ML Aggarwal Factorisation Exe-4.4 Class 9 ICSE Maths Solutions

ML Aggarwal Factorisation Exe-4.4 Class 9 ICSE Maths Solutions . Step by step solutions of Factorisation problems as council prescribe guideline. Visit official website CISCE for detail information about ICSE Board Class-9.

ML Aggarwal Factorisation Exe-4.4 Class 9 ICSE Maths Solutions

Board	ICSE
Subject	Maths
Class	9th
Chapter-4	Factorisation
Topics	Solution of Exe-4.4 Questions
Academic Session	2024-2025

Solution of Exe-4.4 Questions

ML Aggarwal Factorisation Exe-4.4 Class 9 ICSE Maths Solutions

Question 1.

(i) x² + 5x + 6

(ii) x² – 8x + 7

Answer :

(i) x² + 5x + 6

x² + 3x + 2x + 6

Take out common in all terms we get,

x(x + 3) + 2 (x + 3)

(x + 3) (x + 2)

(ii) x² – 8x + 7

x² – 8x + 7

x² – 7x – x + 7

Take out common in all terms we get,

x(x – 7) – 1(x – 7)

(x – 7) (x – 1)

Question 2.

(i) x² + 6x – 7

(ii) y² + 7y – 18

Answer :

(i)  x² + 6x – 7

x² + 7x – x – 7

Take out common in all terms we get,

x(x + 7) – 1(x + 7)

(x + 7) (x – 1)

(ii) y² + 7y – 18

y² + 7y – 18

y² + 9y – 2y – 18

Take out common in all terms we get,

y(y + 9) – 2(y + 9)

(y + 9) (y – 2)

Question 3.

(i) y² – 7y – 18

(ii) a² – 3a – 54

Answer :

(i) y² – 7y – 18

y² + 2y – 9y – 18

Take out common in all terms we get,

y(y + 2) – 9(y + 2)

(y + 2) (y – 9)

(ii) a² – 3a – 54

a² – 3a – 54

a² + 6a – 9a – 54

Take out common in all terms we get,

a(a + 6) – 9(a + 6)

So, (a + 6) (a – 9)

Question 4.

(i) 2x² – 7x + 6

(ii) 6x² + 13x – 5

Answer :

(i) 2x² – 7x + 6

2x² – 4x – 3x + 6

Take out common in all terms we get,

2x(x – 2) – 3(x – 2)

(x – 2) (2x – 3)

(ii) 6x² + 13x – 5

6x² + 13x – 5

6x² + 15x – 2x – 5

Take out common in all terms we get,

3x(2x + 5) – 1(2x + 5)

(2x + 5) (3x – 1)

Question 5.

(i) 6x² + 11x – 10

(ii) 6x² – 7x – 3

Answer :

(i) 6x² + 11x – 10

6x² + 15x – 4x – 10

Take out common in all terms we get,

3x(2x + 5) – 2(2x + 5)

(2x + 5) (3x – 2)

(ii) 6x² – 7x – 3

6x² – 7x – 3

6x² – 9x + 2x – 3

Take out common in all terms we get,

3x(2x – 3) + 1(2x – 3)

(2x – 3) (3x + 1)

Question 6.

(i) 2x² – x – 6

(ii) 1 – 18y – 63y²

Answer :

(i) 2x² – x – 6

2x² – 4x + 3x – 6

Take out common in all terms we get,

2x(x – 2) + 3(x – 2)

(x – 2) (2x + 3)

(ii) 1 – 18y – 63y²

1 – 18y – 63y²

1 – 21y + 3y – 63y²

Take out common in all terms we get,

1(1 – 21y) + 3y(1 – 21y)

(1 – 21y) (1 + 3y)

Question 7.

(i) 2y² + y – 45

(ii) 5 – 4x – 12x²

Answer :

(i) 2y² + y – 45

2y² + 10y – 9y – 45

Take out common in all terms we get,

2y (y + 5) – 9(y + 5)

(y + 5) (2y – 9)

(ii) 5 – 4x – 12x²

5 – 4x – 12x²

5 – 10x + 6x – 12x²

Take out common in all terms we get,

5(1 – 2x) + 6x(1 – 2x)

(1 – 2x) (5 + 6x)

Question 8.

