1. ALGEBRA 1 Lesson 8-4 Warm-Up

2. ALGEBRA 1 “Multiplying Special Cases” (8-4) What are “squares of a binomial”? How do you square a binomial? Square of a Binomials: a binomial squared Example: (a + b) 2 Method 1: To square a binomial, multiply the binomial by itself. Then, use an area model or the FOILing method. Example: Simplify (a + b) 2. Method 2: RULE: To square a binomial, use the following formulas (the square of the first term plus twice the product of the first and second term plus the square of the last term): (a + b) 2 = a 2 + 2ab + b 2 (a - b) 2 = a 2 - 2ab + b 2 S

3. ALGEBRA 1 a. Find (y + 11) 2. = y 2 + 22y + 121Simplify. b. Find (3w – 6) 2. = (3w) 2 + 2(3w)(-6) + (-6) 2 Substitute a with 3w and b with -6 in the rule a 2 + 2ab + b 2. = 9w 2 – 36w + 36Simplify. Multiplying Special Cases LESSON 8-4 Additional Examples = (y) 2 + (2)(y)(11) + (11) 2 Substitute a with y and b with 11 in the rule a 2 + 2ab + b 2.

4. ALGEBRA 1 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, and a guinea pig with two recessive genes (WW) will have white fur. The probability that a guinnea pig will have black or white fur is modeled by the espresssion (B + W) 2. What is the probability of each 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. 1414 1414 BBBWBWWW BBBWBWWW BWBW BWBW Multiplying Special Cases LESSON 8-4 Additional Examples ( B + W ) 2 = ( B ) 2 – 2 ( B )( W ) + ( W ) 2 Square the binomial. Use B for a and W for b in (a + b) 2 = a 2 + 2ab + b 2 1212 1212 1212 1212 1212 1212 1212 1212 = B 2 + BW + W 2 Simplify. 1414 1212 1414

5. ALGEBRA 1 (continued) The expressions B 2 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 model. 1414 1414 1414 1212 1212 Multiplying Special Cases LESSON 8-4 Additional Examples

6. ALGEBRA 1 a. Find 81 2 using mental math. 81 2 = (80 + 1) 2 = (80) 2 + 2(80)(1) + (1) 2 Substitute 80 for a and 1 for b in the rule (a + b) 2 = a 2 + 2ab + b 2. b. Find 59 2 using mental math. 59 2 = (60 – 1) 2 = 60 2 + 2(60)(-1) + (-1) 2 Substitute 80 for a and 1 for b in the rule (a + b) 2 = a 2 + 2ab + b 2. = 3600 – 120 + 1 = 3481Simplify. Multiplying Special Cases LESSON 8-4 Additional Examples = 6400 + 160 + 1 = 6561Simplify.

7. ALGEBRA 1 “Special Cases” (8-4) What is the “difference of squares”? How do you find the “difference of squares”? Difference of Squares: the product of the sum and difference of the same two terms Example: (a + b)(a - b) Rule: To find the difference of squares, use an area model or the FOILing method. The FOILing method is used below. (a + b)(a - b) = a(a) + a(-b) + b(a) + b(-b) = a 2 – ab + ab - b 2 =a 2 - b 2 Notice that the products of the Outside and Inside of FOIL cancel each other out (in other words, equal zero). So, the product of the sum and difference of the same two terms is the difference of their squares. S

8. ALGEBRA 1 Find (p 4 – 8)(p 4 + 8). (p 4 – 8)(p 4 + 8) = (p 4 ) 2 – (8) 2 Use (a + b)(a - b) = a 2 - b 2 where a = p 4 and b = 8. = p 8 – 64Simplify. Multiplying Special Cases LESSON 8-4 Additional Examples

9. ALGEBRA 1 Find 43 37. 43 37 = (40 + 3)(40 – 3)Express each factor using 40 and 3. This models (a + b)(a - b) = a 2 - b 2 where a = 40 and b = 3. = (40) 2 – (3) 2 Use (a + b)(a - b) = a 2 - b 2 = 1600 – 9 = 1591Simplify. Multiplying Special Cases LESSON 8-4 Additional Examples

10. ALGEBRA 1 Find each square. 1.(y + 9) 2 2.(2h – 7) 2 3.41 2 4.29 2 5.Find (p 3 – 7)(p 3 + 7).6.Find 32 28. y 2 + 18y + 814h 2 – 28h + 49 1681841 p 6 – 49896 Multiplying Special Cases LESSON 8-4 Lesson Quiz