# Worksheet on Factorization by Regrouping | Factoring by Regrouping Worksheet

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## How to Solve Factorization Problems by Regrouping?

1. Factorize each of the following by regrouping

(i) a2 + ab + 9a + 9b
(ii) 6ab – 4b + 6 – 9a
(iii) 10xy + 6x + 5y +3
(iv) a3 + a2 + a + 1
(v) m2 – b + mb – m
(vi) a3 – a2b + 5a – 5b
(vii) x (x + 3) – x – 3
(viii) 3mx + 3my – 2nx – 2ny
(ix) a (a + b – c ) – bc
(x) x3 – x2 + xy2 – x2y2

Solution:

(i) Given expression is a2 + ab + 9a + 9b.
Rearrange the terms
a2 + ab + 9a + 9b.
Group the first two terms and last two terms.
The first two terms are a2 + ab and the second two terms are 9a + 9b.
Take a common from the first two terms.
a (a + b)
Take 9 common from the second two terms.
9 (a + b)
a (a + b) + 9 (a + b)
Then, take (a + b) common from the above expression.
(a + b) (a + 9)

The final answer is (a + b) (a + 9).

(ii) Given expression is 6ab – 4b + 6 – 9a.
Rearrange the terms
6ab – 9a – 4b + 6
Group the first two terms and last two terms.
The first two terms are 6ab – 9a and the second two terms are – 4b + 6.
Take 3a common from the first two terms.
3a (2b – 3)
Take -2 common from the second two terms.
-2 (2b  – 3)
3a (2b – 3) – 2 (2b – 3)
Then, take (2b – 3) common from the above expression.
(2b – 3) (3a – 2)

The final answer is (2b – 3) (3a – 2).

(iii) Given expression is 10xy + 6x + 5y +3.
Rearrange the terms
10xy + 5y + 6x +3
Group the first two terms and last two terms.
The first two terms are 10xy + 5y and the second two terms are 6x +3.
Take 5y common from the first two terms.
5y (2x + 1)
Take 3 common from the second two terms.
3 (2x  + 1)
5y (2x + 1) + 3 (2x  + 1)
Then, take (2x  + 1) common from the above expression.
(2x  + 1) (5y + 3)

The final answer is (2x  + 1) (5y + 3).

(iv) Given expression is a3 + a2 + a + 1.
Group the first two terms and last two terms.
The first two terms are a3 + a2 and the second two terms are a + 1.
Take a2 common from the first two terms.
a2 (a + 1)
Take 1 common from the second two terms.
1 (a + 1)
a2 (a + 1) + 1 (a + 1)
Then, take (a + 1) common from the above expression.
(a + 1) (a2 + 1)

The final answer is (a + 1) (a2 + 1).

(v) Given expression is m2 – b + mb – m.
Rearrange the terms
m2 + mb – b  – m.
Group the first two terms and last two terms.
The first two terms are m2 + mb and the second two terms are – b – m.
Take m common from the first two terms.
m (m + b)
Take -1 common from the second two terms.
-1 (m + b)
m (m + b) -1 (m + b)
Then, take (m + b) common from the above expression.
(m + b) (m – 1)

The final answer is (m + b) (m – 1).

(vi) Given expression is a3 – a2b + 5a – 5b.
Rearrange the terms
a3 + 5a – a2b – 5b.
Group the first two terms and last two terms.
The first two terms are a3 + 5a and the second two terms are – a2b – 5b.
Take a common from the first two terms.
a (a2 + 5)
Take -b common from the second two terms.
– b (a2 + 5)
a (a2 + 5) – b (a2 + 5)
Then, take (m + b) common from the above expression.
(a2 + 5) (a – b)

The final answer is (a2 + 5) (a – b).

(vii) Given expression is x (x + 3) – x – 3.
Rearrange the terms
x2  – x + 3x – 3.
Group the first two terms and last two terms.
The first two terms are x2  – x  and the second two terms are 3x – 3.
Take x common from the first two terms.
x (x – 1)
Take 3 common from the second two terms.
3 (x – 1)
x (x – 1) + 3 (x – 1)
Then, take (x – 1) common from the above expression.
(x – 1) (x + 3)

The final answer is (x – 1) (x + 3).

