1. Adding and Subtracting Polynomials ALGEBRA 1 LESSON 9-1 (For help, go to Lesson 1-7.) Simplify each expression. 1.6t + 13t2.5g + 34g 3.7k – 15k4.2b – 6 + 9b 5.4n 2 – 7n 2 6.8x 2 – x 2 9-1

2. Adding and Subtracting Polynomials ALGEBRA 1 LESSON 9-1 1.6t + 13t = (6 + 13)t = 19t2.5g + 34g = (5 + 34)g = 39g 3.7k – 15k = (7 – 15)k = –8k4.2b – 6 + 9b = (2 + 9)b – 6 = 11b – 6 5.4n 2 – 7n 2 = (4 – 7)n 2 = –3n 2 6.8x 2 – x 2 = (8 – 1)x 2 = 7x 2 Solutions 9-1

3. ALGEBRA 1 LESSON 9-1 Adding and Subtracting Polynomials Find the degree of each monomial. a. 18 b. 3xy 3 c. 6c 9-1 The degree of a nonzero constant is 0.Degree: 0 The exponents are 1 and 3. Their sum is 4.Degree: 4 6c = 6c 1. The exponent is 1.Degree: 1

4. Adding and Subtracting Polynomials ALGEBRA 1 LESSON 9-1 Write each polynomial in standard form. Then name each polynomial by its degree and the number of its terms. a. –2 + 7x linear binomial b. 3x 5 – 2 – 2x 5 + 7x x 5 + 7x – 2Combine like terms. fifth degree trinomial Place terms in order. 3x 5 – 2x 5 + 7x – 2 Place terms in order. 7x – 2 9-1

5. Adding and Subtracting Polynomials ALGEBRA 1 LESSON 9-1 Simplify (6x 2 + 3x + 7) + (2x 2 – 6x – 4). Method 1: Add vertically. Line up like terms. Then add the coefficients. 6x 2 + 3x + 7 2x 2 – 6x – 4 8x 2 – 3x + 3 Method 2: Add horizontally. Group like terms. Then add the coefficients. (6x 2 + 3x + 7) + (2x 2 – 6x – 4) = (6x 2 + 2x 2 ) + (3x – 6x) + (7 – 4) = 8x 2 – 3x + 3 9-1

6. Adding and Subtracting Polynomials Simplify (2x 3 + 4x 2 – 6) – (5x 3 + 2x – 2). ALGEBRA 1 LESSON 9-1 Method 1: Subtract vertically. Line up like terms. Then add the coefficients. (2x 3 + 4x 2 – 6)Line up like terms. –(5x 3 – 2x – 2) 2x 3 + 4x 2 – 6Add the opposite. –5x 3 – 2x + 2 –3x 3 + 4x 2 – 2x – 4 9-1

7. Adding and Subtracting Polynomials (continued) ALGEBRA 1 LESSON 9-1 Method 2: Subtract horizontally. = 2x 3 + 4x 2 – 6 – 5x 3 – 2x + 2Write the opposite of each term in the polynomial being subtracted. = (2x 3 – 5x 3 ) + 4x 2 – 2x + (–6 + 2)Group like terms. = –3x 3 + 4x 2 – 2x – 4Simplify. (2x 3 + 4x 2 – 6) – (5x 3 + 2x – 2) 9-1

8. Adding and Subtracting Polynomials ALGEBRA 1 LESSON 9-1 Simplify each expression. Then name each polynomial by its degree and number of terms. 1.–4 + 3x – 2x 2 2.2b 2 – 4b 3 + 6 3.(2x 4 + 3x – 4) + (–3x + 4 + x 4 ) 4.(–3r + 4r 2 – 3) – (4r 2 + 6r – 2) –2x 2 + 3x – 4; quadratic trinomial –4b 3 + 2b 2 + 6; cubic trinomial 3x 4 ; fourth degree monomial –9r – 1; linear binomial 9-1

9. Multiplying and Factoring ALGEBRA 1 LESSON 9-2 (For help, go to Lesson 1–7.) Multiply. 1.3(302)2.41(7)3.9(504) Simplify each expression. 4.4(6 + 5x)5.–8(2y + 1) 6.(5v – 1)57.7(p – 2) 8.(6 – x)99.–2(4q – 1) 9-2

