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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. 4n2 – 7n26. 8x2 – x2 9-1

2. Adding and Subtracting Polynomials ALGEBRA 1 LESSON 9-1 Solutions 1. 6t + 13t = (6 + 13)t = 19t 2. 5g + 34g = (5 + 34)g = 39g 3. 7k – 15k = (7 – 15)k = –8k4. 2b – 6 + 9b = (2 + 9)b – 6 = 11b – 6 5. 4n2 – 7n2 = (4 – 7)n2 = –3n26. 8x2 – x2 = (8 – 1)x2 = 7x2 9-1

3. Degree: 0 The degree of a nonzero constant is 0. Degree: 4 The exponents are 1 and 3. Their sum is 4. Degree: 1 6c = 6c1. The exponent is 1. Adding and Subtracting Polynomials ALGEBRA 1 LESSON 9-1 Find the degree of each monomial. a. 18 b. 3xy3 c. 6c 9-1

4. Place terms in order. 7x – 2 3x5 – 2x5 + 7x – 2 Place terms in order. x5 + 7x – 2 Combine like terms. 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. 3x5 – 2 – 2x5 + 7x fifth degree trinomial 9-1

5. 6x2 + 3x + 7 2x2 – 6x – 4 8x2 – 3x + 3 Group like terms. Then add the coefficients. (6x2 + 3x + 7) + (2x2 – 6x – 4) = (6x2 + 2x2) + (3x – 6x) + (7 – 4) Adding and Subtracting Polynomials ALGEBRA 1 LESSON 9-1 Simplify (6x2 + 3x + 7) + (2x2 – 6x – 4). Method 1:Add vertically. Line up like terms. Then add the coefficients. Method 2:Add horizontally. = 8x2 – 3x + 3 9-1

6. (2x3 + 4x2 – 6) Line up like terms. –(5x3 – 2x – 2) 2x3 + 4x2 – 6 Add the opposite. –5x3– 2x+ 2 –3x3 + 4x2 – 2x – 4 Adding and Subtracting Polynomials ALGEBRA 1 LESSON 9-1 Simplify (2x3 + 4x2 – 6) – (5x3 + 2x – 2). Method 1:Subtract vertically. Line up like terms. Then add the coefficients. 9-1

7. = (2x3 – 5x3) + 4x2 – 2x + (–6 + 2)Group like terms. = –3x3 + 4x2 – 2x – 4 Simplify. Adding and Subtracting Polynomials ALGEBRA 1 LESSON 9-1 (continued) Method 2: Subtract horizontally. (2x3 + 4x2 – 6) – (5x3 + 2x – 2) = 2x3 + 4x2 – 6 – 5x3– 2x+ 2 Write the opposite of each term in the polynomial being subtracted. 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 – 2x2 2. 2b2 – 4b3 + 6 3. (2x4 + 3x – 4) + (–3x + 4 + x4) 4. (–3r + 4r2 – 3) – (4r2 + 6r – 2) –2x2 + 3x – 4; quadratic trinomial –4b3 + 2b2 + 6; cubic trinomial 3x4; 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)5 7. 7(p – 2) 8. (6 – x)9 9. –2(4q – 1) 9-2

10. Multiplying and Factoring ALGEBRA 1 LESSON 9-2 Solutions 1. 3(302) = 906 2. 41(7) = 287 3. 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 9-2

11. Multiplying and Factoring ALGEBRA 1 LESSON 9-2 Simplify –2g2(3g3 + 6g – 5). –2g2(3g3 + 6g – 5) = –2g2(3g3) –2g2(6g) –2g2(–5) Use the Distributive Property. = –6g2+ 3 – 12g2+ 1 + 10g2Multiply the coefficients and add the exponents of powers with the same base. = –6g5 – 12g3 + 10g2Simplify. 9-2

12. Multiplying and Factoring ALGEBRA 1 LESSON 9-2 Find the GCF of 2x4 + 10x2 – 6x. List the prime factors of each term. Identify the factors common to all terms. 2x4 = 2 • x • x • x • x 10x2 = 2 • 5 • x • x 6x = 2 • 3 • x The GCF is 2 • x, or 2x. 9-2

