Chapter55. Algebraic expansion and simplification

Transcription

1 Chpter55 Algebric expnsion nd simplifiction Contents: A The distributive lw B The product ( + b)(c + d) C Difference of two squres D Perfect squres expnsion E Further expnsion F The binomil expnsion

2 88 ALGEBRAIC EXPANSION AND SIMPLIFICATION (Chpter 5) Opening problem Anton thinks tht to find the squre of the sum of two numbers, you cn just squre ech of the numbers, then dd the results. Things to think bout: Does (5 + 3) 2 = ? b Cn you explin why Anton is incorrect? A THE DISTRIBUTIVE LAW When we write rel life problems in terms of lgebr, we often obtin expressions contining brckets. To solve equtions contining these expressions, we my need to expnd the brckets nd simplify the result. The distributive lw is used to expnd expressions such s 4(x +3). Investigtion 1 Wht to do: 1 Copy nd complete: 2 (5 + 3) =2 8 = :::::: b =10+6 = :::::: The distributive lw 2 Evlute the following expressions: 7 (1 + 4) b c 6 (2 + 8) d e 5 (3 + 4) f Comment on your results. You should hve found tht we cn remove the brckets. We multiply the term in front of the brckets by ech term inside the brckets. 2 (5 + 3) = Since lgebric vribles re used to represent numbers, this rule pplies to vribles s well s numbers. We cll it the distributive lw: To remove set of brckets from n expression, we multiply ech term inside the brckets by the term in front of them. We then dd the results. (b + c) =b + c

3 ALGEBRAIC EXPANSION AND SIMPLIFICATION (Chpter 5) 89 GEOMETRIC DEMONSTRATION b c b+c Exmple 1 The re of the lrge rectngle is (b + c). However, this could lso be found by dding the res of the two smll rectngles. This is b + c. So, (b + c) = b + c. fequting resg Expnd the following: 3(4x + 1) b 2x(5 2x) c 2x(x 3) 3(4x +1) = 3 4x =12x +3 b 2x(5 2x) =2x(5 + 2x) = 2x 5+2x 2x =10x 4x 2 c 2x(x 3) = 2x(x + 3) = 2x x + 2x 3 = 2x 2 +6x EXERCISE 5A 1 Expnd nd simplify: 2(x +1) b 5(x +2) c 4(x 6) d 7(x 1) e 3(2 + b) f 6(x +3y) g 8( b) h 4(m 2 n 2 ) i 3(x +2) j 4(2 1) k ( +5) l 3b(b 4) m (y +2) n 5(x 3) o (2 b) p 4( b) q x(x + y) r x( x y) s (2 1) t 3x(7 4x) 2 Expnd nd simplify: 4(x + y +5) b 2(3 5b +1) c 3x(x 2y +3) d 2( 3b +5c) e 6x(x 2 +5x 10) f 4n(n 2 2n 8) Exmple 2 Expnd nd simplify: 2(3x 1) + 3(5 x) b x(2x 1) 2x(5 x) 2(3x 1) + 3(5 x) = 2 3x x =6x x =3x +13 In b, the minus sign in front of 2x ffects both terms inside the following brcket. b x(2x 1) 2x(5 x) = x 2x + x 1+ 2x 5+ 2x x =2x 2 x 10x +2x 2 =4x 2 11x

