NCERT Solutions for Class 9 Maths Exercise 2.4 Question 4

NCERT Solutions for Class 9 Maths Exercise 2.4 Question 4

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NCERT Solutions for Class 9 Maths Exercise 2.4 Question 4 | AskFarhan

NCERT Solutions for Class 9 Maths Exercise 2.4 Question 4

ncert solutions for Class 9 Maths Chapter 2 Polynomials — Exercise 2.4 Question 4 provides complete step-by-step factorisation of four quadratic polynomials using the splitting the middle term method, fully aligned with the CBSE 2025-26 curriculum. Official NCERT Website

✓ Verified by Experts

Understanding the Question

Class 10 Maths Chapter 4 Exercise 4.1 हल सहित
Read moreClass 10 Maths Chapter 4 Exercise 4.1 हल सहित

In this question, we are asked to factorise four quadratic polynomials of the form &&ax^2 + bx + c&&. The key technique used here is splitting the middle term, which is a fundamental algebraic skill tested in CBSE Class 9 board examinations. Each polynomial is broken down into two binomial factors by finding two numbers whose product is &&a \times c&& and whose sum is &&b&&.

Question 4: Factorise the following expressions:
(i) &&12x^2 – 7x + 1&&
(ii) &&2x^2 + 7x + 3&&
(iii) &&6x^2 + 5x – 6&&
(iv) &&3x^2 – x – 4&&

📚 Expert’s Approach — Strategy Overview

For each quadratic &&ax^2 + bx + c&&, we apply the Splitting the Middle Term method:

  1. Compute the product &&a \times c&&.
  2. Find two integers &&p&& and &&q&& such that &&p \times q = a \times c&& and &&p + q = b&&.
  3. Rewrite the middle term &&bx&& as &&px + qx&&.
  4. Group the four terms into two pairs and factor each pair.
  5. Extract the common binomial factor to obtain the final factorised form.
NCERT Solutions for Class 10 Maths Chapter 7 Exercise 7.1 Questio
Read moreNCERT Solutions for Class 10 Maths Chapter 7 Exercise 7.1 Questio

This method is equivalent to reversing the FOIL expansion and is the standard NCERT-approved approach for Class 9 quadratic factorisation. BYJU’S CueMath TiwaraAcademy

Part (i): Factorise &&12x^2 – 7x + 1&&

Step 1 — Identify coefficients:
Here, &&a = 12&&, &&b = -7&&, &&c = 1&&.
Compute &&a \times c = 12 \times 1 = 12&&.
Step 2 — Find the split:
We need two numbers &&p&& and &&q&& such that:
&&p \times q = 12&& and &&p + q = -7&&.
The numbers are &&-4&& and &&-3&&, because &&(-4) \times (-3) = 12&& and &&(-4) + (-3) = -7&&.
Step 3 — Split the middle term:
&&12x^2 – 7x + 1&&
&&= 12x^2 – 4x – 3x + 1&&
Step 4 — Group and factor:
&&= 4x(3x – 1) – 1(3x – 1)&&
The common binomial factor is &&(3x – 1)&&.
&&= (4x – 1)(3x – 1)&&

✓ Result: &&12x^2 – 7x + 1 = (4x – 1)(3x – 1)&&

⚠ Caution: Do not confuse the signs when grouping. Always check: &&(-1)(3x – 1) = -3x + 1&&, which is correct.

Source: BYJU’S NCERT Solutions Ex 2.4 CueMath Class 9 Ch 2

Part (ii): Factorise &&2x^2 + 7x + 3&&

Step 1 — Identify coefficients:
Here, &&a = 2&&, &&b = 7&&, &&c = 3&&.
Compute &&a \times c = 2 \times 3 = 6&&.
Step 2 — Find the split:
We need &&p \times q = 6&& and &&p + q = 7&&.
The numbers are &&6&& and &&1&&, because &&6 \times 1 = 6&& and &&6 + 1 = 7&&.
Step 3 — Split the middle term:
&&2x^2 + 7x + 3&&
&&= 2x^2 + 6x + x + 3&&
Step 4 — Group and factor:
&&= 2x(x + 3) + 1(x + 3)&&
The common binomial factor is &&(x + 3)&&.
&&= (2x + 1)(x + 3)&&

✓ Result: &&2x^2 + 7x + 3 = (2x + 1)(x + 3)&&

⚠ Caution: Verify by expanding: &&(2x + 1)(x + 3) = 2x^2 + 6x + x + 3 = 2x^2 + 7x + 3&& ✓

Source: BYJU’S NCERT Solutions Ex 2.4 TiwaraAcademy Ch 2 Ex 2.4

Part (iii): Factorise &&6x^2 + 5x – 6&&

Step 1 — Identify coefficients:
Here, &&a = 6&&, &&b = 5&&, &&c = -6&&.
Compute &&a \times c = 6 \times (-6) = -36&&.
Step 2 — Find the split:
We need &&p \times q = -36&& and &&p + q = 5&&.
The numbers are &&9&& and &&-4&&, because &&9 \times (-4) = -36&& and &&9 + (-4) = 5&&.
Step 3 — Split the middle term:
&&6x^2 + 5x – 6&&
&&= 6x^2 + 9x – 4x – 6&&
Step 4 — Group and factor:
&&= 3x(2x + 3) – 2(2x + 3)&&
The common binomial factor is &&(2x + 3)&&.
&&= (3x – 2)(2x + 3)&&

✓ Result: &&6x^2 + 5x – 6 = (3x – 2)(2x + 3)&&

⚠ Caution: When &&c&& is negative, the product &&a \times c&& is negative, so one of &&p, q&& must be negative. Be careful with sign selection.

