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


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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.