10. Multiplying and Factoring ALGEBRA 1 LESSON 9-2 1.3(302) = 9062.41(7) = 2873.9(504) = 4536 4.4(6 + 5x) = 4(6) + 4(5x) = 24 + 20x 5.–8(2y + 1) = (–8)(2y) + (–8)(1) = –16y – 8 6.(5v – 1)5 = (5v)(5) – (1)(5) = 25v – 5 7.7(p – 2) = 7p – 7(2) = 7p – 14 8.(6 – x)9 = 6(9) – 9x = 54 – 9x 9.–2(4q – 1) = (–2)(4q) – (–2)(1) = –8q + 2 Solutions 9-2

11. Multiplying and Factoring ALGEBRA 1 LESSON 9-2 Simplify –2g 2 (3g 3 + 6g – 5). –2g 2 (3g 3 + 6g – 5) = –2g 2 (3g 3 ) –2g 2 (6g) –2g 2 (–5)Use the Distributive Property. = –6g 2 + 3 – 12g 2 + 1 + 10g 2 Multiply the coefficients and add the exponents of powers with the same base. = –6g 5 – 12g 3 + 10g 2 Simplify. 9-2

12. Multiplying and Factoring ALGEBRA 1 LESSON 9-2 Find the GCF of 2x 4 + 10x 2 – 6x. List the prime factors of each term. Identify the factors common to all terms. 2x 4 = 2 x x x x 10x 2 = 2 5 x x 6x = 2 3 x The GCF is 2 x, or 2x. 9-2

13. Multiplying and Factoring ALGEBRA 1 LESSON 9-2 Factor 4x 3 + 12x 2 – 16x. Step 1: Find the GCF. 4x 3 = 2 2 x x x 12x 2 = 2 2 3 x x 16x = 2 2 2 2 x Step 2: Factor out the GCF. 4x 3 + 12x 2 – 16x = 4x(x 2 ) + 4x(3x) + 4x(–4) = 4x(x 2 + 3x – 4) The GCF is 2 2 x, or 4x. 9-2

14. Multiplying and Factoring ALGEBRA 1 LESSON 9-2 1.Simplify –2x 2 (–3x 2 + 2x + 8). 2.Find the GCF of 16b 4 – 4b 3 + 8b 2. 3.Factor 3x 3 + 9x 2. 4.Factor 10y 3 + 5y 2 – 15y. 6x 4 – 4x 3 – 16x 2 4b24b2 3x 2 (x + 3) 5y(2y + 3)(y – 1) 9-2

15. Multiplying Binomials ALGEBRA 1 LESSON 9-3 (For help, go to Lesson 9-2.) Find each product. 1.4r(r – 1)2.6h(h 2 + 8h – 3)3.y 2 (2y 3 – 7) Simplify. Write each answer in standard form. 4.(x 3 + 3x 2 + x) + (5x 2 + x + 1) 5.(3t 3 – 6t + 8) + (5t 3 + 7t – 2) 6.w(w + 1) + 4w(w – 7)7.6b(b – 2) – b(8b + 3) 8.m(4m 2 – 6) + 3m 2 (m + 9)9.3d 2 (d 3 – 6) – d 3 (2d 2 + 4) 9-3

16. Multiplying Binomials ALGEBRA 1 LESSON 9-3 1.4r(r – 1) = 4r(r) – 4r(1) = 4r 2 – 4r 2.6h(h 2 + 8h – 3) = 6h(h 2 ) + 6h(8h) – 6h(3) = 6h 3 + 48h 2 – 18h 3.y 2 (2y 3 – 7) = y 2 (2y 3 ) – 7y 2 = 2y 5 – 7y 2 4.x 3 + 3x 2 + x5.3t 3 – 6t + 8 + 5x 2 + x + 1 + 5t 3 + 7t – 2 x 3 + 8x 2 + 2x + 18t 3 + t + 6 6.w(w + 1) + 4w(w – 7)7.6b(b – 2) – b(8b + 3) = w(w) + w(1) + 4w(w) – 4w(7)= 6b(b) – 6b(2) – b(8b) – b(3) = w 2 + w + 4w 2 – 28w= 6b 2 – 12b – 8b 2 – 3b = (1 + 4)w 2 + (1 – 28)w= (6 – 8)b 2 + (–12 – 3)b = 5w 2 – 27w= –2b 2 – 15b Solutions 9-3