(viii) Given expression is 3mx + 3my – 2nx – 2ny.
Rearrange the terms
3mx – 2nx + 3my – 2ny.
Group the first two terms and last two terms.
The first two terms are 3mx – 2nx  and the second two terms are 3my – 2ny.
Take x common from the first two terms.
x (3m – 2n)
Take y common from the second two terms.
y (3m – 2n)
x (3m – 2n) + y (3m – 2n)
Then, take (3m – 2n) common from the above expression.
(3m – 2n) (x + y)

The final answer is (3m – 2n) (x + y).

(ix) Given expression is a (a + b – c ) – bc.
Rearrange the terms
a² – ac + ab – bc.
Group the first two terms and last two terms.
The first two terms are a² – ac and the second two terms are ab – bc.
Take a common from the first two terms.
a (a – c)
Take b common from the second two terms.
b (a – c)
a (a – c) + b (a – c)
Then, take (a – c) common from the above expression.
(a – c) (a + b)

The final answer is (a – c) (a + b).

(x) Given expression is x3 – x2 + xy2 – x2y2.
Rearrange the terms
x3 – x2 – x2y2 + xy2
Group the first two terms and last two terms.
The first two terms are x3 – x2 and the second two terms are – x2y2 + xy2.
Take x2 common from the first two terms.
x2 (x – 1)
Take – xy2 common from the second two terms.
– xy2 (x – 1)
x2 (x – 1) – xy2 (x – 1)
Then, take (x – 1) common from the above expression.
(x – 1) (x2 – xy2)

The final answer is (x – 1) (x2 – xy2).

2. Factor grouping the algebraic expressions

(i) 4ab – 7b + 12a – 21
(ii) 7xy – 5x – 28y + 20
(iii) 5mn – 2m – 5n2 + 2n
(iv) 6a2 – 15ac – 8ba + 20bc
(v) 4mx + 5nx – 12my – 15ny

Solution:

(i) Given expression is 4ab – 7b + 12a – 21.
Rearrange the terms
4ab + 12a – 7b – 21
Group the first two terms and last two terms.
The first two terms are 4ab + 12a and the second two terms are – 7b – 21.
Take 4a common from the first two terms.
4a (b + 3)
Take – 7 common from the second two terms.
– 7 (b + 3)
4a (b + 3) – 7 (b + 3)
Then, take (b + 3) common from the above expression.
(b + 3) (4a – 7)

The final answer is (b + 3) (4a – 7).

(ii) Given expression is 7xy – 5x – 28y + 20.
Rearrange the terms
7xy – 28y – 5x + 20
Group the first two terms and last two terms.
The first two terms are 7xy – 28y and the second two terms are – 5x + 20.
Take 7y common from the first two terms.
7y (x – 4)
Take – 5 common from the second two terms.
– 5 (x – 4)
7y (x – 4) – 5 (x – 4)
Then, take (x – 4) common from the above expression.
(x – 4) (7y – 5)

The final answer is (x – 4) (7y – 5).

(iii) Given expression is 5mn – 2m – 5n2 + 2n.
Rearrange the terms
5mn – 5n2 – 2m + 2n
Group the first two terms and last two terms.
The first two terms are 5mn – 5n2 and the second two terms are – 2m + 2n.
Take n common from the first two terms.
5n (m – n)
Take – 2 common from the second two terms.
– 2 (m – n)
5n (m – n) – 2 (m – n)
Then, take (m – n) common from the above expression.
(m – n) (5n – 2)

The final answer is (m – n) (5n – 2).

(iv) Given expression is 6a2 – 15ac – 8ba + 20bc.
Rearrange the terms
6a2 – 8ba – 15ac + 20bc
Group the first two terms and last two terms.
The first two terms are 6a2 – 8ba and the second two terms are – 15ac + 20bc.
Take 2a common from the first two terms.
2a (3a – 4b)
Take – 5c common from the second two terms.
– 5c (3a – 4b)
2a (3a – 4b) – 5c (3a – 4b)
Then, take (3a – 4b) common from the above expression.
(3a – 4b) (2a – 5c)

The final answer is (3a – 4b) (2a – 5c).

(v) Given expression is 4mx + 5nx – 12my – 15ny.
Rearrange the terms
4mx – 12my + 5nx – 15ny
Group the first two terms and last two terms.
The first two terms are 4mx – 12my and the second two terms are + 5nx – 15ny.
Take 4m common from the first two terms.
4m (x – 3y)
Take 5n common from the second two terms.
5n (x – 3y)
4m (x – 3y) + 5n (x – 3y)
Then, take (x – 3y) common from the above expression.
(x – 3y) (4m + 5n)

The final answer is (x – 3y) (4m + 5n).

