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Dubai

Solving Algebraic Equation

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Published in: Mathematics
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Students will learn to solve algebraic equations in the form ax^2+bx+c=0 using middle term factor.

Shanu / Abu Dhabi

8 years of teaching experience

Qualification: Computer Science Engineering and MBA

Teaches: Maths

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  1. 8-7 Solving axA2 + bx + c = 0 Factor each polynomial, if possible. If the polynomial cannot be factored using integers, writeprime . 1.3x +17x+10 SOLUTION: In this trinomial, a = 3, b = 17 and c = 10, so m +p is positive and mp is positive. Therefore, m andp must both be positive. List the positive factors of 3(10) or 30 and identify the factors with a sum of 17. Factors of 30 1, 30 2, 15 Sum 31 17 30 The correct factors are 2 and 15. 31.- + 17 x + 10 3x +mx+px+10 x(3x + 2) + 5(3x + 2) (31 + 2) (x + 5) Write the pattern. Group term * w / com.factors. Factor the CCF 31+2 the com.factar. so, 3x +17x+ 10 = (3x +2) (x +5). eSolutions Manual - Powered by Cognero Page 1
  2. 8-7 Solving axA2 + bx + c 2. u +22x+56 SOLUTION: The GCF of the terms 2x , 22x, and 56 is 2. Factor this first. 2x +22x + 56 = 2(x + 1 Ix + 28) Distributive Property In the trinomial, a = 1, b = 11 and c = 28, so m +p is positive and mp is positive. Therefore, m andp must both be positive. List the positive factors of 1(28) or 28 and identify the factors with a sum of 11. 4) Factors of 28 1, 28 2, 14 Sum 29 16 11 The correct factors are 4 and 7. --1—1 28 + —I— PX 2.8 —x Write the pattern. andp=7 + + (71 + 28) Group terms w I com.factors. F actor CCF (x +9 common factor. so,2x +22x + + + 7). 3. 5x 3x+4 SOLUTION: In this trinomial, a = 5, b = 3 and c = 4, so m +p is negative and mp is positive. Therefore, m andp must have different si ns. List the factors of 5(4) or 20 and identify the factors with a sum of 3. Factors of 20 1, -20 -1, 20 2, -10 2, 10 Sum —19 19 8 8 -1 1 There are no factors of 20 with a sum of 3. This trinomial is prime. eSolutions Manual - Powered by Cognero Page 2
  3. 8-7 Solving axA2 + bx + c 4.3x -lix 20 SOLUTION: In this trinomial, a = 3, b = —Il and c — -1, 60 2, 30 3, 20 4, -15 4, 15 5, -12 6, -10 20, so m +p is negative and mp is negative. Therefore, m andp must have different signs. List the factors of 3(—20) or —60 and identify the factors with a sum of —11. Factors of 1, 2, 30 3, 20 5, 12 6, 10 Sum 59 59 28 28 — 17 17 -11 11 7 7 4 4 The correct factors are 4 and —15 31 — 1 b: —2.0 31 + 4 —15x (1 +20) 31 + 4x x(3x + 4) 5(3x + 4) (31 + 4) (x 5) so,3x - -20 = (3x eSolutions Manual - Powered by Cognero Write the pattern. Group terms w ccm.factors. Factor the GCF (3"+4) is the com.factor. 5). Page 3
  4. 8-7 Solving axA2 + bx + c = 0 Solve each equation. Confirm your answers using a graphing calculator. 5.21 +9x+9=0 SOLUTION: Factor the trinomial. +31+61+9 + + (61 + 9) X (2.1 + 3) + + 3) Solve the equation using the Zero Product Property. 2x43-o 21--3 3 2 The roots are or 3 and 3. 2 x+3-o Confirm the roots using a graphing calculator. Let YI CALC menu to find the points intersection. = 2x + 9x + 9 and Y2 Intersection -1.5 [-5, Il OS by SCI: 1 1-5, 1] SCI: OS by 51 SCI: 1 3 and 3. Thus, the solutions are 2 eSolutions Manual - Powered by Cognero = 0. Use the intersect option from the Page 4
  5. 8-7 Solving axA2 + bx + c 6.3x +17x+20=o SOLUTION: Factor the trinomial. 31 + 17x+20 31 + 121 + + 20 12.1) + (5x 2. O) 3x(x +4) + + 4) Solve the equation using the Zero Product Property. 3x+5-o 3x--5 3 The roots are or 5 and 4 or about —1.667 and —4. 3 Confirm the roots using a graphing calculator. Let YI the CALC menu to find the points intersection. +17x +20 and Y2- — 0. Use the intersect option from Intersection [-5, 1] SCI: OS by 51 SCI: 1 11 OS by 51 SCI: 1 5 Thus, the solutions are and 4. 3 eSolutions Manual - Powered by Cognero Page 5
  6. 8-7 Solving axA2 + bx + c 7.3x -10x+8=o SOLUTION: Factor the trinomial. 3x2 -lox +8-0 3.v2 —6x—4x +8=0 (3x2 — 6x) — (4x — 8) O Solve the equation using the Zero Product Property. 37 = 4 or 4 3 4 x-2=Ü The roots are — and 2 or about 1.333 and 2. Confirm the roots using a graphing calculator. Let YI = 3x CALC menu to find the points intersection. —10x + 8 and Y2 — — 0. Use the intersect option from the [0, 51 OS by SCI: 1 Thus, the solutions are — and 2. eSolutions Manual - Powered by Cognero 10, 51 SCI: OS by 51 SCI: 1 Page 6