17. Multiplying Binomials ALGEBRA 1 LESSON 9-3 Solutions (continued) 8.m(4m 2 – 6) + 3m 2 (m + 9) = m(4m 2 ) – m(6) + 3m 2 (m) + 3m 2 (9) = 4m 3 – 6m + 3m 3 + 27m 2 = (4 + 3)m 3 + 27m 2 – 6m = 7m 3 + 27m 2 – 6m 9.3d 2 (d 3 – 6) – d 3 (2d 2 + 4) = 3d 2 (d 3 ) – 3d 2 (6) – d 3 (2d 2 ) – d 3 (4) = 3d 5 – 18d 2 – 2d 5 – 4d 3 = (3 – 2)d 5 – 4d 3 – 18d 2 = d 5 – 4d 3 – 18d 2 9-3

18. Multiplying Binomials ALGEBRA 1 LESSON 9-3 Simplify (2y – 3)(y + 2). (2y – 3)(y + 2) = (2y – 3)(y) + (2y – 3)(2)Distribute 2y – 3. = 2y 2 – 3y + 4y – 6Now distribute y and 2. = 2y 2 + y – 6Simplify. 9-3

19. Multiplying Binomials ALGEBRA 1 LESSON 9-3 Simplify (4x + 2)(3x – 6). The product is 12x 2 – 18x – 12. Last (2)(–6)+ Outer (4x)(–6) + Inner (2)(3x) + 24x6x6x12 –+– = 12x 2 18x – 12 – First = (4x)(3x) (4x + 2)(3x – 6) 9-3

20. Multiplying Binomials ALGEBRA 1 LESSON 9-3 Find the area of the shaded region. Simplify. area of outer rectangle = (3x + 2)(2x – 1) area of hole = x(x + 3) area of shaded region = area of outer rectangle – area of hole = (3x + 2)(2x – 1)–x(x + 3)Substitute. = 6x 2 – 3x + 4x – 2 –x 2 – 3xUse FOIL to simplify (3x + 2) (2x – 1) and the Distributive Property to simplify x(x + 3). = 6x 2 – x 2 – 3x + 4x – 3x – 2Group like terms. = 5x 2 – 2x – 2Simplify. 9-3

21. Multiplying Binomials ALGEBRA 1 LESSON 9-3 Simplify the product (3x 2 – 2x + 3)(2x + 7). 21x 2 – 14x + 21 Multiply by 7. 6x 3 – 4x 2 + 6xMultiply by 2x. Method 1: Multiply using the vertical method. 3x 2 – 2x + 3 2x + 7 6x 3 + 17x 2 – 8x + 21Add like terms. 9-3

22. Multiplying Binomials ALGEBRA 1 LESSON 9-3 (continued) Method 2: Multiply using the horizontal method. = 6x 3 – 4x 2 + 6x + 21x 2 – 14x + 21 = 6x 3 + 17x 2 – 8x + 21 The product is 6x 3 + 17x 2 – 8x + 21. = (2x)(3x 2 ) – (2x)(2x) + (2x)(3) + (7)(3x 2 ) – (7)(2x) + (7)(3) (2x + 7)(3x 2 – 2x + 3) 9-3

23. Multiplying Binomials ALGEBRA 1 LESSON 9-3 Simplify each product using any method. 1.(x + 3)(x – 6)2.(2b – 4)(3b – 5) 3.(3x – 4)(3x 2 + x + 2) 4.Find the area of the shaded region. x 2 – 3x – 186b 2 – 22b + 20 9x 3 – 9x 2 + 2x – 8 2x 2 + 3x – 1 9-3

24. Multiplying Special Cases ALGEBRA 1 LESSON 9-4 (For help, go to Lessons 8–4 and 9-3.) Simplify. 1.(7x) 2 2.(3v) 2 3.(–4c) 2 4.(5g 3 ) 2 Use FOIL to find each product. 5.(j + 5)(j + 7)6.(2b – 6)(3b – 8) 7.(4y + 1)(5y – 2)8.(x + 3)(x – 4) 9.(8c 2 + 2)(c 2 – 10)10.(6y 2 – 3)(9y 2 + 1) 9-4