4 90 ALGEBRAIC EXPANSION AND SIMPLIFICATION (Chpter 5) 3 Expnd nd simplify: 2+3(x +1) b 15 2(x +5) c 4(x 1) + 7 d 6(x 2) 8x e ( 2) + 2 f 3m(4 m)+m 2 g 3(b )+4 2 h 5b 2b(b 2) i 6 2 3( +4) 4 Expnd nd simplify: 2(x 3) + 3(x +4) b 4b +( b) c 4b ( b) d 3(x + 2) + 5(4 x) e 6(m 2) 3(2m +1) f 7n 5(3 2n) g 5(y x)+6(x y) h ( +2)+5( 3) i x(x +5) 3(x 4) j 2 + ( +3) k 2 ( 1) l x(x + y) y(x + y) m 3(x 6) (2 x) n 4(3x 2) (3x +1) o 2x(x 5) 3x(2 x) B THE PRODUCT ( + b)(c + d) Consider the product ( + b)(c + d). It hs two fctors, ( + b) nd (c + d). We cn evlute this product by using the distributive lw severl times. ( + b)(c + d) =(c + d)+b(c + d) = c + d + bc + bd This is sometimes So, ( + b)(c + d) =c + d + bc + bd clled the FOIL rule. The finl result contins four terms: c is the product of the First terms of ech brcket. d is the product of the Outer terms of ech brcket. bc is the product of the Inner terms of ech brcket. bd is the product of the Lst terms of ech brcket. Exmple 3 Expnd nd simplify: (x + 3)(x +2) b (2x + 1)(3x 2) (x + 3)(x +2) = x x + x 2+3 x = x 2 +2x +3x +6 = x 2 +5x +6 With prctice you should not need the second line of working. b (2x + 1)(3x 2) = 2x 3x + 2x 2+1 3x =6x 2 4x +3x 2 =6x 2 x 2

5 ALGEBRAIC EXPANSION AND SIMPLIFICATION (Chpter 5) 91 EXERCISE 5B c d 1 Consider the figure longside. Give n expression for the re of: rectngle 1 b rectngle 2 c rectngle 3 d rectngle 4 e the overll rectngle. Wht cn you conclude? b 1 3 c+d 2 4 +b 2 Use the rule ( + b)(c + d) =c + d + bc + bd to expnd nd simplify: (x + 2)(x +5) b (x + 4)(x 6) c (x 3)(x +7) d (x + 5)(x 5) e (x 8)(x +2) f (3x + 2)(4x +1) g (1 3x)(2x +1) h (6 x)(2x +5) i (4x 3)(1 + 3x) j (4 x)(4 + 5x) k (8 x)(2x +5) l (3x + 2)(3x +2) Exmple 4 Expnd nd simplify: (x + 3)(x 3) b (3x 5)(3x +5) Wht do you notice bout the two middle terms? (x + 3)(x 3) = x 2 3x +3x 9 = x 2 9 b (3x 5)(3x +5) =9x 2 +15x 15x 25 =9x Expnd nd simplify: (x + 2)(x 2) b ( + 4)( 4) c (6 + x)(6 x) d (3x + 1)(3x 1) e (5 + 2)(5 2) f (7 3)(7 +3) Exmple 5 Expnd nd simplify: (3x +1) 2 b (2x 3) 2 Wht do you notice bout the two middle terms? (3x +1) 2 =(3x + 1)(3x +1) =9x 2 +3x +3x +1 =9x 2 +6x +1 b (2x 3) 2 =(2x 3)(2x 3) =4x 2 6x 6x +9 =4x 2 12x +9 4 Expnd nd simplify: (x +3) 2 b (x 4) 2 c (1 x) 2 d (2 + x) 2 e (2x 1) 2 f (2 3x) 2 g (3 5x) 2 h (2x y) 2

6 92 ALGEBRAIC EXPANSION AND SIMPLIFICATION (Chpter 5) Exmple 6 Expnd nd simplify: (x + 2)(x 3) + 5(x +4) (x + 2)(x 3) + 5(x +4)=x 2 3x +2x 6+5x +20 = x 2 +4x Expnd nd simplify: (x + 5)(x +1)+2(x 2) b 4(x +2)+(x 3)(x +6) c (2 + 5)( 6) + ( +7) d (3x + 5)(2x 3) x(x +3) e (2x + 3)(x +2)+(x + 6)(x 5) f (y + 7)(y 4) (y + 1)(y +3) g (k + 2)(k 5) (2k + 1)(k 3) h (x + 2)(x 2) + (x + 6)(x 6) C DIFFERENCE OF TWO SQUARES Consider the product ( + b)( b). Using the FOIL rule to expnd this product, ( + b)( b) = 2 b {z + b } b2 = 2 b 2 the middle two terms dd to zero Thus, ( + b)( b) = 2 b 2 This is clled the difference of two squres expnsion, becuse the expression on the right hnd side is the difference between the two squres 2 nd b 2. GEOMETRIC DEMONSTRATION In the figure longside, the shded re = re of lrge squre re of smll squre = 2 b 2 -b -b b -b If the rectngle on the right hnd side is rotted nd plced on top of the remining shded re, we form new rectngle. The shded re =( + b)( b) ) ( + b)( b) = 2 b 2 fequting resg b DEMO -b

