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## How to Solve Factorization Problems by Regrouping?

1. Factorize each of the following by regrouping

(i) a^{2} + ab + 9a + 9b

(ii) 6ab – 4b + 6 – 9a

(iii) 10xy + 6x + 5y +3

(iv) a^{3} + a^{2} + a + 1

(v) m^{2} – b + mb – m

(vi) a^{3} – a^{2}b + 5a – 5b

(vii) x (x + 3) – x – 3

(viii) 3mx + 3my – 2nx – 2ny

(ix) a (a + b – c ) – bc

(x) x^{3} – x^{2} + xy^{2} – x^{2}y^{2}

## Solution:

(i) Given expression is a^{2} + ab + 9a + 9b.

Rearrange the terms

a^{2} + ab + 9a + 9b.

Group the first two terms and last two terms.

The first two terms are a^{2} + 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 a^{3} + a^{2} + a + 1.

Group the first two terms and last two terms.

The first two terms are a^{3} + a^{2} and the second two terms are a + 1.

Take a^{2} common from the first two terms.

a^{2 }(a + 1)

Take 1 common from the second two terms.

1 (a + 1)

a^{2} (a + 1) + 1 (a + 1)

Then, take (a + 1) common from the above expression.

(a + 1) (a^{2} + 1)

The final answer is (a + 1) (a^{2} + 1).

(v) Given expression is m^{2} – b + mb – m.

Rearrange the terms

m^{2} + mb – b – m.

Group the first two terms and last two terms.

The first two terms are m^{2} + 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 a^{3} – a^{2}b + 5a – 5b.

Rearrange the terms

a^{3} + 5a – a^{2}b – 5b.

Group the first two terms and last two terms.

The first two terms are a^{3} + 5a and the second two terms are – a^{2}b – 5b.

Take a common from the first two terms.

a^{ }(a^{2} + 5)

Take -b common from the second two terms.

– b (a^{2} + 5)

a^{ }(a^{2} + 5) – b (a^{2} + 5)

Then, take (m + b) common from the above expression.

(a^{2} + 5) (a – b)

The final answer is (a^{2} + 5) (a – b).

(vii) Given expression is x (x + 3) – x – 3.

Rearrange the terms

x^{2} – x + 3x – 3.

Group the first two terms and last two terms.

The first two terms are x^{2} – 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 x^{3} – x^{2} + xy^{2} – x^{2}y^{2}.

Rearrange the terms

x^{3} – x^{2} – x^{2}y^{2 }+ xy^{2}

Group the first two terms and last two terms.

The first two terms are x^{3} – x^{2} and the second two terms are – x^{2}y^{2 }+ xy^{2}.

Take x^{2} common from the first two terms.

x^{2} (x – 1)

Take – xy^{2} common from the second two terms.

– xy^{2} (x – 1)

x^{2} (x – 1) – xy^{2} (x – 1)

Then, take (x – 1) common from the above expression.

(x – 1) (x^{2} – xy^{2})

The final answer is (x – 1) (x^{2} – xy^{2}).

2. Factor grouping the algebraic expressions

(i) 4ab – 7b + 12a – 21

(ii) 7xy – 5x – 28y + 20

(iii) 5mn – 2m – 5n^{2} + 2n

(iv) 6a^{2} – 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 – 5n^{2} + 2n.

Rearrange the terms

5mn – 5n^{2} – 2m + 2n

Group the first two terms and last two terms.

The first two terms are 5mn – 5n^{2} 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 6a^{2} – 15ac – 8ba + 20bc.

Rearrange the terms

6a^{2} – 8ba – 15ac + 20bc

Group the first two terms and last two terms.

The first two terms are 6a^{2} – 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) a^{2} + ab(b + 1) + b^{3
}(iii) yx^{2} – 2x(1 – y) – 4

(iv) m^{2} – 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 a^{2} + ab(b + 1) + b^{3}

Rearrange the terms

a^{2} + ab + ab^{2} + b^{3}

Group the first two terms and last two terms.

The first two terms are a^{2} + ab and the second two terms are ab^{2} + b^{3}.

Take a common from the first two terms.

a (a + b)

Take b^{2} common from the second two terms.

b^{2} (a + b)

a (a + b) + b^{2} (a + b)

Then, take (a + b) common from the above expression.

(a + b) (a + b^{2})

The final answer is (a + b) (a + b^{2}).

(iii) Given expression is yx^{2} – 2x(1 – y) – 4

Rearrange the terms

yx^{2} + 2xy – 2x – 4

Group the first two terms and last two terms.