Source: BYJU’S NCERT Solutions Ex 2.4 CueMath Class 9 Ch 2

Part (iv): Factorise &&3x^2 – x – 4&&

Step 1 — Identify coefficients:
Here, &&a = 3&&, &&b = -1&&, &&c = -4&&.
Compute &&a \times c = 3 \times (-4) = -12&&.
Step 2 — Find the split:
We need &&p \times q = -12&& and &&p + q = -1&&.
The numbers are &&-4&& and &&3&&, because &&(-4) \times 3 = -12&& and &&(-4) + 3 = -1&&.
Step 3 — Split the middle term:
&&3x^2 – x – 4&&
&&= 3x^2 – 4x + 3x – 4&&
Step 4 — Group and factor:
&&= x(3x – 4) + 1(3x – 4)&&
The common binomial factor is &&(3x – 4)&&.
&&= (x + 1)(3x – 4)&&

✓ Result: &&3x^2 – x – 4 = (x + 1)(3x – 4)&&

⚠ Caution: Verify: &&(x + 1)(3x – 4) = 3x^2 – 4x + 3x – 4 = 3x^2 – x – 4&& ✓

Source: BYJU’S NCERT Solutions Ex 2.4 TiwaraAcademy Ch 2 Ex 2.4 CueMath Class 9 Ch 2

Conclusion and Key Points

In Question 4 of Exercise 2.4, all four quadratic polynomials were successfully factorised using the splitting the middle term method. The final answers are:

  • &&12x^2 – 7x + 1 = (4x – 1)(3x – 1)&&
  • &&2x^2 + 7x + 3 = (2x + 1)(x + 3)&&
  • &&6x^2 + 5x – 6 = (3x – 2)(2x + 3)&&
  • &&3x^2 – x – 4 = (x + 1)(3x – 4)&&
💡 Trick Box — Quick Tip!

To quickly find the split numbers for &&ax^2 + bx + c&&: list all factor pairs of &&|a \times c||&&, then check which pair adds/subtracts to give &&b&&. Always verify your answer by expanding the factors back.

📋 Points to Remember
  • For &&ax^2 + bx + c&&, find &&p, q&& such that &&pq = ac&& and &&p + q = b&&.
  • When &&c > 0&& and &&b < 0&&: both &&p, q&& are negative.
  • When &&c < 0&&: &&p&& and &&q&& have opposite signs.
  • Always verify by expanding the factored form.
  • The Factor Theorem can also be used to verify: if &&x = r&& is a root, then &&(x – r)&& is a factor.

FAQ — Frequently Asked Questions

Q1. What method is used in Exercise 2.4 Question 4?

The splitting the middle term method is used. For &&ax^2 + bx + c&&, we find &&p&& and &&q&& such that &&p \times q = ac&& and &&p + q = b&&, then rewrite and factor by grouping. BYJU’S CueMath

Q2. How do you factorise &&12x^2 – 7x + 1&&?

Split &&-7x&& as &&-4x – 3x&&. Then: &&12x^2 – 4x – 3x + 1 = 4x(3x – 1) – 1(3x – 1) = (4x – 1)(3x – 1)&&. BYJU’S

Q3. How do you factorise &&2x^2 + 7x + 3&&?

Split &&7x&& as &&6x + x&&. Then: &&2x^2 + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)&&. TiwaraAcademy

Q4. How do you factorise &&6x^2 + 5x – 6&&?

Since &&a \times c = -36&&, split &&5x&& as &&9x – 4x&&. Then: &&3x(2x + 3) – 2(2x + 3) = (3x – 2)(2x + 3)&&. BYJU’S CueMath

Q5. How do you factorise &&3x^2 – x – 4&&?

Since &&a \times c = -12&&, split &&-x&& as &&-4x + 3x&&. Then: &&x(3x – 4) + 1(3x – 4) = (x + 1)(3x – 4)&&. BYJU’S TiwaraAcademy

Q6. What is the Factor Theorem and how is it related?

The Factor Theorem states that &&(x – a)&& is a factor of &&p(x)&& if &&p(a) = 0&&. It can be used to verify factors found by splitting the middle term. CueMath

Q7. Can we verify factorisation by expanding?

Yes. Expand &&(4x – 1)(3x – 1) = 12x^2 – 3x – 4x + 1 = 12x^2 – 7x + 1&& ✓. Always expand to confirm. BYJU’S

Q8. How many questions are in Exercise 2.4 of Class 9 Maths?

Exercise 2.4 has 5 questions (with subparts), covering factorisation using Factor Theorem and splitting the middle term. CueMath TiwaraAcademy

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