13. Step 2: Factor out the GCF. 4x3 + 12x2 – 16x Multiplying and Factoring ALGEBRA 1 LESSON 9-2 Factor 4x3 + 12x2 – 16x. Step 1: Find the GCF. 4x3 = 2 • 2 • x • x • x 12x2 = 2 • 2 • 3 • x • x 16x = 2 • 2 • 2 • 2 • x = 4x(x2) + 4x(3x) + 4x(–4) = 4x(x2 + 3x – 4) The GCF is 2 • 2 • x, or 4x. 9-2

14. Multiplying and Factoring ALGEBRA 1 LESSON 9-2 1. Simplify –2x2(–3x2 + 2x + 8). 2. Find the GCF of 16b4 – 4b3 + 8b2. 3. Factor 3x3 + 9x2. 4. Factor 10y3 + 5y2 – 15y. 6x4 – 4x3 – 16x2 4b2 3x2(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(h2 + 8h – 3) 3.y2(2y3 – 7) Simplify. Write each answer in standard form. 4. (x3 + 3x2 + x) + (5x2 + x + 1) 5. (3t3 – 6t + 8) + (5t3 + 7t – 2) 6.w(w + 1) + 4w(w – 7) 7. 6b(b – 2) – b(8b + 3) 8.m(4m2 – 6) + 3m2(m + 9) 9. 3d2(d3 – 6) – d3(2d2 + 4) 9-3

16. Multiplying Binomials ALGEBRA 1 LESSON 9-3 1. 4r(r – 1) = 4r(r) – 4r(1) = 4r 2 – 4r 2. 6h(h2 + 8h – 3) = 6h(h2) + 6h(8h) – 6h(3)= 6h3 + 48h2 – 18h 3.y2(2y3 – 7) = y2(2y3) – 7y2 = 2y5 – 7y2 4.x3 + 3x2 + x5. 3t3 – 6t + 8 + 5x2 + x + 1 + 5t3 + 7t – 2 x3 + 8x2 + 2x + 1 8t3 + t + 6 Solutions 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) = w2 + w + 4w2 – 28w = 6b2 – 12b – 8b2 – 3b = (1 + 4)w2 + (1 – 28)w = (6 – 8)b2 + (–12 – 3)b = 5w2 – 27w = –2b2 – 15b 9-3

17. Multiplying Binomials ALGEBRA 1 LESSON 9-3 Solutions (continued) 8.m(4m2 – 6) + 3m2(m + 9) = m(4m2) – m(6) + 3m2(m) + 3m2(9) = 4m3 – 6m + 3m3 + 27m2 = (4 + 3)m3 + 27m2 – 6m = 7m3 + 27m2 – 6m 9. 3d2(d3 – 6) – d3(2d2 + 4) = 3d2(d3) – 3d2(6) – d3(2d2) – d3(4) = 3d5 – 18d2 – 2d5 – 4d3 = (3 – 2)d5 – 4d3 – 18d2 = d5 – 4d3 – 18d2 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. = 2y2 – 3y + 4y – 6Now distribute y and 2. = 2y2 + y – 6 Simplify. 9-3

19. First Outer Inner Last (4x + 2)(3x – 6) = (4x)(3x) (4x)(–6) (2)(3x) + (2)(–6) + + = 12x2 24x 6x 12 – + – = 12x2 18x 12 – – Multiplying Binomials ALGEBRA 1 LESSON 9-3 Simplify (4x + 2)(3x – 6). The product is 12x2 – 18x – 12. 9-3

20. = 6x2 – x2 – 3x + 4x – 3x – 2 Group like terms. = 5x2 – 2x – 2 Simplify. 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. = 6x2 – 3x + 4x – 2 –x2 – 3xUse FOIL to simplify (3x + 2) (2x – 1) and the Distributive Property to simplify x(x + 3). 9-3