25. Multiplying Special Cases ALGEBRA 1 LESSON 9-4 1.(7x) 2 = 7 2 x 2 = 49x 2 2.(3v) 2 = 3 2 v 2 = 9v 2 3.(–4c) 2 = (–4) 2 c 2 = 16c 2 4.(5g 3 ) 2 = 5 2 (g 3 ) 2 = 25g 6 5.(j + 5)(j + 7) = (j)(j) + (j)(7) + (5)(j) + (5)(7) = j 2 + 7j + 5j + 35 = j 2 + 12j + 35 6.(2b – 6)(3b – 8) = (2b)(3b) + (2b)(–8) + (–6)(3b) + (–6)(–8) = 6b 2 – 16b – 18b + 48 = 6b 2 – 34b + 48 Solutions 9-4

26. Multiplying Special Cases 7.(4y + 1)(5y – 2)) = (4y)(5y) + (4y)(–2) + (1)(5y) + (1)(–2) = 20y 2 – 8y + 5y – 2 = 20y 2 – 3y – 2 ALGEBRA 1 LESSON 9-4 8.(x + 3)(x – 4) = (x)(x) + (x)(-4) + (3)(x) + (3)(–4) = x 2 – 4x + 3x – 12 = x 2 – x – 12 9.(8c 2 + 2)(c 2 – 10) = (8c 2 )(c 2 ) + (8c 2 )(–10) + (2)(c 2 ) + (2)(–10) = 8c 4 – 80c 2 + 2c 2 – 20 = 8c 4 – 78c 2 – 20 10.(6y 2 – 3)(9y 2 + 1) = (6y 2 )(9y 2 ) + (6y 2 )(1) + (–3)(9y 2 ) + (–3)(1) = 54y 4 + 6y 2 – 27y 2 – 3 = 54y 4 – 21y 2 – 3 Solutions (continued) 9-4

27. Multiplying Special Cases ALGEBRA 1 LESSON 9-4 a. Find (y + 11) 2. (y + 11) 2 = y 2 + 2y(11) + 7 2 Square the binomial. = y 2 + 22y + 121Simplify. b. Find (3w – 6) 2. (3w – 6) 2 = (3w) 2 –2(3w)(6) + 6 2 Square the binomial. = 9w 2 – 36w + 36Simplify. 9-4

28. Multiplying Special Cases ALGEBRA 1 LESSON 9-4 Among guinea pigs, the black fur gene (B) is dominant and the white fur gene (W) is recessive. This means that a guinea pig with at least one dominant gene (BB or BW) will have black fur. A guinea pig with two recessive genes (WW) will have white fur. The Punnett square below models the possible combinations of color genes that parents who carry both genes can pass on to their offspring. Since WW is of the outcomes, the probability that a guinea pig has white fur is. 1414 1414 BBBW BWWW BWBW BWBW 9-4

29. Multiplying Special Cases ALGEBRA 1 LESSON 9-4 (continued) You can model the probabilities found in the Punnett square with the expression ( B + W) 2. Show that this product gives the same result as the Punnett square. 1212 1212 ( B + W ) 2 = ( B ) 2 – 2 ( B )( W ) + ( W ) 2 Square the binomial. 1212 1212 1212 1212 1212 1212 = B 2 + BW + W 2 Simplify. 1414 1212 1414 The expressions B 2 and W 2 indicate the probability that offspring will have either two dominant genes or two recessive genes is. The expression BW indicates that there is chance that the offspring will inherit both genes. These are the same probabilities shown in the Punnett square. 1414 1414 1414 1212 1212 9-4

30. Multiplying Special Cases ALGEBRA 1 LESSON 9-4 a. Find 81 2 using mental math. 81 2 = (80 + 1) 2 = 80 2 + 2(80 1) + 1 2 Square the binomial. = 6400 + 160 + 1 = 6561Simplify. b. Find 59 2 using mental math. 59 2 = (60 – 1) 2 = 60 2 – 2(60 1) + 1 2 Square the binomial. = 3600 – 120 + 1 = 3481Simplify. 9-4

31. Multiplying Special Cases ALGEBRA 1 LESSON 9-4 Find (p 4 – 8)(p 4 + 8). (p 4 – 8)(p 4 + 8) = (p 4 ) 2 – (8) 2 Find the difference of squares. = p 8 – 64Simplify. 9-4

