Products of Binomials 7 8 Special Products of

Products of Binomials 7 -8 Special Products of Binomials Warm Up Lesson Presentation Lesson Quiz Holt Algebra 11

7 -8 Special Products of Binomials Warm Up Simplify. 2. 72 49 1. 42 16 3. (– 2)2 4 4. (x)2 x 2 5. –(5 y 2) – 25 y 2 6. (m 2)2 m 4 7. 2(6 xy) 12 xy 8. 2(8 x 2) 16 x 2 Holt Algebra 1

7 -8 Special Products of Binomials Objective Find special products of binomials. Holt Algebra 1

7 -8 Special Products of Binomials Vocabulary perfect-square trinomial difference of two squares Holt Algebra 1

7 -8 Special Products of Binomials Imagine a square with sides of length (a + b): The area of this square is (a + b) or (a + b)2. The area of this square can also be found by adding the areas of the smaller squares and the rectangles inside. The sum of the areas inside is a 2 + ab + b 2. Holt Algebra 1

7 -8 Special Products of Binomials This means that (a + b)2 = a 2+ 2 ab + b 2. You can use the FOIL method to verify this: F L (a + b)2 = (a + b) = a 2 + ab + b 2 I = a 2 + 2 ab + b 2 O A trinomial of the form a 2 + 2 ab + b 2 is called a perfect-square trinomial. A perfect-square trinomial is a trinomial that is the result of squaring a binomial. Holt Algebra 1

7 -8 Special Products of Binomials Example 1: Finding Products in the Form (a + b)2 Multiply. A. (x +3)2 (a + b)2 = a 2 + 2 ab + b 2 (x + 3)2 = x 2 + 2(x)(3) + 32 = x 2 + 6 x + 9 Use the rule for (a + b)2. Identify a and b: a = x and b = 3. Simplify. B. (4 s + 3 t)2 (a + b)2 = a 2 + 2 ab + b 2 Use the rule for (a + b)2. Identify a and b: a = 4 s and b = 3 t. 2 2 2 (4 s + 3 t) = (4 s) + 2(4 s)(3 t) + (3 t) = 16 s 2 + 24 st + 9 t 2 Holt Algebra 1 Simplify.

7 -8 Special Products of Binomials Example 1 C: Finding Products in the Form (a + b)2 Multiply. C. (5 + m 2)2 (a + b)2 = a 2 + 2 ab + b 2 Use the rule for (a + b)2. Identify a and b: a = 5 and b = m 2. (5 + m 2)2 = 52 + 2(5)(m 2) + (m 2)2 = 25 + 10 m 2 + m 4 Holt Algebra 1 Simplify.

7 -8 Special Products of Binomials Check It Out! Example 1 Multiply. A. (x + 6)2 (a + b)2 = a 2 + 2 ab + b 2 (x + 6)2 = x 2 + 2(x)(6) + 62 = x 2 + 12 x + 36 B. (5 a + b)2 Use the rule for (a + b)2. Identify a and b: a = x and b = 6. Simplify. Use the rule for (a + b)2 = a 2 + 2 ab + b 2 Identify a and b: a = 5 a and b = b. (5 a + b)2 = (5 a)2 + 2(5 a)(b) + b 2 = 25 a 2 + 10 ab + b 2 Holt Algebra 1 Simplify.

7 -8 Special Products of Binomials Check It Out! Example 1 C Multiply. (1 + c 3)2 (a + b)2 = a 2 + 2 ab + b 2 Use the rule for (a + b)2. Identify a and b: a = 1 and b = c 3. (1 + c 3)2 = 12 + 2(1)(c 3) + (c 3)2 = 1 + 2 c 3 + c 6 Holt Algebra 1 Simplify.

7 -8 Special Products of Binomials You can use the FOIL method to find products in the form of (a – b)2. F L (a – b)2 = (a – b) = a 2 – ab + b 2 I O = a 2 – 2 ab + b 2 A trinomial of the form a 2 – ab + b 2 is also a perfect-square trinomial because it is the result of squaring the binomial (a – b). Holt Algebra 1

7 -8 Special Products of Binomials Example 2: Finding Products in the Form (a – b)2 Multiply. A. (x – 6)2 (a – b)2 = a 2 – 2 ab + b 2 (x – 6)2 = x 2 – 2 x(6) + (6)2 = x 2 – 12 x + 36 Use the rule for (a – b)2. Identify a and b: a = x and b = 6. Simplify. B. (4 m – 10)2 Use the rule for (a – b)2. Identify a and b: a = 4 m (a – b)2 = a 2 – 2 ab + b 2 and b = 10. (4 m – 10)2 = (4 m)2 – 2(4 m)(10) + (10)2 = 16 m 2 – 80 m + 100 Holt Algebra 1 Simplify.