(i) x(12x + 7) – 10

(ii) (4 – x)² – 2x

Answer :

(i) x(12x + 7) – 10

Above terms can be written as,

12x² + 7x – 10

12x² + 15x – 8x – 10

Take out common in all terms we get,

3x(4x + 5) – 2(4x + 5)

(4x + 5) (3x – 2)

(ii) (4 – x)² – 2x

(4 – x)² – 2x

We know that, (a – b)² = a² – 2ab + b²

So, (4² – (2 × 4 × x) + x²) – 2x

16 – 8x + x² – 2x

x² – 10x + 16

x² – 8x – 2x + 16

Take out common in all terms we get,

x(x – 8) – 2(x – 8)

(x – 8) (x -2)

Question 9.

(i) 60x² – 70x – 30

(ii) x² – 6xy – 7y²

Answer :

(i) 60x² – 70x – 30

Take out common in all terms we get,

10(6x² – 7x – 3)

10(6x² – 9x + 2x – 3)

Again, take out common in all terms we get,

10(3x(2x – 3) + 1(2x – 3))

10(2x – 3) (3x + 1)

(ii) x² – 6xy – 7y²

x² – 6xy – 7y²

x² – 7xy + xy – 7y²

Take out common in all terms we get,

x(x – 7y) + y(x – 7y)

(x – 7y) (x + y)

Question 10.

(i) 2x² + 13xy – 24y²

(ii) 6x² – 5xy – 6y²

Answer :

(i) 2x² + 13xy – 24y²

2x² + 16xy – 3xy – 24y²

Take out common in all terms we get,

2x(x + 8y) – 3y(x + 8y)

(x + 8y) (2x – 3y)

(ii) 6x² – 5xy – 6y²

6x² – 5xy – 6y²

6x² – 9xy + 4xy – 6y²

Take out common in all terms we get,

3x(2x – 3y) + 2y (2x – 3y)

(2x – 3y) (3x + 2y)

Question 11.

(i) 5x² + 17xy – 12y²

(ii) x²y² – 8xy – 48

Answer :

(i) 5x² + 17xy – 12y²

5x² + 20xy – 3xy – 12y²

Take out common in all terms we get,

5x(x + 4y) – 3y(x + 4y)

(x + 4y) (5x – 3y)

(ii) x²y² – 8xy – 48

x²y² – 8xy – 48

x²y² – 12xy + 4xy – 48

Take out common in all terms we get,

xy(xy – 12) + 4(xy – 12)

(xy – 12) (xy + 4)

Question 12.

(i) 2a²b² – 7ab – 30

(ii) a(2a – b) – b²

Answer :

(i) 2a²b² – 7ab – 30

2a²b² – 12ab + 5ab – 30

Take out common in all terms we get,

2ab(ab – 6) + 5 (ab – 6)

(ab – 6) (2ab + 5)

(ii) a(2a – b) – b²

a(2a – b) – b²

Above terms can be written as,

2a² – ab – b²

2a² – 2ab + ab – b²

Take out common in all terms we get,

2a(a – b) + b(a – b)

(a – b) (2a + b)

Question 13.

(i) (x – y)² – 6(x – y) + 5

(ii) (2x – y)² – 11(2x – y) + 28

Answer :

(i) (x – y)² – 6(x – y) + 5

Above terms can be written as,

(x – y)² – 5(x – y) – (x – y) + 5

(x – y) (x – y – 5) – 1(x – y – 5)

Then,

(x – y – 5) (x – y – 1)

(ii) (2x – y)² – 11(2x – y) + 28

(2x – y)² – 11(2x – y) + 28

Above terms can be written as,

(2x – y)² – 7(2x – y) – 4(2x – y) + 28

(2x – y) (2x – y – 7) – 4(2x – y – 7)

(2x – y – 7) (2x – y – 4)

Question 14.