  7. 8-7 Solving axA2 + bx + c 8. u -17x+30=o SOLUTION: Factor the trinomial. —171+30 —12 -121 (5x -30) — 2x-(x 6) —5(x — 6) Solve the equation using the Zero Product Property. 2x-5-o 2x-5 2 or x —6—0 5 The roots are — and 6. Confirm the roots using a graphing calculator. Let YI the CALC menu to find the points intersection. Intersectin -17x + 30 and Y2 - — 0. Use the intersect option from [0, 101 1 by [-10, 10] SCI: 1 5 Thus, the solutions are — and 6. [0, 101 1 by [-10, 1 9. CCSS MODELING Ken throws the discus at a school meet. a. What is the initial height of the discus? b. After how many seconds does the discus hit the ground? SOLUTION: eSolutions Manual - Powered by Cognero Page 7
  8. 8-7 Solving axA2 + bx + c = 0 a. The initial height of the discus is for the time t = 0. —1 h —16t 6 (0) + 38(0) +5 —1 0+0+5 5 The initial height of the discus is 5 feet. b. When the discus hits the ground, the height will be zero, so h = 0. Substitute 0 for h in the equation and factor the trinomial. h —16t —16t 5 5 2. or Equation fortheheight Multiply each sideby — 1 m =2andp FactortheGCF + listhe comtnon factor. O ZeroProduct Property Solv e each equation 8 The solutions to the equation are — and — Since time cannot be negative, the discus will hit the ground after 2.5 seconds. Check by substituting 2.5 in for x in the original equation. O O —16t +38+5 -16(2 5) +38(2 5) +5 25) +95+5 0 —100 +95+5 Therefore, the discus will hit the ground after 2.5 seconds. eSolutions Manual - Powered by Cognero — or Page 8
  9. 8-7 Solving axA2 + bx + c = 0 Factor each polynomial, if possible. If the polynomial cannot be factored using integers, writeprime . 10. 5x +34x+24 SOLUTION: In this trinomial, a = 5, b = 34 and c = 24, so m +p is positive and mp is positive. Therefore, m andp must both be positive. List the positive factors of 5(24) or 120 and identify the factors with a sum of 34. Factors of 120 1, 120 2, 60 3, 40 4, 30 5, 24 6, 20 8, 15 10, 12 Sum 121 62 43 34 29 26 23 22 The correct factors are 4 and 30. 5X 34.1 24 5X PX + 24 + + 3 + 2.4 + + 2.4 x(5x + 4) + +4) Write the pattern. andp=30 Group terms w f com.factors. (51+4) is the com.factor. SO, 5X + 34X + (5X + + 6). eSolutions Manual - Powered by Cognero Page 9
  10. 8-7 Solving axA2 + bx + c I I. u + 19x + 24 SOLUTION: In this trinomial, a = 2, b = 19 and c = 24, so m +p is positive and mp is positive. Therefore, m andp must both be positive. List the positive factors of 2(24) or 48 and identify the factors with a sum of 19. Factors of 48 I, 48 2,24 3, 16 4, 12 Sum 49 26 19 16 14 The correct factors are 3 and 16. 2.1 1 9 X 2.4 + +24 mx PX — 16.1 2.4 + (1 + 24) X (2.1 + 3) + 8(2.X- + 3) Write the pattern. andp=16 Group terms w f com. factors. F CCF (2m +3) is the com. factor. SO, u + 19X + (2X + + 8). eSolutions Manual - Powered by Cognero Page 10
  11. 8-7 Solving axA2 + bx + c 12. 4x +22x+10 SOLUTION: The GCF of the terms 4x , 22x, and 10 is 2. Factor this first. 4x + 22x + 10 = 2(2x + 1 Ix + 5) Distributive Property In the trinomial, a = 2, b = 11 and c = 5, so m +p is positive and mp is positive. Therefore, m andp must both be positive. List the factors of 2(5) or 10 and identify the factors with a sum of I l. Factors of 10 1, 10 Sum 11 7 The correct factors are I and 10. 2x +mx+px+5 2.x(x + 5) + + 5) Write the pattern. 111=1 0 anclp=l Group terms w com.factors Factor the CCF (x +5) is the com.factor. so,4x +22x + 2(2x + + 5). eSolutions Manual - Powered by Cognero Page 11
  12. 8-7 Solving axA2 + bx + c 13. 4x +38x+70 SOLUTION: The GCF of the terms 4x , 38x, and 70 is 2. Factor this first. 4x +38x + 70 = 2(2x + 19x + 35) Distributive Property In the trinomial, a = 2, b = 19 and c = 35, so m +p is positive and mp is positive. Therefore, m andp must both be positive. List the positive factors of 2(35) or 70 and identify the factors with a sum of 19. Factors of 70 1, 70 2, 35 5, 14 7, 10 Sum 71 37 19 17 The correct factors are 5 and 14. — & +mx+px+35 — 14x 3 5 21 + 141) + 35) — 2.x(x +7) +7) Write the pattern. andp=14 Group terms w I com. factors. Factor the GCF (x +7) the com.factor. so, 4x +38x + 2(2x + + 7). eSolutions Manual - Powered by Cognero Page 12
  13. 8-7 Solving axA2 + bx + c 14. 2x2 3x 9 SOLUTION: In this trinomial, a = 2, b = 3 and c — 9, so m +p is negative and mp is negative. Therefore, m andp must have different signs. List the factors of 2(—9) or —18 and identify the factors with a sum of 3. Factors of —18 1, -18 -1, 18 2, 9 3, 6 Sum — 17 17 7 7 3 3 The correct factors are 3 and —31—9 21 + mx = 21 + 31 +31 So, 2x 3x 6-9 3(2X + 3) 3) 6. Write the pattern. Group terms w 1' com.factors. Factor the GCF +3) is the com.factor 3). eSolutions Manual - Powered by Cognero Page 13
