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Polynomials – CBSE NCERT Study Resources

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9th

9th - Mathematics

Polynomials

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Chapter Overview: Polynomials

This chapter introduces the concept of polynomials, their types, degrees, and operations such as addition, subtraction, multiplication, and factorization. Students will learn about the Remainder Theorem, Factor Theorem, and algebraic identities to simplify polynomial expressions.

Polynomial: An algebraic expression consisting of variables and coefficients, involving only the operations of addition, subtraction, multiplication, and non-negative integer exponents.

Key Topics Covered

  • Introduction to Polynomials
  • Types of Polynomials (Linear, Quadratic, Cubic)
  • Degree of a Polynomial
  • Zeroes of a Polynomial
  • Remainder Theorem
  • Factor Theorem
  • Algebraic Identities
  • Factorization of Polynomials

Detailed Explanation

1. Introduction to Polynomials

A polynomial is an expression of the form: P(x) = anxn + an-1xn-1 + ... + a1x + a0, where an, an-1, ..., a0 are coefficients and n is a non-negative integer.

Degree of a Polynomial: The highest power of the variable in the polynomial. For example, in 3x2 + 5x + 2, the degree is 2.

2. Types of Polynomials

  • Linear Polynomial: Degree 1 (e.g., 2x + 3)
  • Quadratic Polynomial: Degree 2 (e.g., x2 + 5x + 6)
  • Cubic Polynomial: Degree 3 (e.g., 4x3 - x2 + 7)

3. Zeroes of a Polynomial

A real number k is called a zero of the polynomial P(x) if P(k) = 0. For example, 2 is a zero of P(x) = x - 2.

4. Remainder Theorem

If a polynomial P(x) is divided by (x - a), the remainder is P(a).

Factor Theorem: (x - a) is a factor of polynomial P(x) if and only if P(a) = 0.

5. Algebraic Identities

Some important identities include:

  • (a + b)2 = a2 + 2ab + b2
  • (a - b)2 = a2 - 2ab + b2
  • a2 - b2 = (a + b)(a - b)
  • (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca

6. Factorization of Polynomials

Methods include splitting the middle term, using algebraic identities, and grouping terms.

All Question Types with Solutions – CBSE Exam Pattern

Explore a complete set of CBSE-style questions with detailed solutions, categorized by marks and question types. Ideal for exam preparation, revision and practice.

Very Short Answer (1 Mark) – with Solutions (CBSE Pattern)

These are 1-mark questions requiring direct, concise answers. Ideal for quick recall and concept clarity.

Question 1:
What is the degree of the polynomial 3x² + 5x + 7?
Answer:
Numeric answer:
2
Question 2:
Identify the coefficient of x³ in 2x³ - 4x + 1.
Answer:
Numeric answer:
2
Question 3:
What is the zero of the polynomial p(x) = x - 5?
Answer:
Numeric answer:
5
Question 4:
Classify 4x + 9 as linear or quadratic polynomial.
Answer:
Linear polynomial
Question 5:
Find the value of p(2) for p(x) = x² - 3x + 2.
Answer:
Numeric answer:
0
Question 6:
Is x² + 1/x a polynomial? Justify.
Answer:
No, negative exponent
Question 7:
What is the standard form of 5 + 3x² - x?
Answer:
3x² - x + 5
Question 8:
How many terms are in 7x³ - 2x² + 4x - 1?
Answer:
Numeric answer:
4
Question 9:
If p(x) = x² - 4, find p(0).
Answer:
Numeric answer:
-4
Question 10:
Name the type of polynomial 2x + 3.
Answer:
Linear polynomial
Question 11:
What is the remainder when x³ - 1 is divided by x - 1?
Answer:
Numeric answer:
0
Question 12:
Factorize: x² - 5x + 6.
Answer:
(x - 2)(x - 3)
Question 13:
Define a polynomial with one variable.
Answer:

A polynomial in one variable is an algebraic expression of the form aₙxⁿ + aₙ₋₁xⁿ⁻¹ + ... + a₁x + a₀, where aₙ, aₙ₋₁, ..., a₀ are constants (coefficients), x is the variable, and n is a non-negative integer (degree).

Question 14:
What is the degree of the polynomial 4x³ - 2x² + 7x - 1?
Answer:

The degree of the polynomial 4x³ - 2x² + 7x - 1 is 3, because the highest power of the variable x is 3.

Question 15:
Identify whether x² + √2x + 3 is a polynomial. Justify your answer.
Answer:

Yes, x² + √2x + 3 is a polynomial because all the exponents of x are non-negative integers, and the coefficients (1, √2, 3) are real numbers.

Question 16:
Find the zero of the polynomial p(x) = 3x - 5.
Answer:

To find the zero of p(x) = 3x - 5, set p(x) = 0:
3x - 5 = 0
3x = 5
x = 5/3.
Thus, the zero is 5/3.

Question 17:
Classify the polynomial 5x + 2 based on its degree.
Answer:

The polynomial 5x + 2 is a linear polynomial because its highest degree is 1.

Question 18:
If p(x) = x² - 4x + 4, evaluate p(2).
Answer:

Substitute x = 2 in p(x):
p(2) = (2)² - 4(2) + 4
= 4 - 8 + 4
= 0.
Thus, p(2) = 0.

Question 19:
What is the remainder when x³ - 2x² + x + 1 is divided by x - 1?
Answer:

Using the Remainder Theorem, substitute x = 1 in the polynomial:
p(1) = (1)³ - 2(1)² + (1) + 1
= 1 - 2 + 1 + 1
= 1.
The remainder is 1.

Question 20:
Give an example of a quadratic polynomial with no real zeros.
Answer:

An example is x² + 1. It has no real zeros because the equation x² + 1 = 0 gives x² = -1, which has no real solutions.

