2.2.1 Factorisation and Algebraic Fractions, PT3 Practice


2.2.1 Factorisation and Algebraic Fractions, PT3 Practice

Question 1:
(a)(i) Factorise 18a + 3
(a)(ii) Expand –3 (–y + 5)
(b) Express 5 6 y 3 x 5 12 y as a single fraction in its simplest form.

Solution:
(a)(i) 18+ 3 = 3(6a + 1)
 
(a)(ii) –3 (–+ 5) = 3y – 15
 
(b)
5 6 y 3 x 5 12 y = 5 × 2 6 y × 2 ( 3 x 5 ) 12 y = 10 3 x + 5 12 y = 15 3 x 12 y = 3 ( 5 x ) 4 12 y = 5 x 4 y


Question 2:
(a) Expand:
(i) 3 (–a + c)
(ii) –5 (c)
(b) Factorise 4+ 2
(c) Simplify: 3 x + 6 x 2 4 ÷ x + 2 x 2   

Solution:
(a)(i) 3 (–+ c) = –3a + 3c
 
(a)(ii) –5 (c) = –5a + 5c
 
(c)
3 x + 6 x 2 4 ÷ x + 2 x 2 = 3 ( x + 2 ) ( x + 2 ) ( x 2 ) × x 2 x + 2 = 3 x + 2



Question 3:
(a) Factorise:
(i) 5m + 25
(ii) 7x + 9xy
(b) Simplify: 4 x 12 4 y ÷ x 2 9 y z

Solution:
(a)(i)5m + 25 = 5 (m + 5)
 
(a)(ii)7x + 9xy = x (7 + 9y)

(b) 4 x 12 4 y ÷ x 2 9 y z = 4 ( x 3 ) 4 y × y z ( x + 3 ) ( x 3 ) = z x + 3


Question 4:
  (a) Factorise completely:
  4 – 100n2
(b) Express 4 5 x 7 10 y 15 x  as a single fraction in its simplest form.

Solution:
(a)
4 – 100n2= (2 + 10n)(2 – 10n)
(b)
4 5 x 7 10 y 15 x = 4 × 3 5 x × 3 ( 7 10 y ) 15 x = 12 7 + 10 y 15 x = 5 + 10 y 15 x = 5 ( 1 + 2 y ) 3 15 x = 1 + 2 y 3 x

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