  14. 8-7 Solving axA2 + bx + c 15. 4x -13x+10 SOLUTION: In this trinomial, a = 4, b = —13 and c = 10, so m +p is negative and mp is positive. Therefore, m andp must both be negative. List the negative factors of 4(10) or 40 and identify the factors with a sum of —13. Factors of 40 — I , 40 2, 20 4, -10 The correct factors are x(4x — 5) (4x 5) (x so,4x - 13x+ 16. u + 3x + 6 SOLUTION: Sum 41 22 — 14 -13 5 and O) 5) 8. Write the pattern. Group term * w com-factors. F the CCF (4x —5) is the com.factor. 2). In this trinomial, a = 2, b = 3, and c = 6, so m +p is positive and mp is positive. Therefore, m andp must both be ositive. List the ositive factors of 2(6) or 12 and identify the factors with a sum of 3. Factors of 12 1, 12 Sum 13 8 7 There are no factors of 12 with a sum of 3. So,2x +3x +6 is prime. eSolutions Manual - Powered by Cognero Page 14
  15. 8-7 Solving axA2 + bx + c 17. 5x +3x+4 SOLUTION: In this trinomial, a = 5, b = 3, and c = 4, so m +p is positive and mp is positive. Therefore, m andp must both be ositive. List the ositive factors of 5(4) or 20 and identify the factors with a sum of 3. 4, 15 Factors of 20 1, 20 2, 10 Sum 21 12 9 There are no factors of 20 with a sum of 3. So, 5x + 3x +4 is prime. 18. 12x +69x+45 SOLUTION: The GCF of the terms 122x , 69x, and 45 is 3. Factor this first. 12x +69x + 45 = 3(4x + 23x + 15) Distributive Property In the trinomial, a = 4, b = 23 and c = 15, so m +p is positive and mp is positive. Therefore, m andp must both be positive. List the positive factors of 4(15) or 60 and identify the factors with a sum of 23. Factors of 60 1,60 2, 30 3,20 5, 12 6, 10 Sum 61 32 23 19 17 The correct factors are 3 and 20. +231+15 = 4x + 31+201+45 Write the pattern. m anclp Group terms w J' com.factors. Factor the GCF (x +5) is the cam.factor. so, 12x +69x +45= 3(4x + + 5). eSolutions Manual - Powered by Cognero Page 15
  16. 8-7 Solving axA2 + bx + c 19. 4x -5x+7 SOLUTION: In this trinomial, a = 4, b = 5, and c = 7, so m +p is negative and mp is positive. Therefore, m andp must both be ne ative. List the ne ative factors of 4(7) or 28 and identify the factors with a sum of —5. Factors of 28 -1, 28 2, — 14 Sum 29 -16 -11 There are no factors of 28 with a sum of So, 4x —5x+7 is prime. 20. 5x +23x +24 SOLUTION: -11. In this trinomial, a = 5, b = 23 and c = 24, so m +p is positive and mp is positive. Therefore, m andp must both be positive. List the positive factors of 5(24) or 120 and identify the factors with a sum of 23. Factors of 120 1, 120 2, 60 3, 40 4, 30 5, 24 6, 20 8, 15 10, 12 Sum 121 62 43 34 29 26 23 22 The correct factors are 8 and 15. 5x 24 5X + PX + 24 + + 1 5 + 2.4 x(5x + 8) + 3(5x + 8) Wiite the pattern. 112=8 anc1p=15 Group terms w j/' com.factors. Factor the GCF (5x +8) the cam.factar. so, 5x +23x + (5x + + 3). eSolutions Manual - Powered by Cognero Page 16
  17. 8-7 Solving axA2 + bx + c 21.3x -8x+15 SOLUTION: In this trinomial, a = 3, b = 8, and c = 15, so m +p is negative and mp is positive. Therefore, m andp must both be ne ative. List the factors of 3(15) or 45 and identify the factors with a sum of —8. Factors of 45 -1, 45 3, -15 Sum 46 -18 — 14 There are no factors of 45 with a sum of —8. So,3x —8x+ 15 is prime. 22. SHOT PUT An athlete throws a shot put with an initial upward velocity of 29 feet per second and from an initial height of 6 feet. a. Write an equation that models the height of the shot put in feet with respect to time in seconds. b. After how many seconds will the shot put hit the ground? SOLUTION: a. A model for the vertical motion of a projected object is given by h = —16t + Vt + ho, where h is the height in feet, t is the time in seconds, v is the initial velocity in feet per second, and ho is the initial height in feet. The initial velocity of the shot put, v, is 29 feet per second. The initial height of the shot put, ho, is 6 feet. So, the equation h = —16t + 29t + 6 models the height of the shot put in feet with respect to the time in seconds. b. When the shot put hits the ground, the height will be zero, so h = 0. 11=-16/2+29/+6 0-16/2-29/-6 0-16"-32/+3/-6 o- +30-2) 16/43-0 16t —3 3 116 The roots are or 3 1-2-0 and 2. Because time cannot be negative, the shot put will hit the ground after 2 seconds. 16 eSolutions Manual - Powered by Cognero Page 17