21. Method 1: Multiply using the vertical method. 3x2  –   2x  +  3 2x  +  7 6x3  + 17x2  –   8x  +  21 Add like terms. Multiplying Binomials ALGEBRA 1 LESSON 9-3 Simplify the product (3x2 – 2x + 3)(2x + 7). 21x2  –  14x  +  21Multiply by 7. 6x3  –  4x2  +   6xMultiply by 2x. 9-3

22. (2x + 7)(3x2 – 2x + 3) = (2x)(3x2) – (2x)(2x) + (2x)(3) + (7)(3x2) – (7)(2x) + (7)(3) = 6x3 – 4x2 + 6x + 21x2 – 14x + 21 = 6x3 + 17x2 – 8x + 21 Multiplying Binomials ALGEBRA 1 LESSON 9-3 (continued) Method 2: Multiply using the horizontal method. The product is 6x3 + 17x2 – 8x + 21. 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)(3x2 + x + 2) 4. Find the area of the shaded region. x2 – 3x – 18 6b2 – 22b + 20 9x3 – 9x2 + 2x – 8 2x2 + 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)22. (3v)23. (–4c)24. (5g3)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. (8c2 + 2)(c2 – 10) 10. (6y2 – 3)(9y2 + 1) 9-4

25. Multiplying Special Cases ALGEBRA 1 LESSON 9-4 Solutions 1. (7x)2 = 72 • x2 = 49x22. (3v)2 = 32 • v2 = 9v2 3. (–4c)2 = (–4)2 • c2 = 16c24. (5g3)2 = 52 • (g3)2 = 25g6 5. (j + 5)(j + 7) = (j)(j) + (j)(7) + (5)(j) + (5)(7) = j2 + 7j + 5j + 35 = j2 + 12j + 35 6. (2b – 6)(3b – 8) = (2b)(3b) + (2b)(–8) + (–6)(3b) + (–6)(–8) = 6b2 – 16b – 18b + 48 = 6b2 – 34b + 48 9-4

26. Multiplying Special Cases ALGEBRA 1 LESSON 9-4 Solutions (continued) 7. (4y + 1)(5y – 2)) = (4y)(5y) + (4y)(–2) + (1)(5y) + (1)(–2) = 20y2 – 8y + 5y – 2 = 20y2 – 3y – 2 8. (x + 3)(x – 4) = (x)(x) + (x)(-4) + (3)(x) + (3)(–4) = x2 – 4x + 3x – 12 = x2 – x – 12 9. (8c2 + 2)(c2 – 10) = (8c2)(c2) + (8c2)(–10) + (2)(c2) + (2)(–10) = 8c4 – 80c2 + 2c2 – 20 = 8c4 – 78c2 – 20 10. (6y2 – 3)(9y2 + 1) = (6y2)(9y2) + (6y2)(1) + (–3)(9y2) + (–3)(1) = 54y4 + 6y2 – 27y2 – 3 = 54y4 – 21y2 – 3 9-4

27. Multiplying Special Cases ALGEBRA 1 LESSON 9-4 a. Find (y + 11)2. (y + 11)2 = y2 + 2y(11) + 72Square the binomial. = y2 + 22y + 121 Simplify. b. Find (3w – 6)2. (3w – 6)2 = (3w)2 –2(3w)(6) + 62Square the binomial. = 9w2 – 36w + 36 Simplify. 9-4

28. 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 . 1 4 1 4 B W B W BB BW BW WW 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. 9-4

29. 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. 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 (B + W)2 = (B)2 – 2(B)(W) + (W)2Square the binomial. 1 4 1 4 1 4 1 2 1 4 The expressions B2 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. = B2 + BW + W 2Simplify. 1 4 1 2 1 2 Multiplying Special Cases ALGEBRA 1 LESSON 9-4 (continued) 9-4

30. = 802 + 2(80 • 1) + 12Square the binomial. = 6400 + 160 + 1 = 6561 Simplify. Multiplying Special Cases ALGEBRA 1 LESSON 9-4 a. Find 812 using mental math. 812 = (80 + 1)2 b. Find 592 using mental math. 592 = (60 – 1)2 = 602 – 2(60 • 1) + 12Square the binomial. = 3600 – 120 + 1 = 3481 Simplify. 9-4

