# ML Aggarwal Factorisation Exe-4.2 Class 9 ICSE Maths Solutions

ML Aggarwal Factorisation Exe-4.2 Class 9 ICSE Maths Solutions. Step by step solutions of all questions as council prescribe guideline. Visit official website CISCE for detail information about ICSE Board Class-9.

## ML Aggarwal Factorisation Exe-4.2 Class 9 ICSE Maths Solutions

 Board ICSE Subject Maths Class 9th Chapter-4 Factorisation Topics Solution of Exe-4.2 Questions Academic Session 2024-2025

### Solution of Exe-4.2 Questions

ML Aggarwal Factorisation Exe-4.2 Class 9 ICSE Maths Solutions

#### Question 1.

(i) x2 + xy – x – y

(ii) y2 – yz – 5y + 5z

##### (i) x2 + xy – x – y

Take out common in all terms,

x(x + y) – 1(x + y)

(x + y) (x – 1)

##### (ii) y2 – yz – 5y + 5z

y2 – yz – 5y + 5z

Take out common in all terms,

y(y – z) – 5(y – z)

(y – z) (y – 5)

#### Question 2.

(i) 5xy + 7y – 5y2 – 7x

(ii) 5p2 – 8pq – 10p + 16q

##### (i) 5xy – 7x – 5y2 + 7y

Take out common in all terms,

x(5y – 7) – y(5y – 7)

(5y – 7) (x – y)

##### (ii) 5p2 – 8pq – 10p + 16q

5p2 – 8pq – 10p + 16q

Take out common in all terms,

p(5p – 8q) – 2(5p – 8q)

(5p – 8q) (p – 2)

#### Question 3.

(i) a2b – ab2 + 3a – 3b

(ii) x3 – 3x2 + x – 3

##### (i) a2b – ab2 + 3a – 3b

Take out common in all terms,

ab(a – b) + 3(a – b)

(a – b) (ab + 3)

##### (ii) x3 – 3x2 + x – 3

x3 – 3x2 + x – 3

Take out common in all terms,

x2 (x – 3) + 1(x – 3)

(x – 3) (x2 + 1)

#### Question 4.

(i) 6xy2 – 3xy – 10y + 5

(ii) 3ax – 6ay – 8by + 4bx

##### (i) 6xy2 – 3xy – 10y + 5

Take out common in all terms,

3xy(2y – 1) – 5(2y – 1)

(2y – 1) (3xy – 5)

##### (ii) 3ax – 6ay – 8by + 4bx

3ax – 6ay – 8by + 4bx

Take out common in all terms,

3a(x – 2y) + 4b (x – 2y)

(x – 2y) (3a + 4b)

#### Question 5.

(i) 1 – a – b + ab

(ii) a(a – 2b – c) + 2bc

##### (i) 1 – a – b + ab

Take out common in all terms,

1(1 – a) – b(1 – a)

(1 – a) (1 – b)

##### (ii) a(a – 2b – c) + 2bc

a(a – 2b – c) + 2bc

Above question can be written as,

a2 – 2ab – ac + 2bc

Take out common in all terms,

a(a – 2b) – c(a + 2b)

(a – 2b) (a – c)

#### Question 6.

(i) x2 + xy (1 + y) + y3

(ii) y2 – xy (1 – x) – x3

##### (i) x2 + xy (1 + y) + y3

Above question can be written as,

x2 + xy + xy2 + y3

Take out common in all terms,

x(x + y) + y2(x + y)

(x + y) (x + y2)

##### (ii) y2 – xy (1 – x) – x3

y2 – xy (1 – x) – x3

Above question can be written as,

y2 – xy + x2y – x3

Take out common in all terms,

y(y – x) + x2 (y – x)

(y – x) (y + x2)

#### Question 7.

(i) ab2 + (a – 1)b – 1

(ii) 2a – 4b – xa + 2bx

##### (i) ab2 + (a – 1)b – 1

Above question can be written as,

ab2 + ab – b – 1

Take out common in all terms,

ab(b + 1) – 1(b + 1)