  18. 8-7 Solving axA2 + bx + c = 0 Solve each equation. Confirm your answers using a graphing calculator. 23. u +9x 18=0 SOLUTION: Factor the trinomial. 2.1 — 1 8 +12 x —31—18 +12 O 0 6 Orig. equation 12 and p = -3 Group terms with com. factors. Factor the CCF +6 is the com. factor. x O 3 3 2 The roots are or and 6. O 6 ZaoProc1uctPropa•ty Solve each equatim. Confirm the roots using a graphing calculator. Let YI CALC menu to find the points intersection. = 2x +9x 18 and Y2 - — 0. Use the intersect option from the [-10, 101 SCI: 1 by [-40, 5 and Thus, the solutions are eSolutions Manual - Powered by Cognero [-10, SCI: 1 by [-40, 20] SCI: 5 6. Page 18
  19. 8-7 Solving axA2 + bx + c 24. 4x +17x+15=o SOLUTION: Factor the trinomial then solve the equation using the Zero Product Property. 4x- +171+15 4x + 5x-+12x+15 4x- + (121+15) x-(4x + 5) + + 5) 5 4 O O 0 equation m = 4 and p Group term with com. factorsg. Factor the GCF 4x +5 the com. factor. x O —3 ZaoProductProp. Solve each equatim. The roots are 5 and 3. 4 Confirm the roots using a graphing calculator. Let YI the CALC menu to find the points intersection. +17x + 15 and Y2 Intersection [-5,51 SCI: 1 by [-10, 101 SCI: 1 Intersectin 51 1 by [-10, 101 SCI: 1 Thus, the solutions are 5 and 3. 4 eSolutions Manual - Powered by Cognero = 0. Use the intersect option from Page 19
  20. 8-7 Solving axA2 + bx + c 25. 3x +26x=16 SOLUTION: Factor the trinomial, then solve the equation using the Zero Product Property. — 31 6x 6 — + 2.4x- —16 2.41) + —16) — 3x(x — 8) 31 + 2) (x 8) 6 O O 0 Oriæ-lal equation Subtract 16 from each side. 24 and p Group the terms w / com. factors. Factor the GCF x —8isthe cmummfactm•. o 2 3 or O 8 ZaoProductProp. Solve each equatim. The roots are — and 8. Confirm the roots using a graphing calculator. Let YI = -3x +26x and Y2 — — 16. Use the intersect option from the CALC menu to find the points intersection. Intersection *z.sssssss.• [0, 101 1 by [-10, 10 2 Thus, the solutions are — and 8. eSolutions Manual - Powered by Cognero Intersectin [0, 101 SCI: 1 by [-10, 100] SCI: 10 Page 20
  21. 8-7 Solving axA2 + bx + c 26. 2x +13x=15 SOLUTION: Factor the trinomial, then solve the equation using the Zero Product Property. +13x +131 —2.x- I ox) + (-31 — 2.x(x — 5) + 5 5) 5) — 2.1 + 3 or 5 Oriælal equation O Subtract 15 from each side. O m —10 and p = 3 O Group terms with com. factore. o Factor the GCF O x —5 is the comm on factor. —5— O ZeroProductProp. = 5 Solve each equation. 3 3 2 3 The roots are — and 5. Confirm the roots using a graphing calculator. Let YI = CALC menu to find the points intersection. -u +13x and Y2- — 15. Use the intersect option from the Intersection [0, 101 SCI: 1 by [-10, SCI: 5 Ir.termcti#tl [0, 101 1 by [-10, 30] SCI: 5 Thus, the solutions are and 5. eSolutions Manual - Powered by Cognero Page 21
  22. 8-7 Solving axA2 + bx + c 27. 3x +5x= 2 SOLUTION: Factor the trinomial, then solve the equation using the Zero Product Property. — 2 equation 31 + 2 — —31 + 61 —3x(x— ) O o o o o Add 2 to each side. m = 6 and p Group terms with com. factors. Factor the GCF x —2 is the common factor. O -31=1 The roots are or 1 3 1 and 2. 3 O ZeroProductProp Solve each equatim. Confirm the roots using a graphing calculator. Let YI — CALC menu to find the points intersection. +5x and Y2 = -2. Use the intersect option from the 51 SCI: 1 by [-20, 101 SCI: 2 eSolutions Manual - Powered by Cognero 51 1 by [-20, 10] 2 Thus, the solutions are 1 and 2. 3 Page 22
  23. 8-7 Solving axA2 + bx + c 28. 4x +19x- SOLUTION: 30 Factor the trinomial, then solve the equation using the Zero Product Property. 9x 30 —41 +24 x —51+30 (5x —4x +24 x 4x(x —6) — o or —6) 6) x o o o o o —6 O Original equation Add 30 to each side. m —24 anclp -5 Group term with com. factors. F the GCF x —6 is the comm on factor. ZeroProductProp. Solve each equation. -4x — 5 The roots are 5 4 5 and 6. 4 Confirm the roots using a graphing calculator. Let YI = -4x +19x and Y2 CALC menu to find the points intersection. Intersectin 1-5, 101 1 by [-50, 10 Intersectin 101 SCI: 1 by [-50, 401 SCI: 10 Thus, the solutions are 5 and 6. 4 eSolutions Manual - Powered by Cognero = -30. Use the intersect option from the Page 23
  24. 8-7 Solving axA2 + bx + c = 0 29. BASKETBALL When Jerald shoots a free throw, the ball is 6 feet from the floor and has an initial upward velocity of 20 feet per second. The hoop is 10 feet from the floor. a. Use the vertical motion model to determine an equation that models Jerald's free throw. b. How long is the basketball in the air before it reaches the hoop? c. Raymond shoots a free throw that is 5 foot 9 inches from the floor with the same initial upward velocity. Will the ball be in the air more or less time? Explain. SOLUTION: a. A model for the vertical motion of a projected object is given by h = —16t + Vt + ho, where h is the height in feet, t is the time in seconds, v is the initial velocity in feet per second, and ho is the initial height in feet. The initial velocity of the basketball, v, is 20 feet per second. The initial height of the basketball, ho, is 6 feet. The height of the hoop, h, is 10 feet. So, the equation 10 = —16t + 20t + 6 models Jerald's free throw. —1) —1=0 4t=1 b. o —16t —16t —44t 5t+1 Orig equation. —10 from each side. Factor CCF of -4. m = —4 andp Group term with com. factors. Factor the GCF ) the comm on factor. 