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3.2.1 Algebraic Expressions II, PT3 Focus Practice

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3.2.1 Algebraic Expressions II, PT3 Focus Practice
 
Question 1:
Calculate the product of each of the following pairs of algebraic terms.
(a) 2rs × 4r2s3t
(b) 1 2 5 a 2 b 2 × 5 14 a b 3 c 2 (c) 1 1 2 x y × 4 9 x 2 z  

Solution:
(a)
2rs × 4r2s3t = 2 × r × s× 4 × r × r × s × s × s × t = 8 r3s4t

(b) 1 2 5 a 2 b 2 × 5 14 a b 3 c 2 = 7 1 5 1 a a b b × 5 1 14 2 a b b b c c = 1 2 a 3 b 5 c 2

(c) 1 1 2 x y × 4 9 x 2 z = 3 1 2 1 x y × 4 2 9 3 x x z = 2 3 x 3 y z


Question 2:
Find the quotients of each of the following pairs of algebraic terms.
(a) − 48 a 2 b 3 c 4 − 16 a b 2 c 2  
(b) − 15 e f 3 g 2 ÷ ( − 40 e f 2 g ) (c) 5 a 3 c 2 ÷ 1 2 a c  

Solution:
(a) − 48 a 2 b 3 c 4 − 16 a b 2 c 2 = − 48 3 a × a × b × b × b × c × c × c × c − 16 1 a × b × b × c × c = 3 a c 2
(b) − 15 e f 3 g 2 ÷ ( − 40 e f 2 g ) = − 15 3 e × f × f × f × g × g − 40 8 e × f × f × g = 3 8 f g

(c) 5 a 3 c 2 ÷ 1 2 a c = 5 a 3 2 c 2 × 2 a c = 10 a 2 c


Question 3:
(– 4a2b) ÷ 3b2c2× 9ac =

Solution:
( − 4 a 2 b ) ÷ 3 b 2 c 2 × 9 a c = − 4 a 2 b 3 b 2 1 c 2 1 × 9 3 a c = − 12 a 3 b c



Question 4:
2a2b × 14b3c ÷ 56ab2c2 =

Solution:
2 a 2 b × 14 b 3 c ÷ 56 a b 2 c 2 = 2 a 2 1 b × 14 b 3 1 c 56 2 a b 2 c 2 1 = a b 2 2 c


Question 5:
− 2 3 ( 3 a − 6 b + 3 4 c ) =  

Solution:
− 2 3 ( 3 a − 6 b + 3 4 c ) = − 2 3 × 3 a − 2 3 ( − 6 2 b ) − 2 3 ( 3 4 2 c ) = − 2 a + 4 b − 1 2 c