The first two terms are yx^{2} + 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 m^{2} – m(a + 4b) + 4ab

Rearrange the terms

m^{2} – 4mb – ma + 4ab

Group the first two terms and last two terms.

The first two terms are m^{2} – 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) (a^{2} + 2a)^{2} – 7 (a^{2} + 2a) – y (a^{2} + 2a) +7y

(iv) m^{4}x + m^{3} (2x – y ) – m(2my + z) – 2z

(v) x^{3} – 2x^{2}y + 3xy^{2} – 6y^{3
}(vi) m^{2} + n – mn – m

(vii) 5ab – b^{2} + 15ca – 3bc

(viii) xy^{2} – yz^{2} – xy + z^{2}

## 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 (a^{2} + 2a)^{2} – 7 (a^{2} + 2a) – y (a^{2} + 2a) +7y

Rearrange the terms

(a^{2} + 2a)^{2} – y (a^{2} + 2a) – 7 (a^{2} + 2a) + 7y

Group the first two terms and last two terms.

The first two terms are (a^{2} + 2a)^{2} – y (a^{2} + 2a) and the second two terms are – 7 (a^{2} + 2a) + 7b.

Take (a^{2} + 2a) common from the first two terms.

(a^{2} + 2a) (a^{2} + 2a – y)

Take – 7 common from the second two terms.

– 7 (a^{2} + 2a – y)

(a^{2} + 2a) (a^{2} + 2a – y) – 7 (a^{2} + 2a – y)

Then, take (a^{2} + 2a – y) common from the above expression.

(a^{2} + 2a – y) (a^{2} + 2a – 7)

The final answer is (a^{2} + 2a – y) (a^{2} + 2a – 7).

(iv) Given expression is m^{4}x + m^{3} (2x – y ) – m(2my + z) – 2z

Rearrange the terms

m^{4}x + 2xm^{3} – ym^{3} – 2m^{2}y – zm – 2z

Group the first two terms, middle terms, and last two terms.

The first two terms are m^{4}x + 2xm^{3}, the middle terms are – ym^{3} – 2m^{2}y, and the second two terms are – zm – 2z.

Take xm^{3} common from the first two terms.

xm^{3} (m + 2)

Take – m^{2}y common from the middle two terms.

– m^{2}y (m + 2)

Take – z common from the middle two terms.

– z (m + 2)

xm^{3} (m + 2) – m^{2}y (m + 2) – z (m + 2)

Then, take (m + 2) common from the above expression.

(m + 2) (xm^{3} – m^{2}y – z)

The final answer is (m + 2) (xm^{3} – m^{2}y – z).

(v) Given expression is x^{3} – 2x^{2}y + 3xy^{2} – 6y^{3}

Rearrange the terms

x^{3} – 2x^{2}y + 3xy^{2} – 6y^{3}

Group the first two terms, and last two terms.

The first two terms are x^{3} – 2x^{2}y, and the second two terms are 3xy^{2} – 6y^{3}.

Take x^{2 }common from the first two terms.

x^{2} (x – 2y)

Take 3y^{2} common from the middle two terms.

3y^{2} (x – 2y)

x^{2} (x – 2y) + 3y^{2} (x – 2y)

Then, take (x – 2y) common from the above expression.

(x – 2y) (x^{2} + 3y^{2})

The final answer is (x – 2y) (x^{2} + 3y^{2}).

(vi) Given expression is m^{2} + n – mn – m

Rearrange the terms

m^{2} – m + n – mn

Group the first two terms, and last two terms.

The first two terms are m^{2} – 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 – b^{2} + 15ca – 3bc

Rearrange the terms

5ab + 15ca – b^{2} – 3bc

Group the first two terms, and last two terms.

The first two terms are 5ab + 15ca, and the second two terms are – b^{2} – 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 xy^{2} – yz^{2} – xy + z^{2}

Rearrange the terms

xy^{2} – xy – yz^{2} + z^{2}

Group the first two terms, and last two terms.

The first two terms are xy^{2} – xy, and the second two terms are – yz^{2} + z^{2}.

Take xy^{ }common from the first two terms.

xy (y – 1)

Take -z^{2} common from the middle two terms.

-z^{2} (y – 1)

xy (y – 1) – z^{2} (y – 1)

Then, take (y – 1) common from the above expression.

(y – 1) (xy – z^{2})

The final answer is (y – 1) (xy – z^{2}).