Introduction to Linear Polynomials: Important Questions and Answers for Class 9

Important Questions on Introduction to Linear Polynomials for Class 9 are provided here to help CBSE students prepare effectively for their exams. These NCERT-based practice questions from Class 9 Maths Chapter 2: Introduction to Linear Polynomials are designed to strengthen the conceptual understanding and to improve problem-solving skills.

Students can use these CBSE Class 9 Maths important questions to understand the exam pattern, revise key concepts, and build confidence in solving different types of polynomial problems.

Table of Contents

• Key Concepts of Class 9 Polynomials
• Important Questions on Polynomials
• Important Questions on Remainder Theorem
• Important Questions on Factor Theorem
• Important Questions on Algebraic Identities
• HOTS & Previous Questions on Polynomials for Class 9

Key Concepts of Class 9 Polynomials

Linear Polynomial: A linear polynomial is a polynomial of degree 1. Its general form is:

p(x) = ax + b, where a ≠ 0

The graph of a linear polynomial is always a straight line. This is where the name ‘linear’ comes from.

Zero of a Polynomial: The zero of a polynomial p(x) is the value of x for which p(x) = 0.

For a linear polynomial p(x) = ax + b:

Zero = −b/a

A linear polynomial has exactly one zero.

Remainder Theorem: If a polynomial p(x) is divided by (x − a), then the remainder = p(a).

Factor Theorem: (x − a) is a factor of p(x) if and only if p(a) = 0.

If the remainder is zero, the divisor is a factor.

Key Algebraic Identities: 

These are used extensively in polynomial problems:

These important questions on Introduction to Linear Polynomials for Class 9 help students strengthen their understanding of polynomial concepts, improve problem-solving skills, and prepare effectively for CBSE Class 9 Maths examinations.

Important Questions on Polynomials

Q1. What is the degree of the polynomial 5x³ − 4x² + 7x − 2?

Answer: The highest power of x is 3, so the degree is 3 (it is a cubic polynomial).

Q2: What is the value of the polynomial p(x) = x² − 3x + 2 at x = 1?

Answer: p(1) = (1)² − 3(1) + 2 = 1 − 3 + 2 = 0

Q3: Is √2 a polynomial?

Answer: Yes. √2 is a constant polynomial of degree 0. It can be written as p(x) = √2, which is a real number with no variable term.

Q4: Find the zero of the linear polynomial p(x) = 5x + 15. Verify your answer.

Answer: Set p(x) = 0:

⇒ 5x + 15 = 0

⇒ 5x = −15

⇒ x = −3

Verification: p(−3) = 5(−3) + 15 = −15 + 15 = 0 

The zero of p(x) = 5x + 15 is −3.

Important Questions on Remainder Theorem 

Q1. Find the remainder when p(x) = x³ − ax² + 6x − a is divided by (x − a).

Answer: By the Remainder Theorem, remainder = p(a):

p(a) = a³ − a(a)² + 6(a) − a

= a³ − a³ + 6a − a

= 5a

The remainder is 5a.

Q2. If p(x) = x³ − 3x² + kx + 7 gives a remainder of 3 when divided by (x − 2), find the value of k.

Answer: By the Remainder Theorem, p(2) = 3:

⇒ (2)³ − 3(2)² + k(2) + 7 = 3

⇒ 8 − 12 + 2k + 7 = 3

⇒ 3 + 2k = 3

⇒ 2k = 0

⇒ k = 0

Q3: If the polynomial cx² + 5x − 3 gives a remainder of 1 when divided by (x − 1), find c.

Answer: p(1) = 1:

⇒ c(1)² + 5(1) − 3 = 1

⇒ c + 2 = 1

c = −1

Q4: By the Remainder Theorem, the remainder when p(x) = x² + 3x + 5 is divided by (x + 1) is:

(A) 5   (B) 3   (C) 9   (D) 7

Answer: (B) 3

p(−1) = (−1)² + 3(−1) + 5 = 1 − 3 + 5 = 3.

Important Questions on Factor Theorem 

Q1. Using the Factor Theorem, show that (x − 5) is a factor of x³ − 125.

Answer: p(5) = (5)³ − 125 = 125 − 125 = 0

Since p(5) = 0, by the Factor Theorem, (x − 5) is a factor of x³ − 125.

Q2. Determine whether (2x + 1) is a factor of p(x) = 2x³ + 5x² − x − 6.

Answer: 2x + 1 = 0 ⇒ x = −1/2

p(−1/2) = 2(−1/8) + 5(1/4) − (−1/2) − 6

= −1/4 + 5/4 + 1/2 − 6

= 4/4 + 1/2 − 6

= 1 + 0.5 − 6

= −4.5 ≠ 0

Since p(−1/2) ≠ 0, (2x + 1) is NOT a factor of p(x).

Q3: If both x − 2 and x − 1/2 are factors of px² + 5x + r, show that p = r.

Answer: Let f(x) = px² + 5x + r.

Since (x − 2) is a factor, f(2) = 0:

⇒ p(4) + 5(2) + r = 0

⇒ 4p + r = −10 ... (i)

Since (x − 1/2) is a factor, f(1/2) = 0.

p(1/4) + 5(1/2) + r = 0

⇒ p/4 + 5/2 + r = 0

Multiplying by 4: p + 10 + 4r = 0

p + 4r = −10 ... (ii)

From (i) and (ii): 4p + r = p + 4r

⇒ 3p = 3r

∴ p = r (Hence Proved)

Q 4: Find the value of k such that (x − 1) is a factor of p(x) = kx² − 3x + k.