32. Multiplying Special Cases ALGEBRA 1 LESSON 9-4 Find 43 37. 43 37 = (40 + 3)(40 – 3)Express each factor using 40 and 3. = 40 2 – 3 2 Find the difference of squares. = 1600 – 9 = 1591Simplify. 9-4

33. Multiplying Special Cases ALGEBRA 1 LESSON 9-4 Find each square. 1.(y + 9) 2 2.(2h – 7) 2 3.41 2 4.29 2 5.Find (p 3 – 7)(p 3 + 7).6.Find 32 28. y 2 + 18y + 814h 2 – 28h + 49 1681841 p 6 – 49896 9-4

34. Factoring Trinomials of the Type x 2 + bx + c ALGEBRA 1 LESSON 9-5 (For help, go to Skills Handbook page 721.) List all of the factors of each number. 1.242.123.544.15 5.366.567.648.96 9-5

35. Factoring Trinomials of the Type x 2 + bx + c ALGEBRA 1 LESSON 9-5 1.Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24 2.Factors of 12: 1, 2, 3, 4, 6, 12 3.Factors of 54: 1, 2, 3, 6, 9, 18, 27, 54 4.Factors of 15: 1, 3, 5, 15 5.Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36 6.Factors of 56: 1, 2, 4, 7, 8, 14, 28, 56 7.Factors of 64: 1, 2, 4, 8, 16, 32, 64 8.Factors of 96: 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96 Solutions 9-5

36. Factoring Trinomials of the Type x 2 + bx + c ALGEBRA 1 LESSON 9-5 Factor x 2 + 8x + 15. Find the factors of 15. Identify the pair that has a sum of 8. Factors of 15Sum of Factors 1 and 1516 3 and 5 8 x 2 + 8x + 15 = (x + 3)(x + 5). = x 2 + 5x + 3x + 15 Check: x 2 + 8x + 15 (x + 3)(x + 5) = x 2 + 8x + 15 9-5

37. Factoring Trinomials of the Type x 2 + bx + c ALGEBRA 1 LESSON 9-5 Factor c 2 – 9c + 20. Since the middle term is negative, find the negative factors of 20. Identify the pair that has a sum of –9. c 2 – 9c + 20 = (c – 5)(c – 4) Factors of 20Sum of Factors –1 and –20–21 –2 and –10–12 –4 and –5–9 9-5

38. Factoring Trinomials of the Type x 2 + bx + c ALGEBRA 1 LESSON 9-5 a. Factor x 2 + 13x – 48. Identify the pair of factors of –48 that has a sum of 13. b. Factor n2 – 5n – 24. Identify the pair of factors of –24 that has a sum of –5. x 2 + 13x – 48 = (x + 16)(x – 3) n 2 – 5n – 24 = (n + 3)(n – 8) Factors of –48Sum of Factors 1 and –48–47 48 and –1 47 2 and –24–22 24 and –2 22 3 and –16–13 16 and –3 13 Factors of –24Sum of Factors 1 and –24–23 24 and–1 23 2 and–12–10 12 and–2 10 3 and–8 –5 9-5

39. Factoring Trinomials of the Type x 2 + bx + c Factor d 2 + 17dg – 60g 2. ALGEBRA 1 LESSON 9-5 Factors of –60Sum of Factors 1 and –60–59 60 and –1 59 2 and –30–28 30 and –2 28 3 and –20–17 20 and –3 17 Find the factors of –60. Identify the pair that has a sum of 17. d 2 + 17dg – 60g 2 = (d – 3g)(d + 20g) 9-5

40. Factoring Trinomials of the Type x 2 + bx + c Factor each expression. 1.c 2 + 6c + 92.x 2 – 11x + 183.g 2 – 2g – 24 4.y 2 + y – 1105.m 2 – 2mn + n 2 ALGEBRA 1 LESSON 9-5 (c + 3)(c + 3)(x – 2)(x – 9)(g – 6)(g + 4) (y + 11)(y – 10)(m – n)(m – n) 9-5

41. Factoring Trinomials of the Type ax 2 + bx + c ALGEBRA 1 LESSON 9-6 (For help, go to Lessons 9-2 and 9-5.) Find the greatest common factor. 1.12x 2 + 6x2.28m 2 – 35m + 143.4v 3 + 36v 2 + 10 Factor each expression. 4.x 2 + 5x + 45.y 2 – 3y – 286.t 2 – 11t + 30 9-6