31. Multiplying Special Cases ALGEBRA 1 LESSON 9-4 Find (p4 – 8)(p4 + 8). (p4 – 8)(p4 + 8) = (p4)2 – (8)2Find the difference of squares. = p8 – 64 Simplify. 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. = 402 – 32Find the difference of squares. = 1600 – 9 = 1591 Simplify. 9-4

33. Multiplying Special Cases ALGEBRA 1 LESSON 9-4 Find each square. 1. (y + 9)22. (2h – 7)2 3. 4124. 292 5. Find (p3 – 7)(p3 + 7). 6. Find 32 • 28. y2 + 18y + 81 4h2 – 28h + 49 1681 841 p6 – 49 896 9-4

34. Factoring Trinomials of the Type x2 + bx + c ALGEBRA 1 LESSON 9-5 (For help, go to Skills Handbook page 721.) List all of the factors of each number. 1. 24 2. 12 3. 54 4. 15 5. 36 6. 56 7. 64 8. 96 9-5

35. Factoring Trinomials of the Type x2 + bx + c ALGEBRA 1 LESSON 9-5 Solutions 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 9-5

36. Factors of 15 Sum of Factors 1 and 15 16 3 and 5 8 Check: x2 + 8x + 15 (x + 3)(x + 5) = x2 + 5x + 3x + 15 = x2 + 8x + 15 Factoring Trinomials of the Type x2 + bx + c ALGEBRA 1 LESSON 9-5 Factor x2 + 8x + 15. Find the factors of 15. Identify the pair that has a sum of 8. x2 + 8x + 15 = (x + 3)(x + 5). 9-5

37. Factors of 20 Sum of Factors –1 and –20 –21 –2 and –10 –12 –4 and –5 –9 Factoring Trinomials of the Type x2 + bx + c ALGEBRA 1 LESSON 9-5 Factor c2 – 9c + 20. Since the middle term is negative, find the negative factors of 20. Identify the pair that has a sum of –9. c2 – 9c + 20 = (c – 5)(c – 4) 9-5

38. Factors of –48 Sum 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 –24 Sum of Factors 1 and –24 –23 24 and –1 23 2 and –12 –10 12 and –2 10 3 and –8 –5 Factoring Trinomials of the Type x2 + bx + c ALGEBRA 1 LESSON 9-5 a. Factor x2 + 13x – 48. b. Factor n2 – 5n – 24. Identify the pair of factors of –48 that has a sum of 13. Identify the pair of factors of –24 that has a sum of –5. x2 + 13x – 48 = (x + 16)(x – 3) n2 – 5n – 24 = (n + 3)(n – 8) 9-5

39. Find the factors of –60. Identify the pair that has a sum of 17. Factors of –60 Sum 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 Factoring Trinomials of the Type x2 + bx + c ALGEBRA 1 LESSON 9-5 Factor d2 + 17dg – 60g2. d2 + 17dg – 60g2 = (d – 3g)(d + 20g) 9-5

40. Factoring Trinomials of the Type x2 + bx + c ALGEBRA 1 LESSON 9-5 Factor each expression. 1.c2 + 6c + 9 2.x2 – 11x + 18 3.g2 – 2g – 24 4.y2 + y – 110 5.m2 – 2mn + n2 (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 ax2 + bx + c ALGEBRA 1 LESSON 9-6 (For help, go to Lessons 9-2 and 9-5.) Find the greatest common factor. 1. 12x2 + 6x2. 28m2 – 35m + 14 3. 4v3 + 36v2 + 10 Factor each expression. 4.x2 + 5x + 4 5.y2 – 3y – 28 6.t2 – 11t + 30 9-6

