Adding and Subtracting Polynomials Lesson 9-1

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

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 Place terms in order. 7x – 2 linear binomial b. 3x5 – 2 – 2x5 + 7x Place terms in order. 3x5 – 2x5 + 7x – 2 x5 + 7x – 2 Combine like terms. fifth degree trinomial

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. 6x2 + 3x + 7 2x2 – 6x – 4 8x2 – 3x + 3 Method 2: Add horizontally. Group like terms. Then add the coefficients. (6x2 + 3x + 7) + (2x2 – 6x – 4) = (6x2 + 2x2) + (3x – 6x) + (7 – 4) = 8x2 – 3x + 3

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

Multiplying and Factoring Lesson 9-2

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 + 10g2 Multiply the coefficients and add the exponents of powers with the same base. = –6g5 – 12g3 + 10g2 Simplify.

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.

Multiplying and Factoring Lesson 9-2 Factor 4x3 – 8x2 + 12x. Step 1: Find the GCF. Step 2: Factor out the GCF. 4x3 – 8x2 + 12x 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.

Multiplying Binomials Lesson 9-3

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 – 6 Now distribute y and 2. = 2y2 + y – 6 Simplify.

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

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

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

Multiplying Binomials Lesson 9-3 (continued) Method 2: Multiply using the horizontal method. = (2x)(3x2) – (2x)(2x) + (2x)(3) + (7)(3x2) – (7)(2x) + (7)(3) (2x + 7)(3x2 – 2x + 3) = 6x3 – 4x2 + 6x + 21x2 – 14x + 21 = 6x3 + 17x2 – 8x + 21 The product is 6x3 + 17x2 – 8x + 21.

Multiplying Special Cases Lesson 9-4

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

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. 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 . BB BW BW WW B W B W 1 4

Multiplying Special Cases 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. 1 2 ( B + W)2 = ( B)2 – 2( B)( W) + ( W)2 Square the binomial. 1 2 = B2 + BW + W 2 Simplify. 1 4 2 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 4

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

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

Multiplying Special Cases Lesson 9-4 Find 43 • 37. 43 • 37 = (40 + 3)(40 – 3) Express each factor using 40 and 3. = 402 – 32 Find the difference of squares. = 1600 – 9 = 1591 Simplify.

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

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. Factors of 15 Sum of Factors 1 and 15 16 3 and 5 8 x2 + 8x + 15 = (x + 3)(x + 5). = x2 + 5x + 3x + 15 Check: x2 + 8x + 15 (x + 3)(x + 5) = x2 + 8x + 15

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. Factors of 20 Sum of Factors –1 and –20 –21 –2 and –10 –12 –4 and –5 –9 c2 – 9c + 20 = (c – 5)(c – 4)

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. 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 x2 + 13x – 48 = (x + 16)(x – 3) n2 – 5n – 24 = (n + 3)(n – 8)

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

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

Factoring Trinomials of the Type ax2 + bx + c Lesson 9-6 Factor 20x2 + 17x + 3. 20x2 + 17x + 3 F O I L 1 • 20 1 • 3 + 1 • 20 = 23 1 • 3 1 • 1 + 3 • 20 = 61 3 • 1 factors of a factors of c 2 • 10 2 • 3 + 1 • 10 = 16 1 • 3 2 • 1 + 3 • 10 = 32 3 • 1 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 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) (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 18x2 + 33x – 30 completely. 18x2 + 33x – 30 = 3(6x2 + 11x – 10) Factor out the GCF. 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 Special Cases Lesson 9-7

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

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 + 52 Write 16h2 as (4h)2 and 25 as 52. = (4h)2 + 2(4h)(5) + 52 Does the middle term equal 2ab? 40h = 2(4h)(5) = (4h + 5)2 Write the factors as the square of a binomial. The side of the square has a length of (4h + 5) in.

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

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

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). Check: Use FOIL to multiply the binomials. Then multiply by the GCF. 5(x + 4)(x – 4) 5(x2 – 16) 5x2 – 80

Factoring by Grouping Lesson 9-8

= (2x + 1)(3x2 – 2) Factor out (2x + 1). 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). = 6x3 – 4x + 3x2 – 2 Use FOIL. Check: 6x3 + 3x2 – 4x – 2 (2x + 1)(3x2 – 2) = 6x3 + 3x2 – 4x – 2 Write in standard form.

= 4t[t2(2t + 3) + 2(2t + 3)] Factor by grouping. 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.

Step 1: 24h2 + 10h – 6 = 2(12h2 + 5h – 3) Factor out the GCF, 2. 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 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. 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.

Step 1: 3x(12x2 + 17x + 6) Factor out the GCF, 3x. 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. 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.

Step 4: 3x(12x2 + 8x + 9x + 6) Rewrite the trinomial. 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).