7 ALGEBRAIC EXPANSION AND SIMPLIFICATION (Chpter 5) 93 Exmple 7 Expnd nd simplify: (x + 5)(x 5) b (3 y)(3 + y) (x + 5)(x 5) = x = x 2 25 b (3 y)(3 + y) =3 2 y 2 =9 y 2 EXERCISE 5C 1 Expnd nd simplify using the rule ( + b)( b) = 2 b 2 : (x + 3)(x 3) b (x 3)(x +3) c (3 + x)(3 x) d (3 x)(3 + x) e (x + 2)(x 2) f (2 x)(2 + x) g (x + 6)(x 6) h ( + 4)( 4) i (b 1)(b +1) j (p q)(p + q) k (5 + n)(5 n) l (7 y)(7 + y) Exmple 8 Expnd nd simplify: (2x 3)(2x +3) b (5 3y)(5 + 3y) (2x 3)(2x +3) =(2x) =4x 2 9 b (5 3y)(5 + 3y) =5 2 (3y) 2 =25 9y 2 2 Expnd nd simplify: (2x + 1)(2x 1) b (5x + 2)(5x 2) c (4 + 3)(4 3) d (3b + 5)(3b 5) e (4x + 1)(4x 1) f (1 4x)(1 + 4x) g (7 2y)(7 + 2y) h (3 2x)(3 + 2x) i (3x + 2)(3x 2) Exmple 9 Expnd nd simplify: (3x +4y)(3x 4y) (3x +4y)(3x 4y) =(3x) 2 (4y) 2 =9x 2 16y 2 3 Expnd nd simplify: (3 + b)(3 b) b ( 3b)( +3b) c (6x + y)(6x y) d (5x +2y)(5x 2y) e (4x +5y)(4x 5y) f (2x 7y)(2x +7y)

8 94 ALGEBRAIC EXPANSION AND SIMPLIFICATION (Chpter 5) Investigtion 2 The product of three consecutive integers Con ws trying to multiply without clcultor. Aimee told him to cube the middle integer nd then subtrct the middle integer to get the nswer. Wht to do: 1 Find using clcultor. 2 Find using clcultor. Does Aimee s rule seem to work? 3 Check tht Aimee s rule works for the following products: b c Let the middle integer be x, so the other integers must be (x 1) nd (x +1). Find the product (x 1) x (x +1) by expnding nd simplifying. Hve you proved Aimee s rule? Hint: Use the difference of two squres expnsion. D PERFECT SQUARES EXPANSION ( + b) 2 nd ( b) 2 re clled perfect squres. Now ( + b) 2 =( + b)( + b) = 2 + b + b + b 2 fusing FOIL g = 2 +2b + b 2 Notice tht the middle two terms re identicl. Thus, we cn stte the perfect squre expnsion rule: ( + b) 2 = 2 +2b + b 2 We cn remember this rule s follows: Step 1: Squre the first term. Step 2: Add twice the product of the first nd lst terms. Step 3: Add on the squre of the lst term. Notice tht ( b) 2 =( +( b)) 2 We cn hence stte: = 2 +2( b)+( b) 2 = 2 2b + b 2 ( b) 2 = 2 2b + b 2 Once gin, we hve the squre of the first term, twice the product of the first nd lst terms, nd the squre of the lst term.