3. Factorize by regrouping the terms

(i) 7mn – 21nr – 7mx + 21xr
(ii) a2 + ab(b + 1) + b3
(iii) yx2 – 2x(1 – y) – 4
(iv) m2 – m(a + 4b) + 4ab
(v) a – 9 – (a – 9)2 + ab – 9b

Solution:

(i) Given expression is 7mn – 21nr – 7mx + 21xr
Rearrange the terms
7mn – 7mx – 21nr + 21xr
Group the first two terms and last two terms.
The first two terms are 7mn – 7mx and the second two terms are – 21nr + 21xr.
Take 7m common from the first two terms.
7m (n – x)
Take -21r common from the second two terms.
-21r (n – x)
7m (n – x) – 21r (n – x)
Then, take (n – x) common from the above expression.
(n – x) (7m – 21r)

The final answer is (n – x) (7m – 21r).

(ii) Given expression is a2 + ab(b + 1) + b3
Rearrange the terms
a2 + ab + ab2 + b3
Group the first two terms and last two terms.
The first two terms are a2 + ab and the second two terms are ab2 + b3.
Take a common from the first two terms.
a (a + b)
Take b2 common from the second two terms.
b2 (a + b)
a (a + b) + b2 (a + b)
Then, take (a + b) common from the above expression.
(a + b) (a + b2)

The final answer is (a + b) (a + b2).

(iii) Given expression is yx2 – 2x(1 – y) – 4
Rearrange the terms
yx2 + 2xy – 2x – 4
Group the first two terms and last two terms.
The first two terms are yx2 + 2xy and the second two terms are – 2x – 4.
Take xy common from the first two terms.
xy (x + 2)
Take -2 common from the second two terms.
-2 (x + 2)
xy (x + 2) – 2 (x + 2)
Then, take (x + 2) common from the above expression.
(x + 2) (xy – 2)

The final answer is (x + 2) (xy – 2).

(iv) Given expression is m2 – m(a + 4b) + 4ab
Rearrange the terms
m2 – 4mb – ma + 4ab
Group the first two terms and last two terms.
The first two terms are m2 – 4mb and the second two terms are – ma + 4ab.
Take m common from the first two terms.
m (m – 4b)
Take -a common from the second two terms.
-a (m – 4b)
m (m – 4b) – a (m – 4b)
Then, take (m – 4b) common from the above expression.
(m – 4b) (m – a)

The final answer is (m – 4b) (m – a).

(v) Given expression is a – 9 – (a – 9)2 + ab – 9b
Rearrange the terms
a – 9 – (a – 9)2 + ab – 9b
Group the first two terms and last two terms.
The first two terms are a – 9 – (a – 9)2  and the second two terms are ab – 9b.
Take m common from the first two terms.
(a – 9) (1 – a + 9) = (a – 9) (10 – a)
Take b common from the second two terms.
b (a – 9)
(a – 9) (10 – a) + b (a – 9)
Then, take (a – 9) common from the above expression.
(a – 9) (10 – a + b)

The final answer is (a – 9) (10 – a + b).

4. Factorize by grouping the following expressions

(i) (a – 4) – (a – 4)2 + 12 – 3a
(ii) b (c – d )2 – a (d – c) + 3c – 3d
(iii) (a2 + 2a)2 – 7 (a2 + 2a) – y (a2 + 2a) +7y
(iv) m4x + m3 (2x – y ) – m(2my + z) – 2z
(v) x3 – 2x2y + 3xy2 – 6y3
(vi) m2 + n – mn – m
(vii) 5ab – b2 + 15ca – 3bc
(viii) xy2 – yz2 – xy + z2

Solution:

(i) Given expression is (a – 4) – (a – 4)2 + 12 – 3a
Rearrange the terms
(a – 4) – (a – 4)2 – 3a + 12
Group the first two terms and last two terms.
The first two terms are (a – 4) – (a – 4)2  and the second two terms are – 3a + 12.
Take (a – 4) common from the first two terms.
(a – 4) (1 – a + 4) = (a – 4) (5 – a)
Take -3 common from the second two terms.
-3 (a – 4)
(a – 4) (5 – a) – 3 (a – 4)
Then, take (a – 4) common from the above expression.
(a – 4) (5 – a – 3) = (a – 4) (2 – a)

The final answer is (a – 4) (2 – a).

