Class 10 Maths Chapter 2 Polynomials – Exercise 2.4 NCERT Solutions

Introduction

Exercise 2.4 focuses on cubic polynomials and the application of the Factor Theorem. You’ll learn how to find all zeros of a cubic polynomial by first identifying one zero, then reducing it to a quadratic polynomial, and finally solving for the remaining zeros.

Formula Used

• Factor Theorem: If $f(a)=0$, then $(x-a)$ is a factor of $f(x)$.

• Division Algorithm for Polynomials:

$$f(x)=g(x)\cdot q(x)+r(x)$$

where remainder $r(x)=0$ if $g(x)$ is a factor.

• Quadratic Formula (for remaining zeros):

$$\alpha ,\beta =\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$

NCERT Questions with Solutions (10)

Q1. Find all zeros of $x^3-3x^2-4x+12$.

Solution: Check $f(2)=0$. So, $(x-2)$ is a factor. Divide: $x^3-3x^2-4x+12$ ÷ $(x-2)$ = $x^2-x-6$. Factorize: $(x-3)(x+2)$. Zeros = 2, 3, -2.

Q2. Find all zeros of $x^3-6x^2+11x-6$.

Solution: Check $f(1)=0$. So, $(x-1)$ is a factor. Divide: Quotient = $x^2-5x+6$. Factorize: $(x-2)(x-3)$. Zeros = 1, 2, 3.

Q3. Find all zeros of $x^3-4x^2-11x+30$.

Solution: Check $f(2)=0$. So, $(x-2)$ is a factor. Divide: Quotient = $x^2-2x-15$. Factorize: $(x-5)(x+3)$. Zeros = 2, 5, -3.

Q4. Find all zeros of $x^3-7x^2+14x-8$.

Solution: Check $f(2)=0$. So, $(x-2)$ is a factor. Divide: Quotient = $x^2-5x+4$. Factorize: $(x-4)(x-1)$. Zeros = 2, 4, 1.

Q5. Find all zeros of $x^3+x^2-4x-4$.

Solution: Check $f(-1)=0$. So, $(x+1)$ is a factor. Divide: Quotient = $x^2-4$. Factorize: $(x-2)(x+2)$. Zeros = -1, 2, -2.

Q6. Find all zeros of $x^3-3x^2-x+3$.

Solution: Check $f(1)=0$. So, $(x-1)$ is a factor. Divide: Quotient = $x^2-2x-3$. Factorize: $(x-3)(x+1)$. Zeros = 1, 3, -1.

Q7. Find all zeros of $x^3-2x^2-5x+6$.

Solution: Check $f(1)=0$. So, $(x-1)$ is a factor. Divide: Quotient = $x^2-x-6$. Factorize: $(x-3)(x+2)$. Zeros = 1, 3, -2.

Q8. Find all zeros of $x^3-9x^2+26x-24$.

Solution: Check $f(2)=0$. So, $(x-2)$ is a factor. Divide: Quotient = $x^2-7x+12$. Factorize: $(x-3)(x-4)$. Zeros = 2, 3, 4.

Q9. Find all zeros of $x^3-5x^2-2x+24$.

Solution: Check $f(2)=0$. So, $(x-2)$ is a factor. Divide: Quotient = $x^2-3x-12$. Factorize: $(x-6)(x+2)$. Zeros = 2, 6, -2.

Q10. Find all zeros of $x^3+3x^2-4x-12$.

Solution: Check $f(-2)=0$. So, $(x+2)$ is a factor. Divide: Quotient = $x^2+x-6$. Factorize: $(x+3)(x-2)$. Zeros = -2, -3, 2.

FAQs (10 from NCERT)

1. Q: What is a cubic polynomial? A: A polynomial of degree 3.

2. Q: How many zeros does a cubic polynomial have? A: Exactly 3 zeros (real or complex).

3. Q: What is the factor theorem? A: If $f(a)=0$, then $(x-a)$ is a factor of $f(x)$.

4. Q: What is the division algorithm for polynomials? A: $f(x)=g(x)q(x)+r(x)$.

5. Q: How do you find all zeros of a cubic polynomial? A: Find one zero, divide, then solve quadratic factor.

6. Q: What is the relation between coefficients and zeros of a cubic polynomial? A: $\alpha +\beta +\gamma =-\frac{b}{a}$, etc.

7. Q: Can cubic polynomials have repeated zeros? A: Yes, if discriminant conditions allow.

8. Q: What is synthetic division? A: A shortcut method for dividing polynomials.

9. Q: What is the importance of factorization? A: It simplifies solving polynomial equations.

10. Q: Why is this exercise important? A: It builds skills in solving cubic equations and applying factor theorem.

Conclusion

Exercise 2.4 has 10 solved questions and 10 FAQs that strengthen your understanding of cubic polynomials and factor theorem. This builds the foundation for solving higher‑degree polynomial equations in Class 10 Maths.

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