9 ALGEBRAIC EXPANSION AND SIMPLIFICATION (Chpter 5) 95 Exmple 10 Expnd nd simplify: (x +3) 2 b (x 5) 2 (x +3) 2 = x 2 +2 x = x 2 +6x +9 b (x 5) 2 = x 2 +2 x ( 5) + ( 5) 2 = x 2 10x +25 EXERCISE 5D 1 Consider the figure longside. Give n expression for the re of: squre 1 b rectngle 2 c rectngle 3 d squre 4 e the overll squre. Wht cn you conclude? b 1 3 b 2 4 +b 2 Consider the Opening Problem on pge 88. Anton thinks tht (5 + 3) 2 = By evluting ech side of the eqution, show tht Anton is incorrect. b Check tht (5 + 3) 2 = Use the rule ( + b) 2 = 2 +2b + b 2 to expnd nd simplify: (x +2) 2 b (x +5) 2 c ( +4) 2 d (y +6) 2 e (1 + b) 2 f (2 + x) 2 4 Expnd nd simplify: +b (x 4) 2 b (x 2) 2 c ( 7) 2 d (b 8) 2 e (3 x) 2 f (5 y) 2 Exmple 11 Expnd nd simplify using the perfect squre expnsion rule: (5x +1) 2 b (4 3x) 2 (5x +1) 2 =(5x) x =25x 2 +10x +1 b (4 3x) 2 = ( 3x)+( 3x) 2 =16 24x +9x 2 5 Expnd nd simplify: (2x +3) 2 b (4 1) 2 c (3y +5) 2 d (3 4) 2 e (2x 7) 2 f (8 + 3) 2 g (2 + 5b) 2 h (6 5x) 2 i (4 5y) 2

10 96 ALGEBRAIC EXPANSION AND SIMPLIFICATION (Chpter 5) Exmple 12 Expnd nd simplify: (2x 2 +3) 2 b 5 (x +2) 2 (2x 2 +3) 2 =(2x 2 ) x =4x 4 +12x 2 +9 b 5 (x +2) 2 =5 [x 2 +4x +4] =5 x 2 4x 4 =1 x 2 4x The squre brckets in the second line remind us tht the minus in front of the brckets ffects ll terms within them. 6 Expnd nd simplify: (x 2 +3) 2 b (y 2 2) 2 c (2 2 +3) 2 d (1 4x 2 ) 2 e ( 2 + b 2 ) 2 f ( 2 b 2 ) 2 7 Expnd nd simplify: 2x +1 (x +2) 2 b 3x 2+(x +5) 2 c (x + 4)(x 4) + (x +1) 2 d (x + 4)(x 4) (x +1) 2 e (1 3x) 2 +(x 2)(x +3) f (3x +2) 2 (x + 1)(x 5) g (5x + 2)(5x 2) (x +2) 2 h (5x + 4)(x 3) (x 1) 2 i (3 x) 2 +(x +2) 2 j (3 x) 2 (x 2) 2 E FURTHER EXPANSION In this section we expnd more complicted expressions by repeted use of the expnsion lws. Consider the expnsion of ( + b)(c + d + e). Now ( + b)(c + d + e) =( + b)c +( + b)d +( + b)e = c + bc + d + bd + e + be fcompre with (c + d + e) = c + d + eg Notice tht there re 6 terms in this expnsion. Ech term within the first brcket is multiplied by ech term in the second. 2 terms in the first brcket 3 terms in the second brcket! 6 terms in the expnsion. Exmple 13 Expnd nd simplify: (2x + 3)(x 2 +4x +5) (2x + 3)(x 2 +4x +5) =2x 3 +8x 2 +10x fll terms of 2nd brcket 2xg +3x 2 +12x +15 fll terms of 2nd brcket 3g =2x 3 +11x 2 +22x +15 fcollecting like termsg Ech term of the first brcket is multiplied by ech term of the second brcket.

