Class 9 Maths Chapter 2 Polynomials – Exercise 2.2 NCERT Solutions

Introduction

Exercise 2.2 focuses on evaluating polynomials at specific values and verifying standard identities. You’ll practice substituting values into polynomials and checking equalities.

Key Formula Used

Polynomial Evaluation: If $p (x) = a_{n} x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{0}$ , then

p(k)=ankn+an−1kn−1+⋯+a0

Standard Identities: $(x + y)^{2} = x^{2} + 2 x y + y^{2}$ $(x - y)^{2} = x^{2} - 2 x y + y^{2}$ $(x + a) (x + b) = x^{2} + (a + b) x + a b$

Solved Questions (Step by Step)

Q1. Evaluate $p (x) = x^{2} + 2 x + 1$ at $x = 1$ .

Solution:

Substitute $x = 1$ .
$p (1) = (1)^{2} + 2 (1) + 1 = 1 + 2 + 1 = 4$ .

Q2. Evaluate $p (x) = x^{2} - 3 x + 2$ at $x = 2$ .

Solution:

$p (2) = (2)^{2} - 3 (2) + 2 = 4 - 6 + 2 = 0$ .

Q3. Evaluate $p (x) = x^{3} + x^{2} + x + 1$ at $x = - 1$ .

Solution:

$p (- 1) = (- 1)^{3} + (- 1)^{2} + (- 1) + 1 = - 1 + 1 - 1 + 1 = 0$ .

Q4. Verify $(x + 1)^{2} = x^{2} + 2 x + 1$ .

Solution:

LHS = $(x + 1)^{2} = x^{2} + 2 x + 1$ .
RHS = $x^{2} + 2 x + 1$ .
LHS = RHS. Verified.

Q5. Verify $(x - 2)^{2} = x^{2} - 4 x + 4$ .

Solution:

LHS = $(x - 2)^{2} = x^{2} - 4 x + 4$ .
RHS = $x^{2} - 4 x + 4$ .
Verified.

Q6. Verify $(x + 3) (x + 4) = x^{2} + 7 x + 12$ .

Solution:

LHS = $x^{2} + 4 x + 3 x + 12 = x^{2} + 7 x + 12$ .
RHS = $x^{2} + 7 x + 12$ .
Verified.

Q7. Evaluate $p (x) = x^{2} + 5 x + 6$ at $x = - 2$ .

Solution:

$p (- 2) = (- 2)^{2} + 5 (- 2) + 6 = 4 - 10 + 6 = 0$ .

Q8. Evaluate $p (x) = x^{3} - 2 x^{2} + x - 2$ at $x = 2$ .

Solution:

$p (2) = 8 - 8 + 2 - 2 = 0$ .

Q9. Verify $(x + 5) (x - 5) = x^{2} - 25$ .

Solution:

LHS = $x^{2} - 25$ .
RHS = $x^{2} - 25$ .
Verified.

Q10. Evaluate $p (x) = x^{2} + 4 x + 4$ at $x = - 2$ .

Solution:

$p (- 2) = (- 2)^{2} + 4 (- 2) + 4 = 4 - 8 + 4 = 0$ .

Q11. Verify $(x + 2)^{2} = x^{2} + 4 x + 4$ .

Solution:

LHS = $x^{2} + 4 x + 4$ .
RHS = $x^{2} + 4 x + 4$ .
Verified.

Q12. Evaluate $p (x) = x^{3} + 3 x^{2} + 3 x + 1$ at $x = - 1$ .

Solution:

$p (- 1) = - 1 + 3 (1) - 3 + 1 = 0$ .

Q13. Verify $(x - 3) (x + 3) = x^{2} - 9$ .

Solution:

LHS = $x^{2} - 9$ .
RHS = $x^{2} - 9$ .
Verified.

Q14. Evaluate $p (x) = x^{2} - 2 x + 1$ at $x = 1$ .

Solution:

$p (1) = 1 - 2 + 1 = 0$ .

Q15. Verify $(x + a) (x + b) = x^{2} + (a + b) x + a b$ .

Solution:

LHS = $x^{2} + a x + b x + a b = x^{2} + (a + b) x + a b$ .
RHS = same. Verified.

FAQs (10 for Exercise 2.2)

Q: What does it mean to evaluate a polynomial? A: Substituting a given value of $x$ into the polynomial.
Q: How do you check if a polynomial equals zero at a value? A: Substitute the value; if result = 0, that value is a root.
Q: What is a root of a polynomial? A: A value of $x$ for which $p (x) = 0$ .
Q: Why do we verify identities? A: To confirm algebraic formulas hold true for all values of $x$ .
Q: Is $(x + y)^{2}$ always equal to $x^{2} + 2 x y + y^{2}$ ? A: Yes, for all real numbers $x, y$ .
Q: What is the difference between evaluating and verifying? A: Evaluating means substituting; verifying means proving equality.
Q: Can a polynomial have more than one root? A: Yes, depending on its degree.
Q: What is the degree of $x^{2} - 2 x + 1$ ? A: Degree = 2.
Q: Why do many answers in this exercise equal zero? A: Because those values are roots of the polynomial.
Q: Are identities useful beyond exams? A: Yes, they simplify algebraic calculations in higher mathematics.

Conclusion

Exercise 2.2 has 15 questions that strengthen your skills in evaluating polynomials and verifying identities. These basics prepare you for factorization and advanced algebra.

Visit:www.fuzymathacademy.com

FUZY MATH ACADEMY

Search This Blog