(b + 1) (ab – 1)

##### (ii) 2a – 4b – xa + 2bx

2a – 4b – xa + 2bx

Take out common in all terms,

2(a – 2b) – x(a – 2b)

(a – 2b) (2 – x)

#### Question 8.

(i) 5ph – 10qk + 2rph – 4qrk

(ii) x2 – x(a + 2b) + 2ab

##### (i) 5ph – 10qk + 2rph – 4qrk

Re-arranging the given question we get,

5ph + 2rph – 10qk – 4qrk

Take out common in all terms,

ph(5 + 2r) – 2qk(5 + 2r)

(5 + 2r) (ph – 2qk)

##### (ii) x2 – x(a + 2b) + 2ab

x2 – x(a + 2b) + 2ab

Above question can be written as,

x2 – xa – 2xb + 2ab

Take out common in all terms,

x(x – a) – 2b(x – a)

(x – a) (x – 2b)

#### Question 9.

(i) ab(x2 + y2) – xy(a2 + b2)

(ii) (ax + by)2 + (bx – ay)2

##### (i) ab(x2 + y2) – xy(a2 + b2)

Above question can be written as,

abx2 + aby2 – xya2 – xyb2

Re-arranging the above we get,

abx2 – xyb2 + aby2 – xya2

Take out common in all terms,

bx(ax – by) + ay(by – ax)

bx(ax – by) – ay (ax – by)

(ax – by) (bx – ay)

##### (ii) (ax + by)2 + (bx – ay)2

By expanding the give question, we get,

(ax)2 + (by)2 + 2axby + (bx)2 + (ay)2 – 2bxay

a2x2 + b2y2 + b2x2 + a2y2

Re-arranging the above we get,

a2x2 + a2y2 + b2y2 + b2x2

Take out common in all terms,

a2 (x2 + y2) + b2 (x2 + y2)

(x2 + y2) (a2 + b2)

#### Question 10.

(i) a3 + ab(1 – 2a) – 2b2

(ii) 3x2y – 3xy + 12x – 12

##### (i)a3 + ab(1 – 2a) – 2b2

Above question can be written as,

a3 + ab – 2a2b – 2b2

Re-arranging the above we get,

a3 – 2a2b + ab – 2b2

Take out common in all terms,

a2(a – 2b) + b(a – 2b)

(a – 2b) (a2 + b)

##### (ii) 3x2y – 3xy + 12x – 12

3x2y – 3xy + 12x – 12

Take out common in all terms,

3xy(x – 1) + 12(x – 1)

(x – 1) (3xy + 12)

#### Question 11. a2b + ab2 –abc – b2c + axy + bxy

a2b + ab2 –abc – b2c + axy + bxy

Re-arranging the above we get,

a2b – abc + axy + ab2 – b2c + bxy

Take out common in all terms,

a(ab – bc + xy) + b(ab – bc + xy)

(a + b) (ab – bc + xy)

#### Question 12. ax2 – bx2 + ay2 – by2 + az2 – bz2

ax2 – bx2 + ay2 – by2 + az2 – bz2

Re-arranging the above we get,

ax2 + ay2 + az2 – bx2 – by2 – bz2

Take out common in all terms,

a(x2 + y2 + z2) – b(x2 + y2 + z2)

(x2 + y2 + z2) (a – b)

#### Question 13. x – 1 – (x – 1)2 + ax – a

x – 1 – (x – 1)2 + ax – a

By expanding the above we get,

X – 1 – (x2 + 1 – 2x) + ax – a

x – 1 – x2 -1 + 2x + ax – a

2x – x2 + ax – 2 + x – a

Take out common in all terms,

x(2 – x + a) – 1(2 – x + a)

(2 – x + a) (x – 1)

—  : End of ML Aggarwal Factorisation Exe-4.2 Class 9 ICSE Maths Solutions :–

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