11 ALGEBRAIC EXPANSION AND SIMPLIFICATION (Chpter 5) 97 EXERCISE 5E 1 Expnd nd simplify: (x + 2)(x 2 + x +1) b (x + 3)(x 2 + x 2) c (x + 4)(x 2 x +2) d (x + 6)(x 2 x 1) e (3x + 1)(x 2 + x +5) f (5x 2)(x 2 x +4) g (x + 3)(2x 2 + x 2) h (2x 1)(3x 2 x +4) Exmple 14 Expnd nd simplify: (x +2) 3 (x +2) 3 =(x +2) (x +2) 2 =(x + 2)(x 2 +4x +4) fperfect squre expnsiong 2 Expnd nd simplify: = x 3 +4x 2 +4x fll terms in 2nd brcket xg +2x 2 +8x +8 fll terms in 2nd brcket 2g = x 3 +6x 2 +12x +8 fcollecting like termsg (x +1) 3 b (x +3) 3 c (x 1) 3 d (x 2) 3 e (2x +1) 3 f (2x 3) 3 Exmple 15 Expnd nd simplify: x(x + 1)(x +2) b (x + 1)(x 2)(x +2) x(x + 1)(x +2) =(x 2 + x)(x +2) fll terms in first brcket xg = x 3 +2x 2 + x 2 +2x fexpnding remining fctorsg = x 3 +3x 2 +2x fcollecting like termsg b (x + 1)(x 2)(x +2) =(x + 1)(x 2 4) fdifference of two squresg = x 3 4x + x 2 4 fexpnding fctorsg = x 3 + x 2 4x 4 Alwys look for wys to mke your expnsions simpler. In b we cn use the difference of two squres. 3 Expnd nd simplify: x(x + 2)(x +4) b x(x 5)(x +3) c x(x 4)(x 1) d 2x(x + 3)(x 2) e 2x(x + 1)(7 x) f x(x + 3)(x 6) g 5x(2x 1)(x +3) h x(1 4x)(x 3) i 3x 2 (x +2) 2 4 Expnd nd simplify: (x + 1)(x + 2)(x +4) b (x 2)(x + 1)(x 3) c (x 5)(x + 2)(x 1) d (2x 1)(x 4)(x 1) e (3x + 1)(x + 2)(x +3) f (4x 1)(4x + 1)(x +4) g (1 x)(2 3x)(x +2) h (x 5)(1 x)(5x +2)

12 98 ALGEBRAIC EXPANSION AND SIMPLIFICATION (Chpter 5) 5 Expnd nd simply: i x 1)(x 2 + x +1 ii x 1)(x 4 + x 3 + x 2 + x +1 iii x 1)(x 6 + x 5 + x 4 + x 3 + x 2 + x +1 b Predict the simplifiction of x 1)(x n + x n 1 + x n 2 + :::: + x 2 + x +1 where n is positive even integer. F THE BINOMIAL EXPANSION + b is clled binomil becuse it contins two terms. We cn use the FOIL rule to find the expnsion of powers of + b. We cll these binomil expnsions. For exmple, erlier in the chpter we found tht ( + b) 2 = 2 +2b + b 2. Now ( + b) 3 =( + b) 2 ( + b) =( 2 +2b + b 2 )( + b) = b + b b +2b 2 + b 3 = b +3b 2 + b 3 So, ( + b) 3 = b +3b 2 + b 3. Exmple 16 Expnd nd simplify using the rule ( + b) 3 = b +3b 2 + b 3 : (x +2) 3 b (2x 1) 3 We substitute = x nd b =2 ) (x +2) 3 = x 3 +3 x x = x 3 +6x 2 +12x +8 b We substitute =(2x) nd b =( 1) ) (2x 1) 3 =(2x) 3 +3 (2x) 2 ( 1) + 3 (2x) ( 1) 2 +( 1) 3 =8x 3 12x 2 +6x 1 We use brckets to ssist our substitution. EXERCISE 5F 1 Use the binomil expnsion for ( + b) 3 to expnd nd simplify: (x +1) 3 b (x +5) 3 c (3 + ) 3 d (x 1) 3 e (x 2) 3 f (x 3) 3 g (3x +2) 3 h (2x +3y) 3 2 Copy nd complete the following expnsion, giving your nswer in simplest form: ( + b) 4 =( + b)( + b) 3 =( + b)( b +3b 2 + b 3 ) = ::::::