7 -8 Special Products of Binomials Example 2: Finding Products in the Form (a – b)2 Multiply. C. (2 x – 5 y)2 Use the rule for (a – b)2 = a 2 – 2 ab + b 2 Identify a and b: a = 2 x and b = 5 y. 2 2 2 (2 x – 5 y) = (2 x) – 2(2 x)(5 y) + (5 y) = 2 x 2 – 20 xy +25 y 2 D. (7 – r 3)2 (a – b)2 = a 2 – 2 ab + b 2 Simplify. Use the rule for (a – b)2. Identify a and b: a = 7 and 3. b = r 2 (7 – r 3)2 = 72 – 2(7)(r 3) + (r 3) Simplify. = 49 – 14 r 3 + r 6 Holt Algebra 1

7 -8 Special Products of Binomials Check It Out! Example 2 Multiply. a. (x – 7)2 (a – b)2 = a 2 – 2 ab + b 2 (x – 7)2 = x 2 – 2(x)(7) + (7)2 = x 2 – 14 x + 49 b. (3 b – 2 c)2 Use the rule for (a – b)2. Identify a and b: a = x and b = 7. Simplify. Use the rule for (a – b)2 = a 2 – 2 ab + b 2 Identify a and b: a = 3 b and b = 2 c. 2 2 2 (3 b – 2 c) = (3 b) – 2(3 b)(2 c) + (2 c) = 9 b 2 – 12 bc + 4 c 2 Holt Algebra 1 Simplify.

7 -8 Special Products of Binomials Check It Out! Example 2 c Multiply. (a 2 – 4)2 Use the rule for (a – b)2 = a 2 – 2 ab + b 2 Identify a and b: a = a 2 (a 2 – 4)2 = (a 2)2 – 2(a 2)(4) + (4)2 and b = 4. = a 4 – 8 a 2 + 16 Holt Algebra 1 Simplify.

7 -8 Special Products of Binomials You can use an area model to see that (a + b) = a 2 – b 2. The new arrange. Then remove the ment is a rectangle smaller rectangle with length a + b and on the bottom. Turn it and slide it width a – b. Its area is (a + b)(a – b). up next to the top rectangle. So (a + b)(a – b) = a 2 – b 2. A binomial of the form a 2 – b 2 is called a difference of two squares. Begin with a square with area a 2. Remove a square with area b 2. The area of the new figure is a 2 – b 2. Holt Algebra 1

7 -8 Special Products of Binomials Example 3: Finding Products in the Form (a + b)(a – b) Multiply. A. (x + 4)(x – 4) (a + b)(a – b) = a 2 – b 2 (x + 4)(x – 4) = x 2 – 42 = x 2 – 16 B. (p 2 + 8 q)(p 2 – 8 q) Use the rule for (a + b)(a – b). Identify a and b: a = x and b = 4. Simplify. Use the rule for (a + b)(a – b) = a 2 – b 2 Identify a and b: a = p 2 and b = 8 q. (p 2 + 8 q)(p 2 – 8 q) = p 4 – (8 q)2 = p 4 – 64 q 2 Simplify. Holt Algebra 1

7 -8 Special Products of Binomials Example 3: Finding Products in the Form (a + b)(a – b) Multiply. C. (10 + b)(10 – b) Use the rule for (a + b)(a – b) = a 2 – b 2 (10 + b)(10 – b) = 102 – b 2 = 100 – b 2 Holt Algebra 1 Identify a and b: a = 10 and b = b. Simplify.

7 -8 Special Products of Binomials Check It Out! Example 3 Multiply. a. (x + 8)(x – 8) (a + b)(a – b) = a 2 – b 2 (x + 8)(x – 8) = x 2 – 82 = x 2 – 64 b. (3 + 2 y 2)(3 – 2 y 2) (a + b)(a – b) = a 2 – b 2 (3 + 2 y 2)(3 – 2 y 2) = 32 – (2 y 2)2 = 9 – 4 y 4 Holt Algebra 1 Use the rule for (a + b)(a – b). Identify a and b: a = x and b = 8. Simplify. Use the rule for (a + b)(a – b). Identify a and b: a = 3 and b = 2 y 2. Simplify.

7 -8 Special Products of Binomials Check It Out! Example 3 Multiply. c. (9 + r)(9 – r) (a + b)(a – b) = a 2 – b 2 (9 + r)(9 – r) = 92 – r 2 = 81 – r 2 Holt Algebra 1 Use the rule for (a + b)(a – b). Identify a and b: a = 9 and b = r. Simplify.