(i) 4(a – 1)² – 4(a – 1) – 3

(ii) 1 – 2a – 2b – 3(a + b)²

Answer :

(i) 4(a – 1)² – 4(a – 1) – 3

Above terms can be written as,

4(a – 1)² – 6(a – 1) + 2(a – 1) – 3

Take out common in all terms we get,

2(a – 1) [2(a – 1) – 3] + 1[2(a – 1) – 3]

(2(a – 1) – 3) (2(a – 1) + 1)

(2a – 2 – 3) (2a – 2 + 1)

(2a – 5) (2a – 1)

(ii) 1 – 2a – 2b – 3(a + b)²

1 – 2a – 2b – 3(a + b)²

Above terms can be written as,

1 – 2(a + b) – 3(a + b)²

1 – 3(a + b) + (a + b) – 3(a + b)²

Take out common in all terms we get,

1(1 – 3(a + b)) + (a + b) (1 – (a + b))

(1 – 3(a + b)) (1 + (a + b))

(1 – 3a + 3b) (1 + a + b)

Question 15.

(i) 3 – 5a – 5b – 12(a + b)²

(ii) a⁴ – 11a² + 10

Answer :

(i) 3 – 5a – 5b – 12(a + b)²

Above terms can be written as,

3 – 5(a + b) – 12(a + b)²

3 – 9(a + b) + 4(a + b) – 12(a + b)²

Take out common in all terms we get,

3(1 – 3(a + b)) + 4(a + b) (1 – 3(a + b))

(1 – 3(a + b)) (3 + 4(a + b))

(1 – 3a – 3b) (3 + 4a + 4b)

(ii) a⁴ – 11a² + 10

a⁴ – 11a² + 10

Above terms can be written as,

a⁴ – 10a² – a² + 10

Take out common in all terms we get,

a² (a² – 10) – 1(a² – 10)

(a² – 10) (a² – 1)

Question 16.

(i) (x + 4)² – 5xy -20y – 6y²

(ii) (x² – 2x²) – 23(x² – 2x) + 120

Answer :

(i) (x + 4)² – 5xy -20y – 6y²

Above terms can be written as,

(x + 4)² – 5y(x + 4) – 6y²

(x + 4)² – 6y(x + 4) + y(x + 4) – 6y²

Take out common in all terms we get,

(x + 4) (x + 4 – 6y) + y(x + 4 – 6y)

(x – 6y + 4) (x + 4 + y)

(ii) (x² – 2x²) – 23(x² – 2x) + 120

(x² – 2x²) – 23(x² – 2x) + 120

Above terms can be written as,

(x² – 2x)² – 15(x² – 2x) – 8(x² – 2x) + 120

Take out common in all terms we get,

(x² – 2x) (x² – 2x – 15) – 8(x² – 2x – 15)

(x² – 2x – 15) (x² – 2x – 8)

Question 17. 4(2a – 3)² – 3(2a – 3) (a – 1) – 7 (a – 1)²

Answer :

4(2a – 3)² – 3(2a – 3) (a – 1) – 7 (a – 1)²

Let us assume, 2a – 3 = p and a – 1 = q

So, 4p² – 3pq – 7q²

Then, 4p² – 7pq + 4pq – 7q²

Take out common in all terms we get,

P(4p – 7q) + q(4p – 7q)

(4p – 7q) (p + q)

Now, substitute the value of p and q we get,

(4(2a – 3) – 7(a – 1)) (2a – 3 + a – 1)

(8a – 12 – 7a + 7) (3a – 4)

(a – 5) (3a – 4)

Question 18. (2x² + 5x) (2x² + 5x – 19) + 84

Answer :

(2x² + 5x) (2x² + 5x – 19) + 84

Let us assume, 2x² + 5x = p

So, (p) (p – 19) + 84

p² – 19p + 84

p² – 12p – 7p + 84

p(p – 12) – 7(p – 12)

(p – 12) (p – 7)

Now, substitute the value of p we get,

(2x² + 5x – 12) (2x² + 5x – 7)

—  : End of ML Aggarwal Factorisation Exe-4.4 Class 9 ICSE Maths Solutions :–

Return to :-  ML Aggarawal Maths Solutions for ICSE  Class-9

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