4t —1) O = 4(4t — 1) (t 1 4 (t —1 (t or The roots are and l. The basketball takes -recond to reach a height of 10 feet on its way up. The basketball takes 1 second to reach a height of 10 feet on its way down. So, the basketball will be in the air 1 second before it reaches the hoop. c. The ball will be in the air less time because it starts closer to the ground so the shot will not have as far to fall. eSolutions Manual - Powered by Cognero Page 24
  25. 8-7 Solving axA2 + bx + c = 0 30. DIVING Ben dives from a 36-foot platform. The equation h = —16t + 14t + 36 models the dive. How long will it take Ben to reach the water? SOLUTION: When Ben reaches the pool, his height, h will be 0. —2 h = — 16 O —16 9—0 + 14t —I— 36 + 14t —I— 36 —18 —2) -2. or 9 9 8 Equation forheight Divide each sideby —2 —16 and p —9 FactortheGCF t — 2 isthecommon factor 0 ZeroProductPrope1ty Solv e each equation. The roots are _ P and 2. Because time cannot be negative, Ben will reach the water after 2 seconds. 8 31. NUMBER THEORY Six times the square of a number x plus Il times the number equals 2. What are possible values of x? SOLUTION: Let x = a number. Then, 6x +11 x = 2. 61 + 11.1- 61 + 1 fr- 61 12. (61 + 12 x) 6x(x 2) (61 or 61 1 6 The possible values of x are eSolutions Manual - Powered by Cognero O 0 equation Subtract 2 from each side m 12 andp Group term with com. factors. Factor the GCF (x +2) the comm on factor. 1 2 or 6 Page 25
  26. 8-7 Solving axA2 + bx + c = 0 Factor each polynomial, if possible. If the polynomial cannot be factored using integers, writeprime . 32. -6x -23x 20 SOLUTION: First factor the number —1 out of each term in the polynomial. 2 —61 —2 5x —20 — 2.31 20) Then factor the trinomial 6x + 23x + 20. In this trinomial, a = 6, b = 23 and c = 20, so m +p is positive and mp is positive. Therefore, m andp must both be positive. List the positive factors of 6(20) or 120 and identify the factors with a sum of 23. Factors of 120 1, 120 2, 60 3, 40 4, 30 5, 24 6, 20 8, 15 10, 12 The correct factors are 8 and 15. —61 —231-20 — 20) (6x +mx +px+20) — 1 5X 20) — I(2x + 5) (31 + 4) so,-6x 23x +4). eSolutions Manual - Powered by Cognero Sum 121 62 43 34 29 26 23 22 F actor out —1 Write the pattern. andp=15 Group term with comm on factors. Factor the GCF (31+4) is the comm on factor. Page 26
  27. 8-7 Solving axA2 + bx + c 33. 4x -15x 14 SOLUTION: First factor the number —1 out of each term in the polynomial. 4x -15x +15x+14) Then factor the trinomial 4x + 15x + 14. In this trinomial, a = 4, b = 15 and c = 14, so m +p is positive and mp is positive. Therefore, m andp must both be positive. List the positive factors of 4(14) or 56 and identify the factors with a sum of 15. Factors of 56 1, 56 2, 28 4, 14 The correct factors are 7 and 8. —4x —15x — 14 — + 1 5 14 — I + + + 14 —I(4x -1-71+81-+14) — + 14 ) — 1 [x(4x +7) + +7)] Sum 57 30 18 115 F actor out —1 Write the pattern. "1=7 andp=S Group terms w com.factors. F the CCF (4x +7) is the com.factor. so, 4x -15x eSolutions Manual - Powered by Cognero Page 27
  28. 8-7 Solving axA2 + bx + c 34. 5x +18x+8 SOLUTION: First factor the number —1 out of each term in the polynomial. I(5x2 - 18x - 8) 5X + 18X + 8 = Then factor the trinomial 5x — 18x 8. In this trinomial, a = 5, b = —18 and c — 8, so m +p is negative and mp is negative. Therefore, m andp must have different signs. List the factors of 5(—8) or Factors of 40 — , 40 , 40 2, 20 2, -20 4, 10 4, -10 The correct factors are 20 and 2 I(5x-2 —1 —I(5x- +mx+px—8) —201+21 8) —1 —201) + ex 8) 5x(x —4) 2 -l 40 and identify the factors with a sum of —18. Sum 39 39 18 -18 6 6 3 3 Factor our —1 Write the pattern. andp= 2 Group terms with com.factors. F the CCF (x —4) is the common factor. SO, eSolutions Manual - Powered by Cognero Page 28