42. Factoring Trinomials of the Type ax2 + bx + c ALGEBRA 1 LESSON 9-6 1. 12x2 + 6x12x2 = 2 • 2 • 3 • x • x; 6x = 2 • 3 • x;GCF = 2 • 3 • x = 6x 2. 28m2 – 35m + 1428m2 = 2 • 2 • 7 • m • m; 35m = 5 • 7 • m; 14 = 2 • 7; GCF = 7 3. 4v3 + 36v2 + 104v3 = 2 • 2 • v • v • v; 36v2 = 2 • 2 • 3 • 3 • v • v; 10 = 2 • 5; GCF = 2 4. Factors of 4 with a sum of 5: 1 and 4x2 + 5x + 4 = (x + 1)(x + 4) 5. Factors of –28 with a sum of –3: 4 and –7y2 – 3y – 28 = (y + 4)(y – 7) 6. Factors of 30 with a sum of –11: –5 and –6t2 – 11t + 30 = (t – 5)(t – 6) Solutions 9-6

43. 20x2 + 17x + 3 FOIL factors of a 1 • 20 1 • 3 + 1 • 20 = 23 1 • 3 1 • 1 + 3 • 20 = 61 3 • 1 factors of c 4 • 5 4 • 3 + 1 • 5 = 17 1 • 3 20x2 + 17x + 3 = (4x + 1)(5x + 3) Factoring Trinomials of the Type ax2 + bx + c ALGEBRA 1 LESSON 9-6 Factor 20x2 + 17x + 3. 2 • 10 2 • 3 + 1 • 10 = 16 1 • 3 2 • 1 + 3 • 10 = 32 3 • 1 9-6

44. (1)(2) + (–3)(3) = –7 (–3)(2) 3n2 – 7n – 6 = (n– 3)(3n + 2) Factoring Trinomials of the Type ax2 + bx + c ALGEBRA 1 LESSON 9-6 Factor 3n2 – 7n – 6. 3n2 –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) 9-6

45. Factor 6x2 + 11x – 10. 6x2 + 11x –10 (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) 6x2 + 11x – 10 = (2x + 5)(3x– 2) 18x2 + 33x – 30 = 3(2x + 5)(3x – 2) Include the GCF in your final answer. Factoring Trinomials of the Type ax2 + bx + c ALGEBRA 1 LESSON 9-6 Factor 18x2 + 33x – 30 completely. 18x2 + 33x – 30 = 3(6x2 + 11x – 10) Factor out the GCF. 9-6

46. Factoring Trinomials of the Type ax2 + bx + c ALGEBRA 1 LESSON 9-6 Factor each expression. 1. 3x2 – 14x + 11 2. 6t2 + 13t – 63 3. 9y2 – 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)22. (5y)23. (15h2)24. (2ab2)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 Solutions 1. (3x)2 = 32 • x2 = 9x22. (5y)2 = 52 • y2 = 25y2 3. (15h2)2 = 152 • (h2)2 = 225h44. (2ab2)2 = 22 • a2 • (b2)2 = 4a2b4 5. (c – 6)(c + 6) is the difference of squares.(c – 6)(c + 6) = c2 – 62 = c2 – 36 6. (p – 11)(p – 11) is the square of a binomial.(p – 11)2 = p2 – 2p(11) + 112 = p2 – 22p + 121 7. (4d + 7)(4d + 7) is the square of a binomial.(4d + 7)2 = (4d)2 + 2(4d)(7) + 72 = 16d2 + 56d + 49 9-7

49. Factoring Special Cases ALGEBRA 1 LESSON 9-7 Factor m2 – 6m + 9. m2 – 6m + 9 = m • m – 6m + 3 • 3Rewrite first and last terms. = m • m – 2(m • 3) + 3 • 3 Does the middle term equal 2ab? 6m = 2(m • 3) = (m – 3)2Write the factors as the square of a binomial. 9-7

50. = (4h)2 + 2(4h)(5) + 52Does the middle term equal 2ab? 40h = 2(4h)(5) Factoring Special Cases ALGEBRA 1 LESSON 9-7 The area of a square is (16h2 + 40h + 25) in.2. Find the length of a side. 16h2 + 40h + 25 = (4h)2 + 40h + 52Write 16h2 as (4h)2 and 25 as 52. = (4h + 5)2Write the factors as the square of a binomial. The side of the square has a length of (4h + 5) in. 9-7