42. Factoring Trinomials of the Type ax 2 + bx + c ALGEBRA 1 LESSON 9-6 1.12x 2 + 6x 12x 2 = 2 2 3 x x; 6x = 2 3 x; GCF = 2 3 x = 6x 2.28m 2 – 35m + 14 28m 2 = 2 2 7 m m; 35m = 5 7 m; 14 = 2 7; GCF = 7 3.4v 3 + 36v 2 + 10 4v 3 = 2 2 v v v; 36v 2 = 2 2 3 3 v v; 10 = 2 5; GCF = 2 4.Factors of 4 with a sum of 5: 1 and 4 x 2 + 5x + 4 = (x + 1)(x + 4) 5.Factors of –28 with a sum of –3: 4 and –7 y 2 – 3y – 28 = (y + 4)(y – 7) 6.Factors of 30 with a sum of –11: –5 and –6 t 2 – 11t + 30 = (t – 5)(t – 6) Solutions 9-6

43. Factoring Trinomials of the Type ax 2 + bx + c ALGEBRA 1 LESSON 9-6 Factor 20x 2 + 17x + 3. 20x 2 + 17x+ 3 F O I L 2 102 3 + 1 10 = 161 3 2 1 + 3 10 = 323 1 1 201 3 + 1 20 = 231 3 1 1 + 3 20 = 613 1 factors of a factors of c 4 54 3 + 1 5 = 171 3 20x 2 + 17x + 3 = (4x + 1)(5x + 3) 9-6

44. Factoring Trinomials of the Type ax 2 + bx + c ALGEBRA 1 LESSON 9-6 Factor 3n 2 – 7n – 6. 3n 2 –7n–6 (1)(3) (1)(–6) + (1)(3) = –3(1)(–6) (1)(1) + (–6)(3) = –17(–6)(1) (1)(–3) + (2)(3) = 3(2)(–3) (1)(2) + (–3)(3) = –7(–3)(2) 3n 2 – 7n – 6 = (n – 3)(3n + 2) 9-6

45. Factoring Trinomials of the Type ax 2 + bx + c ALGEBRA 1 LESSON 9-6 Factor 18x 2 + 33x – 30 completely. 18x 2 + 33x – 30 = 3(6x 2 + 11x – 10)Factor out the GCF. 18x 2 + 33x – 30 = 3(2x + 5)(3x – 2)Include the GCF in your final answer. Factor 6x 2 + 11x – 10. 6x 2 + 11x–10 6x 2 + 11x – 10 = (2x + 5)(3x – 2) 9-6 (2)(3)(2)(–10) + (1)(3) = –17(1)(–10) (2)(1) + (–10)(3) = –28(–10)(1) (2)(–5) + (2)(3) = –4(2)(–5) (2)(2) + (–5)(3) = –11(–5)(2) (2)(–2) + (5)(3) = 11(5)(–2)

46. Factoring Trinomials of the Type ax 2 + bx + c ALGEBRA 1 LESSON 9-6 Factor each expression. 1.3x 2 – 14x + 11 2.6t 2 + 13t – 63 3.9y 2 – 48y – 36 (x – 1)(3x – 11) (2t + 9)(3t – 7) 3(3y + 2)(y – 6) 9-6

47. Factoring Special Cases ALGEBRA 1 LESSON 9-7 (For help, go to Lessons 8–4 and 9-4.) Simplify each expression. 1.(3x) 2 2.(5y) 2 3.(15h 2 ) 2 4.(2ab 2 ) 2 Simplify each product. 5.(c – 6)(c + 6)6.(p – 11)(p – 11)7.(4d + 7)(4d + 7) 9-7

48. Factoring Special Cases ALGEBRA 1 LESSON 9-7 1.(3x) 2 = 3 2 x 2 = 9x 2 2.(5y) 2 = 5 2 y 2 = 25y 2 3.(15h 2 ) 2 = 15 2 (h 2 ) 2 = 225h 4 4.(2ab 2 ) 2 = 2 2 a 2 (b 2 ) 2 = 4a 2 b 4 5.(c – 6)(c + 6) is the difference of squares. (c – 6)(c + 6) = c 2 – 6 2 = c 2 – 36 6.(p – 11)(p – 11) is the square of a binomial. (p – 11) 2 = p 2 – 2p(11) + 11 2 = p 2 – 22p + 121 7.(4d + 7)(4d + 7) is the square of a binomial. (4d + 7) 2 = (4d) 2 + 2(4d)(7) + 7 2 = 16d 2 + 56d + 49 Solutions 9-7