  29. 8-7 Solving axA2 + bx + c 35.-6x +31x 35 SOLUTION: First factor the number —1 out of each term in the polynomial. -6x +31x 31x+35) Then factor the trinomial 6x — 31x + 35. In this trinomial, a = 6, b = 31 and c = 35, so m +p is negative and mp is positive. Therefore, m andp must both be negative. List the negative factors of 6(35) or 210 and identify the factors with a sum of 31. 2, -105 7, 30 -14, -15 —2 Factors of 210 -1, 3, 5, 6, -10, 210 70 42 35 21 The correct factors are +311--35 21 and -10. —1 —1 2 — I(6x 6x so,-6x +31x 3 fr+35) my —pc + 3 5) fr—10x+35) ( 35) 1 5). Sum 211 -107 73 47 41 37 31 29 F actor out -1. Write the pattern. m=-21 and p = -10 Group terms with com.factors. F the CCF (21-7) the comm on factor. eSolutions Manual - Powered by Cognero Page 29
  30. 8-7 Solving axA2 + bx + c 36. 4x + 5x 12 SOLUTION: First factor the number —1 out of each term in the polynomial. 4x + 5x 5x+ 12) Then factor the trinomial 4x — 5x + 12. In this trinomial, a = 4, b = —5 and c = 12, so m +p is negative and mp is positive. Therefore, m andp must both be ne ative. List the ne ative factors of 4(12) or 48 and identif the factors with a sum of —5. 3, -16 4, -12 2, -120 6, 40 Factors of 48 2, 6, 48 24 8 Sum 49 26 —19 —16 — 14 There are no factors of 48 with a sum of —5. So, 37. -12x +x+20 SOLUTION: 12 is prime. First factor the number —1 out of each term in the polynomial. -I(12x2 x - 20) —12X + X + 20 = Then factor the trinomial 12x x 20. In this trinomial, a = 12, b = —1 and c — 20, so m +p is negative and mp is negative. Therefore, m andp must have different signs. List the factors of 12(—20) or —240 and identify the factors with a sum of —1. Factors of —240 — 240 1, 240 2, 120 3, 80 3, 80 4, 60 4, 60 5, 48 5, 48 6, 40 8, 30 8, 30 -10, 24 10, -24 -12, 20 12, 20 -15, 16 eSolutions Manual - Powered by Cognero Sum 239 239 118 -118 77 77 56 56 43 43 34 34 22 22 14 — 14 8 8 1 Page 30
  31. 8-7 Solving axA2 + bx + c 15, -16 The correct factors are —16 and 15. — 121 X 20 I(12x —x o) I(12x —mx 212 —161-I-15x o) o) —Il 12 —161) + 20) — 1 [4x(3x —4) + 5(3x so, -12x + x +20= (4x 4). -1 Factor out —1 . Write the pattern. andp= 15 Group term with com.factors. Factor the GCF (h —4) is the common factor. 38. URBAN PLANNING The city has commissioned the building of a rectangular park. The area of the park can be expressed as 660x + 524x + 85. Factor this expression to find binomials with integer coefficients that represent possible dimensions of the park. If x = 8, what is a possible perimeter of the park? SOLUTION: 6601* +85 = (660x2 +17) So, (22x + 5)(30x + 17) represent the possible dimensions of the park. Evaluate each dimension when x = 8 to find the possible length and width of the park when x = 8. =176+5 =240+17 = 257 P-2/+2w - 876 A possible perimeter of the park is 876 units. eSolutions Manual - Powered by Cognero Page 31
  32. 8-7 Solving axA2 + bx + c = 0 39. MULTIPLE REPRESENTATIONS In this problem, you will explore factoring a special type of polynomial. a. GEOMETRIC Draw a square and label the sides a. Within this square, draw a smaller square that shares a vertex with the first square. Label the sides b. What are the areas of the two squares? b. GEOMETRIC Cut and remove the small square. What is the area of the remaining region? c. ANALYTICAL Draw a diagonal line between the inside corner and outside corner of the figure, and cut along this line to make two congruent pieces. Then rearrange the two pieces to form a rectangle. What are the dimensions? d. ANALYTICAL Write the area of the rectangle as the product of two binomials. 2 e. VERBAL Complete this statement: a b - SOLUTION: a. b b' The area of a square is found using the formula A = s 2 smaller square is b . b. b Why is this statement true? . So, the area of the larger square is a and the area of the To find the area of the remaining region, subtract the area of the smaller square from the area of the larger square. 2 So, the area of the remaining region is a b . c. b a-b b a-b The width is a d. a-b a-b b b and the length is a + b. e. The figure with area a eSolutions Manual - Powered by Cognero b and the rectangle with area (a b)(a + b) have the same area, so a b b) Page 32
  33. 8-7 Solving axA2 + bx + c = 0 40. CCSS CRITIQUE Zachary and Samantha are solving 6x reasoning. 12 12 $12 13 2.e—3zOF3x SOLUTION: -o 2. Is either of them correct? Explain your Sample answer: By the Zero Product Property, the equation must be set equal to zero before it can be factored. Then each of the factors can be set equal to zero and solved for x. —X ± 12 6x:-x-12-o 2x-3-o 21=3 2 or 4 3 So, Samantha is correct. 41. REASONING A square has an area of9x + 30xy + 25y square inches. The dimensions are binomials with positive integer coefficients. What is the perimeter of the square? Explain. SOLUTION: 2 The area of the square equals 9x + 30xy + 25y = (3x)2 + + (5y)2 = (3x + + 5y) So, the length of each side of the square is (3x + 5y) inches. -4(3x+5y) 12x+20y The perimeter of the square is (12x + 20y) inches. eSolutions Manual - Powered by Cognero Page 33
