Adding and Subtracting Polynomials

1. Adding and Subtracting Polynomials Lesson 9-1

2. 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 Lesson 9-1 Find the degree of each monomial. a. 18 b. 3xy3 c. 6c

3. Place terms in order. 7x – 2 3x5 – 2x5 + 7x – 2 Place terms in order. x5 + 7x – 2 Combine like terms. Adding and Subtracting Polynomials 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

4. 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 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

5. (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 Lesson 9-1 Simplify (2x3 + 4x2 – 6) – (5x3 + 2x – 2). Method 1:Subtract vertically. Line up like terms. Then add the coefficients.

6. = (2x3 – 5x3) + 4x2 – 2x + (–6 + 2)Group like terms. = –3x3 + 4x2 – 2x – 4 Simplify. Adding and Subtracting Polynomials 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.

7. Multiplying and Factoring Lesson 9-2

8. Multiplying and Factoring 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. Multiplying and Factoring 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.

10. Step 2: Factor out the GCF. 4x3 – 8x2 + 12x Multiplying and Factoring Lesson 9-2 Factor 4x3 – 8x2 + 12x. Step 1: Find the GCF. 4x3 = 2 • 2 • x • x • x 8x2 = 2 • 2 • 2 • x • x 12x = 2 • 2 • 3 • x = 4x(x2) + 4x(–2x) + 4x(3) = 4x(x2 – 2x + 3) The GCF is 2 • 2 • x, or 4x.

11. Multiplying Binomials Lesson 9-3

12. Multiplying Binomials 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.

13. 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 Lesson 9-3 Simplify (4x + 2)(3x – 6). The product is 12x2 – 18x – 12.

14. Substitute. = (3x + 2)(2x – 1) –x(x + 3) Use FOIL to simplify (3x + 2) (2x – 1) and the Distributive Property to simplify x(x + 3). = 6x2 – 3x + 4x – 2 –x2 – 3x Group like terms. = 6x2 – x2 – 3x + 4x – 3x – 2 Simplify. = 5x2 – 2x – 2 Multiplying Binomials 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

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

16. (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 Lesson 9-3 (continued) Method 2: Multiply using the horizontal method. The product is 6x3 + 17x2 – 8x + 21.

17. Multiplying Special Cases Lesson 9-4

18. Multiplying Special Cases 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.

19. 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 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.

20. 1 4 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 2 1 4 = B2 + BW + W 2Simplify. 1 4 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. 1 2 1 2 Multiplying Special Cases Lesson 9-4 (continued)

21. = 802 + 2(80 • 1) + 12Square the binomial. = 6400 + 160 + 1 = 6561 Simplify. Multiplying Special Cases 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.

22. Multiplying Special Cases Lesson 9-4 Find (p4 – 8)(p4 + 8). (p4 – 8)(p4 + 8) = (p4)2 – (8)2Find the difference of squares. = p8 – 64 Simplify.

23. Multiplying Special Cases 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.

24. Factoring Trinomials of the Type x2 + bx + c Lesson 9-5

25. 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 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).

26. 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 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)

27. 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 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)

28. 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 Lesson 9-5 Factor d2 + 17dg – 60g2. d2 + 17dg – 60g2 = (d – 3g)(d + 20g)

29. Factoring Trinomials of the Type ax2 + bx + c Lesson 9-6

30. 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 Lesson 9-6 Factor 20x2 + 17x + 3. 2 • 10 2 • 3 + 1 • 10 = 16 1 • 3 2 • 1 + 3 • 10 = 32 3 • 1

31. (1)(2) + (–3)(3) = –7 (–3)(2) 3n2 – 7n – 6 = (n– 3)(3n + 2) Factoring Trinomials of the Type ax2 + bx + c 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)

32. 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 Lesson 9-6 Factor 18x2 + 33x – 30 completely. 18x2 + 33x – 30 = 3(6x2 + 11x – 10) Factor out the GCF.

33. Factoring Special Cases Lesson 9-7

34. Factoring Special Cases 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.

35. = (4h)2 + 2(4h)(5) + 52Does the middle term equal 2ab? 40h = 2(4h)(5) Factoring Special Cases 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.

36. Check: Use FOIL to multiply. (a + 4)(a – 4) a2 – 4a + 4a – 16 a2 – 16 Factoring Special Cases Lesson 9-7 Factor a2 – 16. a2 – 16 = a2 – 42Rewrite 16 as 42. = (a + 4)(a – 4) Factor.

37. Factoring Special Cases Lesson 9-7 Factor 9b2 – 225. 9b2 – 225 = (3b)2 – 152Rewrite 9b2 as (3b)2 and 225 as 152. = (3b + 15)(3b –15) Factor.

38. Check: Use FOIL to multiply the binomials. Then multiply by the GCF. 5(x + 4)(x – 4) 5(x2 – 16) 5x2 – 80 Factoring Special Cases Lesson 9-7 Factor 5x2 – 80. 5x2 – 80 = 5(x2 – 16) Factor out the GCF of 5. = 5(x + 4)(x – 4) Factor (x2 – 16).

39. Factoring by Grouping Lesson 9-8

40. Check: 6x3 + 3x2 – 4x – 2 (2x + 1)(3x2 – 2) = 6x3 – 4x + 3x2 – 2 Use FOIL. = 6x3 + 3x2 – 4x – 2 Write in standard form. Factoring by Grouping Lesson 9-8 Factor 6x3 + 3x2 – 4x – 2. 6x3 + 3x2 – 4x – 2 = 3x2(2x + 1) – 2(2x + 1) Factor the GCF from each group of two terms. = (2x + 1)(3x2 – 2) Factor out (2x + 1).

41. Factoring by Grouping Lesson 9-8 Factor 8t4 + 12t3 + 16t2 + 24t. 8t4 + 12t3 + 16t2 + 24t = 4t(2t3 + 3t2 + 4t + 6) Factor out the GCF, 4. = 4t[t2(2t + 3) + 2(2t + 3)] Factor by grouping. = 4t(2t + 3)(t2 + 2) Factor again.

42. 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. Factoring by Grouping Lesson 9-8 Factor 24h2 + 10h – 6. Step 1:24h2 + 10h – 6 = 2(12h2 + 5h – 3)   Factor out the GCF, 2. Step 2: 12 • –3 = –36 Find the product ac. Step 4:12h2– 4h + 9h – 3 Rewrite the trinomial. Step 5: 4h(3h – 1) + 3(3h – 1) Factor by grouping.   (4h + 3)(3h – 1) Factor again. 24h2 + 10h – 6 = 2(4h + 3)(3h – 1) Include the GCF in your final answer.

43. Step 3: Factors     Sum 4 • 18 4 + 18 = 22 6 • 12 6 + 12 = 18 8 • 9 8 + 9 = 17 Find two factors of ac that have sum b. Use mental math to determine a good place to start. Factoring by Grouping Lesson 9-8 A rectangular prism has a volume of 36x3 + 51x2 + 18x. Factor to find the possible expressions for the length, width, and height of the prism. Factor 36x3 + 51x2 + 18x. Step 1: 3x(12x2 + 17x + 6) Factor out the GCF, 3x. Step 2: 12 • 6 = 72 Find the product ac.

44. Factoring by Grouping Lesson 9-8 (continued) Step 4: 3x(12x2 + 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).