(ii) Given expression is b (c – d )2 – a (d – c) + 3c – 3d
Rearrange the terms
b (c – d )2 + a (c – d) + 3c – 3d
Group the first two terms and last two terms.
The first two terms are b (c – d )2 + a (c – d) and the second two terms are 3c – 3d.
Take (c – d) common from the first two terms.
(c – d) (b (c – d) + a) = (c – d) (bc – bd + a)
Take 3 common from the second two terms.
3 (c – d)
(c – d) (bc – bd + a) + 3 (c – d)
Then, take (a – 4) common from the above expression.
(c – d) (bc – bd + a + 3)

The final answer is (c – d) (bc – bd + a + 3).

(iii) Given expression is (a2 + 2a)2 – 7 (a2 + 2a) – y (a2 + 2a) +7y
Rearrange the terms
(a2 + 2a)2 – y (a2 + 2a) – 7 (a2 + 2a) + 7y
Group the first two terms and last two terms.
The first two terms are (a2 + 2a)2 – y (a2 + 2a) and the second two terms are – 7 (a2 + 2a) + 7b.
Take (a2 + 2a) common from the first two terms.
(a2 + 2a) (a2 + 2a – y)
Take – 7 common from the second two terms.
– 7 (a2 + 2a – y)
(a2 + 2a) (a2 + 2a – y) – 7 (a2 + 2a – y)
Then, take (a2 + 2a – y) common from the above expression.
(a2 + 2a – y) (a2 + 2a – 7)

The final answer is (a2 + 2a – y) (a2 + 2a – 7).

(iv) Given expression is m4x + m3 (2x – y ) – m(2my + z) – 2z
Rearrange the terms
m4x + 2xm3 – ym3 –  2m2y – zm – 2z
Group the first two terms, middle terms, and last two terms.
The first two terms are m4x + 2xm3, the middle terms are – ym3 –  2m2y, and the second two terms are – zm – 2z.
Take xm3 common from the first two terms.
xm3 (m + 2)
Take – m2y common from the middle two terms.
– m2y (m + 2)
Take – z common from the middle two terms.
– z (m + 2)
xm3 (m + 2) – m2y (m + 2) – z (m + 2)
Then, take (m + 2) common from the above expression.
(m + 2) (xm3 – m2y – z)

The final answer is (m + 2) (xm3 – m2y – z).

(v) Given expression is x3 – 2x2y + 3xy2 – 6y3
Rearrange the terms
x3 – 2x2y + 3xy2 – 6y3
Group the first two terms, and last two terms.
The first two terms are x3 – 2x2y, and the second two terms are 3xy2 – 6y3.
Take x2 common from the first two terms.
x2 (x – 2y)
Take 3y2 common from the middle two terms.
3y2 (x – 2y)
x2 (x – 2y) + 3y2 (x – 2y)
Then, take (x – 2y) common from the above expression.
(x – 2y) (x2 + 3y2)

The final answer is (x – 2y) (x2 + 3y2).

(vi) Given expression is m2 + n – mn – m
Rearrange the terms
m2 – m + n – mn
Group the first two terms, and last two terms.
The first two terms are m2 – m, and the second two terms are n – mn.
Take m common from the first two terms.
m (m – 1)
Take -n common from the middle two terms.
-n (m – 1)
m (m – 1) – n (m – 1)
Then, take (m – 1) common from the above expression.
(m – 1) (m – n)

The final answer is (m – 1) (m – n).

(vii) Given expression is 5ab – b2 + 15ca – 3bc
Rearrange the terms
5ab + 15ca – b2 – 3bc
Group the first two terms, and last two terms.
The first two terms are 5ab + 15ca, and the second two terms are – b2 – 3bc.
Take 5a common from the first two terms.
5a (b + 3c)
Take -b common from the middle two terms.
-b (b + 3c)
5a (b + 3c) – b (b + 3c)
Then, take (b + 3c) common from the above expression.
(b + 3c) (5a – b)

The final answer is (b + 3c) (5a – b).

(viii) Given expression is xy2 – yz2 – xy + z2
Rearrange the terms
xy2 – xy – yz2 + z2
Group the first two terms, and last two terms.
The first two terms are xy2 – xy, and the second two terms are – yz2 + z2.
Take xy common from the first two terms.
xy (y – 1)
Take -z2 common from the middle two terms.
-z2 (y – 1)
xy (y – 1) – z2 (y – 1)
Then, take (y – 1) common from the above expression.
(y – 1) (xy – z2)

The final answer is (y – 1) (xy – z2).