Suppose we want to factorise,
2 x y + 3 x + 2y + 3
We note that,
2xy & 2y both have 2
And
3x & 3 both have 3
So, we put them together
(2xy + 2y) + (3𝑥 + 3)
So,
2𝑥y + 3𝑥 + 2y + 3 = (𝑥 + 1) (2y + 3)
Factorise 4ab − 5a − 12b + 15
Factorisation by regrouping terms
Example 3 Important
Ex 12.1, 3 (i)
Ex 12.1, 3 (iii)
Ex 12.1, 3 (ii) Important
Ex 12.1, 3 (iv) Important
Ex 12.1, 3 (v) Important
Ex 12.2, 3 (i)
Ex 12.2, 3 (ii) Important
Ex 12.2, 3 (iii)
Ex 12.2, 3 (v) Important
Ex 12.2, 3 (iv) Important
Ex 12.2, 3 (vi)
Ex 12.2, 3 (vii)
Ex 12.2, 3 (viii) Important
Ex 12.2, 3 (ix)
Last updated at April 16, 2024 by Teachoo
Suppose we want to factorise,
2 x y + 3 x + 2y + 3
We note that,
2xy & 2y both have 2
And
3x & 3 both have 3
So, we put them together
(2xy + 2y) + (3𝑥 + 3)
So,
2𝑥y + 3𝑥 + 2y + 3 = (𝑥 + 1) (2y + 3)
(2𝑥y + 2y) + (3𝑥 + 3) Both have 2 & y as common factor Both have 2 & y as common factor = 2y (𝑥 + 1) + 3 (𝑥 + 1) Now, (𝑥 + 1) is common between them both = (𝑥 + 1) (2y + 3) So, 2𝑥y + 3𝑥 + 2y + 3 = (𝑥 + 1) (2y + 3) Factorise 4ab − 5a + 12b + 15 4ab − 5a + 12b + 15 = (4ab − 12b) + (15 − 5a) Both have 4 & b as common factor Both have 3 as common factor = 4b (a − 3) + 5 (3 − a) Writing (3 − a) = −(a − 3) = 4b (a − 3) − 5 (a − 3) Taking (a − 3) common = (a − 3) (4b − 5)