13 ALGEBRAIC EXPANSION AND SIMPLIFICATION (Chpter 5) 99 3 Use the binomil expnsion ( + b) 4 = b +6 2 b 2 +4b 3 + b 4 to expnd nd simplify: (x +1) 4 b (y +2) 4 c (3 + ) 4 d (b +4) 4 e (x 1) 4 f (y 2) 4 g (3 ) 4 h (b 4) 4 4 Find the binomil expnsion of ( + b) 5 by considering ( + b)( + b) 4. Hence, write down the binomil expnsion for ( b) 5. Review set 5 1 Expnd nd simplify: 3(x 4) b 5(2x 3) c 7+4(x 2) d x(1 x) e 3 2(x +4) f 4(2x 3) (x +1) 2 Expnd nd simplify: (x + 4)(x +7) b (x + 8)(x 2) c (1 x)(5x +6) d (2 3)( 9) e (x + 3)(x +2)+3(x 4) f (n 3)(n +5) (n 2)(n +3) 3 Expnd nd simplify: (x +4) 2 b (x + 9)(x 9) c (x 10) 2 d (2x +5) 2 e (3 4x) 2 f x(4 x)(4 + x) g (3 5x) 2 h x(2x 1) 2 4 Expnd nd simplify: (x + 4)(x 2 + x 7) b (4x 3)(x 2 x 5) c 3x +5 (x 3) 2 d (x + 3)(2x + 1)(x 4) 5 Expnd nd simplify: (x +2) 3 b (3x +1) 3 c (1 x) 4 Prctice test 5A Click on the icon to obtin this printble test. Multiple Choice PRINTABLE TEST Prctice test 5B Short response 1 Expnd nd simplify: 3(x +4) b x(y +3) c 2( 5) 2 Expnd nd simplify: (x + 9)(x 4) b (2x 7)(x +6) c (2 3)( 4) 3 Expnd nd simplify: (y +3) 2 b (3x +2) 2 c (4 b) 2

14 100 ALGEBRAIC EXPANSION AND SIMPLIFICATION (Chpter 5) 4 Expnd nd simplify: (5 x)(5 + x) b (3y + 4)(3y 4) c (6 +5b)(6 5b) 5 Expnd nd simplify: 4(x +2) (x +1) 2 b 5(y 3) + (y + 5)(y 5) 6 Using the rule ( + b) 3 = b +3b 2 + b 3, expnd nd simplify: (x 2) 3 b (x +3) 3 7 Expnd nd simplify: (x +4) 2 (x 2) b (x + 2)(x 3) 2 8 Expnd nd simplify: (x 2 3x + 4)(2 x) b (x 2 +4x 5)(x 4) 9 Expnd nd simplify (2x 9)(2x + 9)(x 5). 10 Expnd nd simplify (x +3) 4 using the rule ( + b) 4 = b +6 2 b 2 +4b 3 + b 4. Prctice test 5C Extended response 1 Emm receives lrge box in the mil. The box is 10 cm longer thn it is wide, nd the height is 6 cm more thn the width. Let x cm represent the width of the box. Write n expression for the i length ii height of the box. b The re of the bse of the box is given by the formul Are = length width. Write n expression for the re of the bse of the box, in terms of x. Expnd nd simplify your nswer. c The volume of the box is given by the formul V = length width height. Write n expression for the volume of the box in terms of x. Expnd nd simplify your nswer. 2 The squre longside hs sides divided into lengths nd b. 1 2 Write n expression for the overll re of the squre. b Find the re of ech section of the squre. 3 4 b c Hence write the expnsion of ( + b) 2. b 3 Use ( + b) 4 = b +6 2 b 2 +4b 3 + b 4 to expnd nd simplify (x +2) 4. b Hence find the expnsion of (x +2) 5. c By substituting x =1 into the expnsion from b, find the vlue of 3 5.