Créer une présentation
Télécharger la présentation

Télécharger la présentation
## Adding and Subtracting Polynomials

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -

**Adding and Subtracting Polynomials**Lesson 9-1**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**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**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**(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.**= (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.**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 + 10g2Multiply the coefficients and add the exponents of powers with the same base. = –6g5 – 12g3 + 10g2Simplify.**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.**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.**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 – 6Now distribute y and 2. = 2y2 + y – 6 Simplify.**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.**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**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.**(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.**Multiplying Special Cases**Lesson 9-4**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.**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.**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)**= 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.**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.**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.**Factoring Trinomials of the Type x2 + bx + c**Lesson 9-5**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).**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)**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)**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)**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**(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)**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.**Factoring Special Cases**Lesson 9-7**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.**= (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.**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.**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.**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).**Factoring by Grouping**Lesson 9-8**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).**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 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.**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.**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).