7 -8 Special Products of Binomials Example 4: Problem-Solving Application Write a polynomial that represents the area of the yard around the pool shown below. Holt Algebra 1

7 -8 Special Products of Binomials Example 4 Continued 1 Understand the Problem The answer will be an expression that shows the area of the yard less the area of the pool. List important information: • The yard is a square with a side length of x + 5. • The pool has side lengths of x + 2 and x – 2. Holt Algebra 1

7 -8 Special Products of Binomials Example 4 Continued 2 Make a Plan The area of the yard is (x + 5)2 less the area of the pool. The area of the pool is (x + 2)(x – 2). You can subtract the area of the pool from the yard to find the area of the yard surrounding the pool. Holt Algebra 1

7 -8 Special Products of Binomials Example 4 Continued 3 Solve Step 1 Find the total area. (x +5)2 = x 2 + 2(x)(5) + 52 = x 2 + 10 x + 25 Use the rule for (a + b)2: a = x and b = 5. Step 2 Find the area of the pool. (x + 2)(x – 2) = x 2 – 2 x + 2 x – 4 Use the rule for (a + b)(a – b): a = x 2 – 4 and b = 2. Holt Algebra 1

7 -8 Special Products of Binomials Example 4 Continued 3 Solve Step 3 Find the area of the yard. Area of yard = a total area – area of pool = x 2 + 10 x + 25 – (x 2 – 4) Identify like = + 10 x + 25 – +4 terms. = (x 2 – x 2) + 10 x + ( 25 + 4) Group like terms = 10 x + 29 together x 2 The area of the yard is 10 x + 29. Holt Algebra 1

7 -8 Special Products of Binomials Example 4 Continued 4 Look Back Suppose that x = 20. Then the total area in the back yard would be 252 or 625. The area of the pool would be 22 18 or 396. The area of the yard around the pool would be 625 – 396 = 229. According to the solution, the area of the pool is 10 x + 29. If x = 20, then 10 x +29 = 10(20) + 29 = 229. Holt Algebra 1

7 -8 Special Products of Binomials Remember! To subtract a polynomial, add the opposite of each term. Holt Algebra 1

7 -8 Special Products of Binomials Check It Out! Example 4 Write an expression that represents the area of the swimming pool. Holt Algebra 1

7 -8 Special Products of Binomials Check It Out! Example 4 Continued 1 Understand the Problem The answer will be an expression that shows the area of the two rectangles combined. List important information: • The upper rectangle has side lengths of 5 + x and 5 – x. • The lower rectangle is a square with side length of x. Holt Algebra 1

7 -8 Special Products of Binomials Check It Out! Example 4 Continued 2 Make a Plan The area of the upper rectangle is (5 + x)(5 – x). The area of the lower square is x 2. Added together they give the total area of the pool. Holt Algebra 1

7 -8 Special Products of Binomials Check It Out! Example 4 Continued 3 Solve Step 1 Find the area of the upper rectangle. (5 + x)(5 – x) = 25 – 5 x + 5 x – x 2 Use the rule for (a + b) (a – b): a = 5 and b = x. = –x 2 + 25 Step 2 Find the area lower square. = x x = x 2 Holt Algebra 1

7 -8 Special Products of Binomials Check It Out! Example 4 Continued 3 Solve Step 3 Find the area of the pool. Area of pool = rectangle area + square area a = = –x 2 + 25 + x 2 = (x 2 – x 2) + 25 = 25 The area of the pool is 25. Holt Algebra 1 + x 2 Identify like terms. Group like terms together

7 -8 Special Products of Binomials Check It Out! Example 4 Continued 4 Look Back Suppose that x = 2. Then the area of the upper rectangle would be 21. The area of the lower square would be 4. The area of the pool would be 21 + 4 = 25. According to the solution, the area of the pool is 25. Holt Algebra 1

7 -8 Special Products of Binomials Holt Algebra 1

7 -8 Special Products of Binomials Lesson Quiz: Part I Multiply. 1. (x + 7)2 x 2 + 14 x + 49 2. (x – 2)2 x 2 – 4 x + 4 3. (5 x + 2 y)2 25 x 2 + 20 xy + 4 y 2 4. (2 x – 9 y)2 4 x 2 – 36 xy + 81 y 2 5. (4 x + 5 y)(4 x – 5 y) 16 x 2 – 25 y 2 6. (m 2 + 2 n)(m 2 – 2 n) Holt Algebra 1 m 4 – 4 n 2

7 -8 Special Products of Binomials Lesson Quiz: Part II 7. Write a polynomial that represents the shaded area of the figure below. x+6 x– 7 x– 6 x– 7 14 x – 85 Holt Algebra 1