When is it 19.6 m high? (Two answers) 18. A rocket is fired upward with an initial speed of 1960 m/s. After how many minutes does it hit the ground? lJ A batter hit a baseball upward with an initial speed of 120 ft/s. How much later did the catcher catch it? 20. Mitch tossed an apple to Kathy, who was on a balcony 40 ft above him, with an initial speed of 56 ft/s. Kathy missed the apple on its way up, but caught it on its way down. How long was the apple in the air? 21. A signal flare is fired upward with initial speed 245 m/s. A stationary balloonist at a height of 1960 m sees the flare pass on the way up. How long after this will the flare pass the balloonist again on the way down? 236 Chapter 5 22. A ball is thrown upward from the top of a 98 m tower with initial speed 39.2 m/s. How much later will it hit the ground? (Hint: Consider the top of the tower as level zero. If h is the height of the ball above the top of the tower, then h = —98 when the ball hits the ground.) 23. A rocket is fired upward with an initial velocity of 160 ft/s. a. When is the rocket 400 ft high? b. How do you know that 400 ft is the greatest height the rocket reaches? Solve. 24. A garden plot 4 m by 12 m has one side along a fence as shown at the right. The area of the garden is to be doubled by digging a border of uniform width on the other three sides. What should the width of the border be? fence 4 in garden 12 m fence 25. Vanessa built a rectangular pen for her dogs. She used an outside wall of the garage for one of the sides of the pen. She had to buy 20 m of fencing in order to build the other sides of the pen. Find the dimensions of the pen if its area is 48 m2 . 26. A rectangular garden 30 m by 40 m has two paths of equal width crossing through it as shown. Find the width of each path if the total area covered by the paths is 325 m2 . 30 m 27. A box has a square bottom and top 40m and is 5 cm high. Find its volume if its total surface area is 192 cm 2 . 28. The bottom and top of a box are rectangles twice as long as they are wide. Find the volume of the box if it is 4 ft high and has a total surface area of_ 220 ft2 . 29. A 50 m by 120 m park consists of a rectangular lawn surrounded by a path of uniform width. Find the dimensions of the lawn if its area is the same as the area of the path. (Hint: Let x = the width of path.) C 30. The Parkhursts used 160 yd of fencing to enclose a rectangular corral and to divide it into two parts by a fence parallel to one of the shorter sides. Find the dimensions of the corral if its area is 1000 yd 2 . 31. Each edge of one cube is 2 cm longer than each edge of another cube. The volumes of the cubes differ by 98 cm 3 . Find the lengths of the edges of each cube. Factoring Polynomials 237 32. A rectangular sheet of metal is 10 cm longer than it is wide. Squares, 5 cm on a side, are cut from the corners of the sheet, and the flaps are bent up to form an opentopped box having volume 6 L. Find the original dimensions of the sheet of metal. You may wish to make a model. (Recall that 1 L = 1000 cm3 .) Mixed Review Exercises Simplify. 1. (8a2b)(2ab2) 2. (5a2)3 3. 3a(4 — 2b) 4. (6r) (3 5. (8)(16n — 24p) 6. (-28x — 14y)(-4) 7. (3a + 2)(2a2 + 5 — 7a) 8. (3b2y) 2 9. 6x(x2 — 8) Factor completely. 10. —28 + 6m + 10m2 13. y4 — y 3 — 12y2 11. 36a3 — 9ab2 14. 15m2 + 26mn + 8n2 21n2 + 22n — 8 5. 3 + 10x2 — 17x Self-Test 4 Vocabulary factor completely (p. 227) converse (p. 230) polynomial equation (p. 231) linear equation (p. 231) quadratic equation (p. 231) cubic equation (p. 231) standard form of a polynomial equation (p. 231) Factor completely. 1. 7r — 3rt + 7s — 3st 3. 18a3 — 12a2 + 2a n2 — 2n + 1 — 100e 4. 21xy — 18x2 — 6y2 Obj. 5-10, p. 224 Obj. 5-11, p. 227 Solve. 5. - 4k = 32 6. 5m2 + 20m + 20 = 0 8. z3 = z2 + 30z 7. a3 = 169a 9. The length of a rectangle is 9 cm more than its width. The area of the rectangle is 90 cm 2 . Find the dimensions of the rectangle. Check your answers with those at the back of the book. 238 Chapter 5 Obj. 5-12, p. 230 Obj. 5-13, p. 234