ICSE Class 9 Maths Chapter 05 Factorisation

Factorisation

Points To Remember

1. \((a + b)^2 = a^2 + b^2 + 2ab\) or \(a^2 + 2ab + b^2\)

2. \((a - b)^2 = a^2 + b^2 - 2ab\) or \(a^2 - 2ab + b^2\)

3. \(a^2 - b^2 = (a + b)(a - b)\)

4. \(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\)

5. \(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\)

6. \((x + a)(x + b) = x^2 + (a + b)x + ab\)

7. \(ax^2 + bx + c = ax^2 + px + qx + c\), now we have to find p and q where \(pq = ac\) and \(p + q = b\).

Exercise 5 (A)

Factorise:

Q. 1. \(5x^2 - 20xy\)

Sol. \(5x^2 - 20xy = 5x(x - 4y)\) Ans.

Q. 2. \(18a^2b - 24abc\)

Sol. \(18a^2b - 24abc = 6ab(3a - 4c)\) Ans.

Q. 3. \(27x^3y^3 - 45x^4y^2\)

Sol. \(27x^3y^3 - 45x^4y^2 = 9x^3y^2(3y - 5x)\) Ans.

Q. 4. \(5a(b + c) - 7b(b + c)\)

Sol. \(5a(b + c) - 7b(b + c) = (b + c)(5a - 7b)\) Ans.

Q. 5. \(2x(p^2 + q^2) + 4y(p^2 + q^2)\)

Sol. \(2x(p^2 + q^2) + 4y(p^2 + q^2) = (p^2 + q^2)(2x + 4y) = (p^2 + q^2)2(x + 2y) = 2(p^2 + q^2)(x + 2y)\) Ans.

Q. 6. \(x(a - 5) + y(5 - a)\)

Sol. \(x(a - 5) + y(5 - a) = x(a - 5) - y(a - 5) = (a - 5)(x - y)\) Ans.

Q. 7. \(4(x + y) - 6(x + y)^2\)

Sol. \(4(x + y) - 6(x + y)^2 = 2(x + y)\{2 - 3(x + y)\} = 2(x + y)(2 - 3x - 3y)\) Ans.

Q. 8. \(6(2a + 3b)^2 - 8(2a + 3b)\)

Sol. \(6(2a + 3b)^2 - 8(2a + 3b) = 2(2a + 3b)\{3(2a + 3b) - 4\} = 2(2a + 3b)(6a + 9b - 4)\) Ans.

Q. 9. \(x(x + y)^3 - 3x^2y(x + y)\)

Sol. \(x(x + y)^3 - 3x^2y(x + y) = x(x + y)\{(x + y)^2 - 3xy\} = x(x + y)\{x^2 + y^2 + 2xy - 3xy\} = x(x + y)(x^2 + y^2 - xy)\) Ans.

Q. 10. \(a^3 + 2a^2 + 5a + 10\)

Sol. \(a^3 + 2a^2 + 5a + 10 = a^2(a + 2) + 5(a + 2) = (a + 2)(a^2 + 5)\) Ans.

Teacher's Note

Factorisation helps us break down complex expressions into simpler parts, much like breaking down a large task into smaller, manageable steps in daily life.

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