Answer: p(1) = 0:

 k(1) − 3(1) + k = 0

⇒ k − 3 + k = 0

⇒ 2k = 3

⇒ k = 3/2

Important Questions on Algebraic Identities 

Q1. Expand (3a − 7b − c)².

Answer: Using (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca,

with a = 3a, b = −7b, c = −c:

= (3a)² + (−7b)² + (−c)² + 2(3a)(−7b) + 2(−7b)(−c) + 2(−c)(3a)

= 9a² + 49b² + c² − 42ab + 14bc − 6ca

Q2. Evaluate 103 × 97 using identities.

Answer: 103 × 97 = (100 + 3)(100 − 3)

Using a² − b² = (a + b)(a − b):

= (100)² − (3)²

= 10000 − 9

= 9991

Q3: Using suitable algebraic identities, evaluate the following:

(i) (102)³  (ii) (999)²  (iii) 9x² + 4y² if xy = 6 and 3x + 2y = 12

Answer:

(i) (102)³ = (100 + 2)³

Using (a + b)³ = a³ + b³ + 3ab(a + b):

= (100)³ + (2)³ + 3(100)(2)(102)

= 1000000 + 8 + 600 × 102

= 1000000 + 8 + 61,200

= 1061208

(ii) (999)² = (1000 − 1)²

Using (a − b)² = a² − 2ab + b²:

= (1000)² − 2(1000)(1) + (1)²

= 1000000 − 2,000 + 1

= 998001

(iii) Given: 3x + 2y = 12 and xy = 6

Squaring both sides of 3x + 2y = 12:

(3x + 2y)² = 144

⇒ 9x² + 12xy + 4y² = 144

⇒ 9x² + 4y² = 144 − 12xy

= 144 − 12(6)

= 144 − 72

= 72

Q4. If a + b + c = 9 and a² + b² + c² = 35, find ab + bc + ca. Also, find the value of a³ + b³ + c³ − 3abc.

Answer: Using (a + b + c)² = a² + b² + c² + 2(ab + bc + ca):

81 = 35 + 2(ab + bc + ca)

⇒ 2(ab + bc + ca) = 46

⇒ ab + bc + ca = 23

Now, using a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca):

= 9 × (35 − 23)

= 9 × 12

= 108

HOTS & Previous Questions on Polynomials for Class 9

Q1: Prove that 2x⁴ − 5x³ + 2x² − x + 2 is divisible by x² − 3x + 2.

Answer: First, factorise the divisor x² − 3x + 2:

x² − 3x + 2 = x² − 2x − x + 2

= x(x − 2) − 1(x − 2)

= (x − 1)(x − 2)

So we need to show x = 1 and x = 2 are both zeros of p(x) = 2x⁴ − 5x³ + 2x² − x + 2.

Checking x = 2:

p(2) = 2(16) − 5(8) + 2(4) − 2 + 2

= 32 − 40 + 8 − 2 + 2

= 0 

⇒ (x − 2) is a factor.

Checking x = 1:

p(1) = 2(1) − 5(1) + 2(1) − 1 + 2

= 2 − 5 + 2 − 1 + 2

= 0 

⇒ (x − 1) is a factor.

Since both (x − 1) and (x − 2) are factors of p(x), their product (x − 1)(x − 2) = x² − 3x + 2 is also a factor.

∴ 2x⁴ − 5x³ + 2x² − x + 2 is divisible by x² − 3x + 2. Hence proved

Q2. Factorise completely: x³ − 23x² + 142x − 120

Answer: Let p(x) = x³ − 23x² + 142x − 120.

Try x = 1: p(1) = 1 − 23 + 142 − 120 = 0

So (x − 1) is a factor.

Divide p(x) by (x − 1):

x³ − 23x² + 142x − 120 = (x − 1)(x² − 22x + 120)

Now factorise x² − 22x + 120:

x² − 22x + 120 = (x − 10)(x − 12)

∴ x³ − 23x² + 142x − 120 = (x − 1)(x − 10)(x − 12)

Q 3: If p(x) = x² − 5x + k and p(2) = p(3), find k. Are x = 2 and x = 3 zeros of p(x)?

Answer: p(2) = 4 − 10 + k = k − 6

p(3) = 9 − 15 + k = k − 6

Since p(2) = p(3) for all values of k, this is true for any k.

Now check: are x = 2 and x = 3 zeros

p(2) = 0 ⇒ k − 6 = 0 ⇒ k = 6

p(3) = 0 ⇒ k − 6 = 0 ⇒ k = 6

Yes, when k = 6, both x = 2 and x = 3 are zeros of p(x) = x² − 5x + 6.

Q 4: The polynomial p(x) = x³ + ax² + bx + c has zeros at x = −1, x = 2, and x = −3. Find a, b, and c.

Answer: Since −1, 2, −3 are zeros:

p(x) = (x + 1)(x − 2)(x + 3)

Expand step by step:

(x + 1)(x − 2) = x² − x − 2

⇒ (x² − x − 2)(x + 3) = x³ + 3x² − x² − 3x − 2x − 6

= x³ + 2x² − 5x − 6

So a = 2, b = −5, c = −6.