Check sibling questions

Suppose we want to factorise,

Β  2 x y + 3 x + 2y + 3

Β 

We note that,

Β  Β 2xy & 2y both have 2

And

Β  3x & 3 both have 3

Β 

So, we put them together

Β 

Β  (2xy + 2y) + (3π‘₯ + 3)

2.jpg

Β 

So,

Β  2π‘₯y + 3π‘₯ + 2y + 3 = (π‘₯ + 1) (2y + 3)

Β 

Factorise 4ab βˆ’ 5a βˆ’ 12b + 15

Factorisation by regrouping terms - Part 2

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Transcript

(2π‘₯y + 2y) + (3π‘₯ + 3) Both have 2 & y as common factor Both have 2 & y as common factor = 2y (π‘₯ + 1) + 3 (π‘₯ + 1) Now, (π‘₯ + 1) is common between them both = (π‘₯ + 1) (2y + 3) So, 2π‘₯y + 3π‘₯ + 2y + 3 = (π‘₯ + 1) (2y + 3) Factorise 4ab βˆ’ 5a + 12b + 15 4ab βˆ’ 5a + 12b + 15 = (4ab βˆ’ 12b) + (15 βˆ’ 5a) Both have 4 & b as common factor Both have 3 as common factor = 4b (a βˆ’ 3) + 5 (3 βˆ’ a) Writing (3 βˆ’ a) = βˆ’(a βˆ’ 3) = 4b (a βˆ’ 3) βˆ’ 5 (a βˆ’ 3) Taking (a βˆ’ 3) common = (a βˆ’ 3) (4b βˆ’ 5)

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.