Lesson 5, Topic 2
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# Complete Factorisation

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To factorize an expression completely, take the H.C.F outside the bracket and then divide each term with the H.C.F.

Find the H.C.F of the following and then factorise;

a. 6x and 3y

$$\scriptsize 6x = 2 \: \times \: 3 \: \times \: x$$

$$\scriptsize 3y = 3 \: \times \: y$$

H.C.F  =  3

Take the H.C.F outside the bracket and then divide each term with the HCF

∴$$\normalsize \frac{6x}{3} \scriptsize =2x \: and \: \normalsize \frac{3y}{3} \scriptsize = y$$

6x and 3y  =  3(2x and y)

b. 10xy and  20xy2

10xy = $$\scriptsize 2 \: \times \: 5 \: \times \: x \: \times \: y$$

20xy2  = $$\scriptsize 2 \: \times \: 2 \: \times \: 5 \: \times \: x \: \times \: y\: \times \: y$$

H.C.F = $$\scriptsize 2 \: \times \: 5 \: \times \: x \: \times \: y = 10xy$$

∴$$\normalsize \frac{10xy}{10xy} \scriptsize =1 \: and \: \normalsize \frac{20xy^2}{10xy} \scriptsize = 2y$$

10xy and  20xy2 = 10xy(1 and 2y)

c. 12y + 8y

Factors of 12y = 2 × 2 × 3 × y

Factors of 8y2 = 2 × 2 × 2 × y × y

H.C.F = 2 × 2 × y = 4y

∴$$\normalsize \frac{12y}{4y} \scriptsize = 3 \: and \: \normalsize \frac{8y^2}{4y} \scriptsize = 2y$$

= 4y  ( 3 + 2y)

d. 36ab  + 12ab2

36ab  =  2 × 2 × 3 × 3 × a × b

12ab2 = 2 × 2 × 3 × a × b × b

HCF =  2 × 2 × 3 × a × b = 12ab

∴$$\normalsize \frac{36ab}{12ab} \scriptsize = 3 \: and \: \normalsize \frac{12ab^2}{12ab} \scriptsize = b$$

= 12ab (3 + b)

e. 12p2 q – 9pq

12p2 q = 2 × 2 × 3 × p × p × q

9pq = 3 × 3 × p × q

HCF  =  3 × p × q =  3pq

∴$$\normalsize \frac{12p^2q}{3pq} \scriptsize = 4p \: and \: \normalsize \frac{9pq}{3pq} \scriptsize = 3$$

= 3pq(4p – 3)

#### Responses

1. Well simplified and explanatory. Thank you

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