49. Factoring Special Cases ALGEBRA 1 LESSON 9-7 Factor m 2 – 6m + 9. m 2 – 6m + 9 = m m – 6m + 3 3Rewrite first and last terms. = m m – 2(m 3) + 3 3Does the middle term equal 2ab? 6m = 2(m 3) = (m – 3) 2 Write the factors as the square of a binomial. 9-7

50. Factoring Special Cases ALGEBRA 1 LESSON 9-7 The area of a square is (16h 2 + 40h + 25) in. 2. Find the length of a side. 16h 2 + 40h + 25 = (4h) 2 + 40h + 5 2 Write 16h 2 as (4h) 2 and 25 as 5 2. = (4h + 5) 2 Write the factors as the square of a binomial. The side of the square has a length of (4h + 5) in. = (4h) 2 + 2(4h)(5) + 5 2 Does the middle term equal 2ab? 40h = 2(4h)(5) 9-7

51. Factoring Special Cases ALGEBRA 1 LESSON 9-7 Factor a 2 – 16. a 2 – 16 = a 2 – 4 2 Rewrite 16 as 4 2. = (a + 4)(a – 4)Factor. Check: Use FOIL to multiply. (a + 4)(a – 4) a 2 – 4a + 4a – 16 a 2 – 16 9-7

52. Factoring Special Cases Factor 9b 2 – 25. ALGEBRA 1 LESSON 9-7 9b 2 – 225 = (3b) 2 – 5 2 Rewrite 9b 2 as (3b) 2 and 25 as 5 2. = (3b + 5)(3b – 5)Factor. 9-7

53. Factoring Special Cases ALGEBRA 1 LESSON 9-7 Factor 5x 2 – 80. 5x 2 – 80 = 5(x 2 – 16)Factor out the GCF of 5. = 5(x + 4)(x – 4)Factor (x 2 – 16). Check: Use FOIL to multiply the binomials. Then multiply by the GCF. 5(x + 4)(x – 4) 5(x 2 – 16) 5x 2 – 80 9-7

54. Factoring Special Cases ALGEBRA 1 LESSON 9-7 Factor each expression. 1.y 2 – 18y + 812.9a 2 – 24a + 16 3.p 2 – 1694.36x 2 – 225 5.5m 2 – 456.2c 2 + 20c + 50 (y – 9) 2 (3a – 4) 2 (p + 13)(p – 13)(6x + 15)(6x – 15) 5(m + 3)(m – 3)2(c + 5) 2 9-7

55. Factoring by Grouping ALGEBRA 1 LESSON 9-8 (For help, go to Lessons 9-2 and 9-3.) Find the GCF of the terms of each polynomial. 1.6y 2 + 12y – 42.9r 3 + 15r 2 + 21r 3.30h 3 – 25h 2 – 40h4.16m 3 – 12m 2 – 36m Find each product. 5.(v + 3)(v 2 + 5)6.(2q 2 – 4)(q – 5) 7.(2t – 5)(3t + 4)8.(4x – 1)(x 2 + 2x + 3) 9-8

56. Factoring by Grouping ALGEBRA 1 LESSON 9-8 1.6y 2 + 12y – 42.9r 3 + 15r 2 + 21r 6y 2 = 2 3 y y; 9r 3 = 3 3 r r r; 12y = 2 2 3 y; 4 = 2 2;15r 2 = 3 5 r r; 21r = 3 7 r; GCF = 2GCF = 3r 3.30h 3 – 25h 2 – 40h4.16m 3 – 12m 2 – 36m 30h 3 = 2 3 5 h h h;16m 3 = 2 2 2 2 m m m; 25h 2 = 5 5 h h;12m 2 = 2 2 3 m m; 40h = 2 2 2 5 h;36m = 2 2 3 3 m; GCF = 5hGCF = 2 2 m = 4m 5.(v + 3)(v 2 + 5) = (v)(v 2 ) + (v)(5) + (3)(v 2 ) + (3)(5) = v 3 + 5v + 3v 2 + 15 = v 3 + 3v 2 + 5v + 15 Solutions 9-8