  34. 8-7 Solving axA2 + bx + c = 0 42. CHALLENGE Find all values of k so that2x + kx + 12 can be factored as two binomials using integers. SOLUTION: The values for k must be the sum of two ne ative factors or two positive factors of 2(12) or 24. Factors of 24 I , 24 2, -12 2, 12 Possible Values for k 25 25 — 14 14 -11 11 -10 10 43. WRITING IN MATH What should you consider when solving a quadratic equation that models a real-world situation? SOLUTION: Sample answer: A quadratic equation may have zero, one, or two solutions. If there are two solutions, you must consider the context of the situation to determine whether one or both solutions answer the given question. For example: Time cannot be negative. A solution stating that a person ran 60 miles per hour is unrealistic. 44. WRITING IN MATH Explain how to determine which values should be chosen for m andp when factoring a polynomial of the form ax + bx + c. SOLUTION: Sample answer: If the trinomial factors, then ax + bx + c = ax + mx + PX + c where m and n are two integers that have a product equal to ac and a sum equal b. For example, given the trinomial 3x + 14x + 15, a = 3, b = 14, and c = 15. Find two integers such that mp = 3(15) or 45 and m +p = 14. Since 5 x 9 = 45 and 5 + 9 — 14, the values of the variables could be m = 5 andp 31 —I— + 1 5 31 + 5x- 1 5 eSolutions Manual - Powered by Cognero = 9. Check to see if the trinomial factors using these values for m andp . m —5 and p Group tamgwith factm•x Factm•theGCF x +3i*theccnummfactT. Page 34
  35. 8-7 Solving axA2 + bx + c = 0 45. GRIDDED RESPONSE Savannah has two sisters. One sister is 8 years older than her, and the other sister is 2 years younger than her. The product of Savannah's sisters' ages is 56. How old is Savannah? SOLUTION: Let s = Savannah's age. Then, s + 8 = the age of her older sister and s — 2 = the age of her younger sister. —2 s —2s s -1-12s —6s 7 7 S (S 12.) (G +72) or O O O o Oriæ-jal equation Multiply Subtract 72 from each Side. 12 anclp = —6 Group terms with emu. factors. F actor the GC ('+12) is the common factor. s The roots are —12 and 6. Savannah's age cannot be negative. So, Savannah is 6 years old. 235 46. What is the product of —a b and 87 5 2 23 5 283 5 2 27 5 SOLUTION: (4 3535 3 a 87 5 Choice A is the correct answer. eSolutions Manual - Powered by Cognero 52 a 2) •b Page 35
  36. 8-7 Solving axA2 + bx + c = 0 47. What is the solution set of x + 2x 24 — SOLUTION: — 24 — — 24 (4x + 24 or The solutions are —6 and 4. 0 O Original equation. m = 6 and p Group terms with com. factors. F actor the GC (x +6) is the common factor. 6 2? Choice J is the correct answer. 48. Which is the solution set of x > c D SOLUTION: Because x is greater than or equal to —2, there should be a closed circle at 2, and the line should be shaded to the right of —2. Choice C is the correct answer. eSolutions Manual - Powered by Cognero Page 36
  37. 8-7 Solving axA2 + bx + c = 0 Factor each polynomial. 49. x —9x+ 14 SOLUTION: In this trinomial, b = —9 and c = 14, so m +p is negative and mp is positive. Therefore, m andp must both be ne ative. List the ne ative factors of 14, and look for the air of factors and identify the factors with a sum of —9. Factors of 14 Sum -15 9 The correct factors are 2 and 7. —Sr + 15 = (r + + p) 50. n -811+15 SOLUTION: In this trinomial, b = —8 and c = 15, so m +p is negative and mp is positive. Therefore, m andp must both be ne ative. List the ne ative factors of 15, and look for the air of factors and identify the factors with a sum of —8. Factors of 15 -1, -15 The correct factors are — + 15 = (n + 51. x 5x-24 SOLUTION: 3 and 5) Sum —16 8 5. 24, so m +p is negative and mp is negative. Therefore, m andp must be opposite In this trinomial, b = —5 and c — Factors of 24 — 24 2, -12 2, 12 The correct factors are 3 and —5x—24 = (x+ m) (x + p) eSolutions Manual - Powered by Cognero si ns. List the factors of —24, and look for the air of factors and identify the factors with a sum of —5. Sum 23 23 -10 10 5 5 2 2 8. Page 37
