# NCERT Solutions for Class 9 Maths Exercise 2.4 Question 4
## Question 4:
**Factorise:**
**(i)** $$12x^2 – 7x + 1$$
**(ii)** $$2x^2 + 7x + 3$$
**(iii)** $$6x^2 + 5x – 6$$
**(iv)** $$3x^2 – x – 4$$
—
## Solutions:
—
### (i) $$12x^2 – 7x + 1$$
We need two numbers whose:
– **Product** = $$12 \times 1 = 12$$
– **Sum** = $$-7$$
Those numbers are **−4** and **−3**
$$12x^2 – 7x + 1$$
$$= 12x^2 – 4x – 3x + 1$$
$$= 4x(3x – 1) – 1(3x – 1)$$
$$= \boxed{(4x – 1)(3x – 1)}$$
—
### (ii) $$2x^2 + 7x + 3$$
We need two numbers whose:
– **Product** = $$2 \times 3 = 6$$
– **Sum** = $$7$$
Those numbers are **+6** and **+1**
$$2x^2 + 7x + 3$$
$$= 2x^2 + 6x + x + 3$$
$$= 2x(x + 3) + 1(x + 3)$$
$$= \boxed{(2x + 1)(x + 3)}$$
—
### (iii) $$6x^2 + 5x – 6$$
We need two numbers whose:
– **Product** = $$6 \times (-6) = -36$$
– **Sum** = $$5$$
Those numbers are **+9** and **−4**
$$6x^2 + 5x – 6$$
$$= 6x^2 + 9x – 4x – 6$$
$$= 3x(2x + 3) – 2(2x + 3)$$
$$= \boxed{(3x – 2)(2x + 3)}$$
—
### (iv) $$3x^2 – x – 4$$
We need two numbers whose:
– **Product** = $$3 \times (-4) = -12$$
– **Sum** = $$-1$$
Those numbers are **−4** and **+3**
$$3x^2 – x – 4$$
$$= 3x^2 – 4x + 3x – 4$$
$$= x(3x – 4) + 1(3x – 4)$$
$$= \boxed{(x + 1)(3x – 4)}$$
—
## 📌 Key Method Used: **Splitting the Middle Term**
| Step | Action |
|——|——–|
| **Step 1** | Multiply coefficient of $$x^2$$ with constant term |
| **Step 2** | Find two numbers with that product and sum = middle term coefficient |
| **Step 3** | Split the middle term using those two numbers |
| **Step 4** | Group and factorise |