57. Factoring by Grouping ALGEBRA 1 LESSON 9-8 Solutions (continued) 7.(2t – 5)(3t + 4) = (2t)(3t) + (2t)(4) + (–5)(3t) + (–5)(4) = 6t 2 + 8t – 15t – 20 = 6t 2 – 7t – 20 6.(2q 2 – 4)(q – 5) = (2q 2 )(q) + (2q 2 )(–5) + (–4)(q) + (–4)(–5) = 2q 3 – 10q 2 – 4q + 20 8.(4x – 1)(x 2 + 2x + 3) = (4x)(x 2 ) + (4x)(2x) + (4x)(3) + (–1)(x 2 ) + (–1)(2x) + (–1)(3) = 4x 3 + 8x 2 + 12x – x 2 – 2x – 3 = 4x 3 + (8 – 1)x 2 + (12 – 2)x – 3 = 4x 3 + 7x 2 + 10x – 3 9-8

58. Factoring by Grouping ALGEBRA 1 LESSON 9-8 Factor 6x 3 + 3x 2 – 4x – 2. 6x 3 + 3x 2 – 4x – 2 = 3x 2 (2x + 1) – 2(2x + 1)Factor the GCF from each group of two terms. = (2x + 1)(3x 2 – 2)Factor out (2x + 1). = 6x 3 – 4x + 3x 2 – 2Use FOIL. Check: 6x 3 + 3x 2 – 4x – 2 (2x + 1)(3x 2 – 2) = 6x 3 + 3x 2 – 4x – 2 Write in standard form. 9-8

59. Factoring by Grouping ALGEBRA 1 LESSON 9-8 Factor 8t 4 + 12t 3 + 16t + 24. 8t 4 + 12t 3 + 16t + 24 = 4(2t 4 + 3t 3 + 4t + 6)Factor out the GCF, 4. = 4[t 3 (2t + 3) + 2(2t + 3)]Factor by grouping. = 4(2t + 3)(t 3 + 2)Factor again. 9-8

60. Factoring by Grouping Factor 24h 2 + 10h – 6. ALGEBRA 1 LESSON 9-8 Step 1: 24h 2 + 10h – 6 = 2(12h 2 + 5h – 3) Factor out the GCF, 2. Step 2: 12 –3 = –36Find the product ac. Step 4: 12h 2 – 4h + 9h – 3Rewrite the trinomial. Step 5: 4h(3h – 1) + 3(3h – 1)Factor by grouping. (4h + 3)(3h – 1)Factor again. 24h 2 + 10h – 6 = 2(4h + 3)(3h – 1)Include the GCF in your final answer. Step 3: Factors Sum –2(18) = –36–2 + 18 = 16 –3(12) = –36–3 + 12 = 9 –4(9) = –36 –4 + 9 = 5 Find two factors of ac that have a sum b. Use mental math to determine a good place to start. 9-8

61. Factoring by Grouping ALGEBRA 1 LESSON 9-8 A rectangular prism has a volume of 36x 3 + 51x 2 + 18x. Factor to find the possible expressions for the length, width, and height of the prism. Factor 36x 3 + 51x 2 + 18x. Step 1: 3x(12x 2 + 17x + 6)Factor out the GCF, 3x. Step 2: 12 6 = 72Find the product ac. Step 3: Factors Sum 4 184 + 18 = 22 6 126 + 12 = 18 8 98 + 9 = 17 Find two factors of ac that have sum b. Use mental math to determine a good place to start. 9-8

62. Factoring by Grouping ALGEBRA 1 LESSON 9-8 (continued) Step 4: 3x(12x 2 + 8x + 9x + 6)Rewrite the trinomial. Step 5: 3x[4x(3x + 2) + 3(3x + 2)]Factor by grouping. 3x(4x + 3)(3x + 2)Factor again. The possible dimensions of the prism are 3x, (4x + 3), and (3x + 2). 9-8

63. Factoring by Grouping ALGEBRA 1 LESSON 9-8 Factor each expression. 1.10p 3 – 25p 2 + 4p – 10 2.36x 4 – 48x 3 + 9x 2 – 12x 3.16a 3 – 24a 2 + 12a – 18 (5p 2 + 2)(2p – 5) 3x(4x 2 + 1)(3x – 4) 2(4a 2 + 3)(2a – 3) 9-8