  38. 8-7 Solving axA2 + bx + c 52.z +15z +36 SOLUTION: In this trinomial, b = 15 and c = 36, so m +p is negative and mp is positive. Therefore, m andp must both be positive. List the factors of 36, and look for the air of factors and identify the factors with a sum of 15. -1, 40 4, -10 Factors of 36 1, 36 2, 18 3, 12 The correct factors are 3 and 12. z: +15z+ 36 = (z + + p) (z +12) 53. r +3r-40 SOLUTION: In this trinomial, b = 3 and c — Factors of 40 Sum 37 20 15 13 12 40, so m +p is positive and mp is negative. Therefore, m andp must be opposite si ns. List the factors of 40, and look for the air of factors and identify the factors with a sum of 3. 1, 40 2, 20 2, 20 4, 10 5, 8 The correct factors are —5 and 8. r2 + 3r —40 = (r + + p) 54. v 16v + 63 SOLUTION: Sum 39 39 -18 18 6 6 3 3 In this trinomial, b = 16 and c = 63, so m +p is positive and mp is positive. Therefore, m andp must be opposite si ns. List the factors of 63, and look for the air of factors and identify the factors with a sum of 16. Factors of 63 1, 63 3, 21 The correct factors are 7 and 9. v: +16v+63 = (v + + p) eSolutions Manual - Powered by Cognero Sum 24 16 Page 38
  39. 8-7 Solving axA2 + bx + c = 0 Solve each equation. Check your solutions. 55. 9) SOLUTION: or 0-9=0 The roots are 0 and 9. Check by substituting 0 and 9 in for a in the original equation. 0=0 and ala —9) O 0=0 The solutions are 0 and 9. 56. (2y + - 1) SOLUTION: 2Y+6=0 or y-l=o The roots are 3 and 1. Check by substituting 3 and 1 in for y in the original equation. 0=0 and The solutions are 3 and 1. eSolutions Manual - Powered by Cognero Page 39
  40. 8-7 Solving axA2 + bx + c 57. lox 20x=o SOLUTION: 10x2-20x=o lox -o or x-2=0 The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation. -20x=o and I Ox 0-0=0 0=0 2-20x=o 40 40=0 0=0 The solutions are 0 and 2. eSolutions Manual - Powered by Cognero Page 40
  41. 8-7 Solving axA2 + bx + c 58. 8b2 - 12b SOLUTION: . Check by substituting 0 and — in for b in the original equation. 4b 2b 4b-O 3) or 2b-3-O 2b -3 The roots are 0 and 81/-126=0 0-0=0 0=0 and 8b: -12b=O -12 18-18-0 0=0 The solutions are 0 and eSolutions Manual - Powered by Cognero Page 41
  42. 8-7 Solving axA2 + bx + c 59. 15a =60a SOLUTION: 1502 600 150(0-4) -o -o 3x--2 150=0 or The roots are 0 and 4. Check by substituting 0 and 4 in for a in the original equation. and = 600 0-0 600 240 - 240 The solutions are 0 and 4. 60. 33x = 22x SOLUTION: +22x -o lix-o The roots are 0 and --22x 0-0 and 33r2- 22x 2 33-2) =-22 3 44 3 or 3x+2-o 2 3 2 3 . Check by substituting 0 and 2 3. 2 — in for x in the original equation. 3 2 3 44 3 The solutions are 0 and eSolutions Manual - Powered by Cognero Page 42
  43. 8-7 Solving axA2 + bx + c = 0 61. ART A painter has 32 units of yellow dye and 54 units of blue dye to make two shades of green. The units needed to make a gallon of light green and a gallon of dark green are shown. Make a graph showing the numbers of gallons of the two greens she can make, and list three possible solutions. h:ght green SOLUTION: 6 Let x = the number of gallons of light green and let y = the number of gallons of dark green. Then, 4x + y < 32 and x +6y < 54. Graph 4x + y = 32 and x + 6y = 54. Both lines are included in the solution of the system and should be solid. Green Paint 45 40 V z 32 10 5 •o 40 sox G eea Some possible solutions are 2 light, 8 dark; 6 light, 8 dark; 7 light, 4 dark. They fall in the shaded region which represents the solution set. Solve each compound inequality. Then graph the solution set. 62. k 12 and k +212 k>lO and k+2S18 kS16 The solution set is {k110 < k < 16}. To graph the solution set, graph k < 16 and graph k > 10. Then find the intersection. 8 9 11 12 13 14 15 1718 63. d ord 43 or The solution set is {dld < 5 or d > 7}. Notice that the graphs do not intersect. To graph the solution set, graph d < 5 and graph d > 7. Then find the union. 012345678910 eSolutions Manual - Powered by Cognero Page 43
  44. 8-7 Solving axA2 + bx + c 64. 3
  45. 8-7 Solving axA2 + bx + c = 0 68. FINANCIAL LITERACY A home security company provides security systems for $5 per week, plus an installation fee. The total cost for installation and 12 weeks of service is $210. Write the point-slope form of an equation to find the total fee y for any number of weeks x. What is the installation fee? SOLUTION: The slope of the equation of the line that represents the weekly cost is 5. The total cost for 12 weeks of service 210 = including installation is $210. So, the line passes through the point (12, 210). The point-slope form of an equation to find the total fee y for any number of weeks x is y Solve the equation for y to find the flat fee for installation. y-210-5(r-12) y-210-5x-60 y = 5x + 150 So, the flat fee for installation is $150. Find the principal square root of each number. 69. 16 SOLUTION: 70. 36 SOLUTION: 71.64 SOLUTION: 72. 81 SOLUTION: 73. 121 SOLUTION: •Ei-JiF eSolutions Manual - Powered by Cognero 12). Page 45
  46. 8-7 Solving axA2 + bx + c 74. 100 SOLUTION: = 10 eSolutions Manual - Powered by Cognero Page 46

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