Class 9 Maths Chapter 2 Polynomials – Exercise 2.4 NCERT Solutions

Introduction

Exercise 2.4 focuses on factorization of polynomials. You’ll learn how to split middle terms, apply standard identities, and use the factor theorem to find roots and factors.

Key Formula Used

Standard Identities: $(x + y)^{2} = x^{2} + 2 x y + y^{2}$ $(x - y)^{2} = x^{2} - 2 x y + y^{2}$ $(x + a) (x + b) = x^{2} + (a + b) x + a b$ $(x + y) (x - y) = x^{2} - y^{2}$
Factor Theorem: If $p (a) = 0$ , then $(x - a)$ is a factor of $p (x)$ .

Solved Questions (Step by Step)

Q1. Factorize $x^{2} + 5 x + 6$ .

Solution:

Split middle term: $x^{2} + 2 x + 3 x + 6$ .
Group: $(x^{2} + 2 x) + (3 x + 6)$ .
Factor: $x (x + 2) + 3 (x + 2)$ .
Final: $(x + 2) (x + 3)$ .

Q2. Factorize $x^{2} - 4$ .

Solution:

Identity: $x^{2} - 4 = (x - 2) (x + 2)$ .

Q3. Factorize $x^{2} + 7 x + 10$ .

Solution:

Split middle term: $x^{2} + 5 x + 2 x + 10$ .
Group: $(x^{2} + 5 x) + (2 x + 10)$ .
Factor: $x (x + 5) + 2 (x + 5)$ .
Final: $(x + 5) (x + 2)$ .

Q4. Factorize $x^{2} - 9$ .

Solution:

Identity: $x^{2} - 9 = (x - 3) (x + 3)$ .

Q5. Factorize $x^{2} + x - 6$ .

Solution:

Split middle term: $x^{2} + 3 x - 2 x - 6$ .
Group: $(x^{2} + 3 x) - (2 x + 6)$ .
Factor: $x (x + 3) - 2 (x + 3)$ .
Final: $(x + 3) (x - 2)$ .

Q6. Factorize $x^{2} - 5 x + 6$ .

Solution:

Split middle term: $x^{2} - 2 x - 3 x + 6$ .
Group: $(x^{2} - 2 x) - (3 x - 6)$ .
Factor: $x (x - 2) - 3 (x - 2)$ .
Final: $(x - 2) (x - 3)$ .

Q7. Factorize $x^{2} + 4 x + 4$ .

Solution:

Identity: $(x + 2)^{2}$ .

Q8. Factorize $x^{2} - 2 x + 1$ .

Solution:

Identity: $(x - 1)^{2}$ .

Q9. Factorize $x^{3} - 1$ .

Solution:

Identity: $x^{3} - 1 = (x - 1) (x^{2} + x + 1)$ .

Q10. Factorize $x^{3} + 1$ .

Solution:

Identity: $x^{3} + 1 = (x + 1) (x^{2} - x + 1)$ .

Q11. Factorize $x^{3} - 8$ .

Solution:

Identity: $x^{3} - 8 = (x - 2) (x^{2} + 2 x + 4)$ .

Q12. Factorize $x^{3} + 27$ .

Solution:

Identity: $x^{3} + 27 = (x + 3) (x^{2} - 3 x + 9)$ .

Q13. Factorize $x^{3} - 6 x^{2} + 11 x - 6$ .

Solution:

Check root: $p (1) = 0$ .
Factor: $(x - 1) (x^{2} - 5 x + 6)$ .
Further: $(x - 1) (x - 2) (x - 3)$ .

Q14. Factorize $x^{3} - 3 x^{2} + 3 x - 1$ .

Solution:

Identity: $(x - 1)^{3}$ .

Q15. Factorize $x^{4} - 1$ .

Solution:

Identity: $x^{4} - 1 = (x^{2} - 1) (x^{2} + 1)$ .
Further: $(x - 1) (x + 1) (x^{2} + 1)$ .

FAQs (10 for Exercise 2.4)

Q: What is factorization? A: Breaking a polynomial into simpler polynomials whose product equals the original.
Q: What is the factor theorem? A: If $p (a) = 0$ , then $(x - a)$ is a factor of $p (x)$ .
Q: Why do we split the middle term? A: To make grouping easier for quadratic polynomials.
Q: What is the difference between identity and factorization? A: Identity is a formula; factorization applies it to simplify polynomials.
Q: Can cubic polynomials be factorized? A: Yes, using identities or factor theorem.
Q: What is a perfect square polynomial? A: A polynomial like $x^{2} + 4 x + 4$ that equals $(x + 2)^{2}$ .
Q: What is the degree of factors compared to the original polynomial? A: The sum of degrees of factors equals the degree of the original.
Q: Why is factorization important? A: It helps solve equations and find roots.
Q: Can all polynomials be factorized? A: Yes, though some factors may be complex.
Q: What is the simplest factorization identity? A: $x^{2} - y^{2} = (x - y) (x + y)$ .

Conclusion

Exercise 2.4 has 15 questions that strengthen your skills in factorization using identities and the factor theorem. This completes Chapter 2 of Class 9 Maths.

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