Question 21:
Factorize the polynomial x² - 5x + 6.
Answer:

To factorize x² - 5x + 6, find two numbers that multiply to 6 and add to -5:
x² - 2x - 3x + 6
= x(x - 2) - 3(x - 2)
= (x - 2)(x - 3).
The factors are (x - 2) and (x - 3).

Question 22:
What is the coefficient of x² in the polynomial 2 - x² + x³?
Answer:

The coefficient of x² in 2 - x² + x³ is -1, as the term is -x².

Very Short Answer (2 Marks) – with Solutions (CBSE Pattern)

These 2-mark questions test key concepts in a brief format. Answers are expected to be accurate and slightly descriptive.

Question 1:
Define a polynomial and give an example of a quadratic polynomial.
Answer:

A polynomial is an algebraic expression consisting of variables and coefficients, combined using addition, subtraction, and multiplication, with non-negative integer exponents.
Example of a quadratic polynomial: 2x² + 3x - 5, where the highest power of the variable is 2.

Question 2:
What is the degree of the polynomial 4x³ - 7x² + 2x + 1?
Answer:

The degree of a polynomial is the highest power of the variable in the expression.
For 4x³ - 7x² + 2x + 1, the highest power is 3.
Thus, the degree is 3.

Question 3:
Find the value of the polynomial p(x) = x² - 5x + 6 at x = 2.
Answer:

Substitute x = 2 into the polynomial:
p(2) = (2)² - 5(2) + 6
= 4 - 10 + 6
= 0.

Question 4:
Identify whether x + 1/x is a polynomial. Justify your answer.
Answer:

x + 1/x is not a polynomial because it contains a term with a negative exponent (1/x = x⁻¹).
Polynomials only allow non-negative integer exponents.

Question 5:
If p(x) = 3x - 4, find p(0) + p(1).
Answer:

First, calculate p(0):
3(0) - 4 = -4.
Next, calculate p(1):
3(1) - 4 = -1.
Now, add them: -4 + (-1) = -5.

Question 6:
Write the standard form of a linear polynomial and give an example.
Answer:

The standard form of a linear polynomial is ax + b, where a ≠ 0.
Example: 5x + 2, where the highest power of x is 1.

Question 7:
Check whether x = -2 is a zero of the polynomial p(x) = x² + 3x + 2.
Answer:

Substitute x = -2 into p(x):
p(-2) = (-2)² + 3(-2) + 2
= 4 - 6 + 2
= 0.
Since the result is 0, x = -2 is a zero of the polynomial.

Question 8:
What is the remainder when the polynomial p(x) = x³ - 4x² + 2x - 5 is divided by (x - 1)?
Answer:

Using the Remainder Theorem, substitute x = 1 into p(x):
p(1) = (1)³ - 4(1)² + 2(1) - 5
= 1 - 4 + 2 - 5
= -6.
Thus, the remainder is -6.

Question 9:
Factorize the polynomial x² - 5x + 6.
Answer:

To factorize x² - 5x + 6, find two numbers that multiply to 6 and add to -5.
These numbers are -2 and -3.
Thus, the factorization is: (x - 2)(x - 3).

Question 10:
If one zero of the polynomial p(x) = 2x² - 5x + k is 2, find the value of k.
Answer:

Since 2 is a zero of p(x), substitute x = 2 and set p(2) = 0:
2(2)² - 5(2) + k = 0
8 - 10 + k = 0
-2 + k = 0
Thus, k = 2.

Short Answer (3 Marks) – with Solutions (CBSE Pattern)

These 3-mark questions require brief explanations and help assess understanding and application of concepts.

Question 1:
Divide the polynomial p(x) = 3x³ - 4x² + 5x - 2 by g(x) = x - 1 and verify the division algorithm.
Answer:

Using synthetic division:
1 | 3 -4 5 -2
_____3 -1 4
Result: 3x² - x + 4 with remainder 2.


Verification:
p(x) = g(x) × q(x) + r(x)
3x³ - 4x² + 5x - 2 = (x - 1)(3x² - x + 4) + 2.
Expanding confirms the equality.

Question 2:
Define a polynomial and give an example of a quadratic polynomial with real coefficients.
Answer:

A polynomial is an algebraic expression consisting of variables, coefficients, and non-negative integer exponents, combined using addition, subtraction, and multiplication.


Example of a quadratic polynomial: 2x² + 3x - 5, where the highest exponent is 2.

Question 3:
Find the zeroes of the polynomial p(x) = x² - 5x + 6 and verify the relationship between the zeroes and coefficients.
Answer:

To find the zeroes, factorize the polynomial:


x² - 5x + 6 = (x - 2)(x - 3)

Zeroes are x = 2 and x = 3.


Verification:
Sum of zeroes = 2 + 3 = 5 = -(-5)/1 (coefficient of x / coefficient of x²).
Product of zeroes = 2 × 3 = 6 = 6/1 (constant term / coefficient of x²).

Question 4:
If one zero of the polynomial 2x² + px + 4 is 2, find the value of p and the other zero.
Answer:

Substitute x = 2 into the polynomial:
2(2)² + p(2) + 4 = 0
8 + 2p + 4 = 0
2p = -12
p = -6.


Now, the polynomial becomes 2x² - 6x + 4.
Sum of zeroes = 6/2 = 3.
If one zero is 2, the other zero is 3 - 2 = 1.

Question 5:
Explain why x + 1 is a factor of the polynomial x³ + x² + x + 1 but not of x³ + x² - x - 1.
Answer:

For x + 1 to be a factor, substituting x = -1 must give zero.


For x³ + x² + x + 1:
(-1)³ + (-1)² + (-1) + 1 = -1 + 1 - 1 + 1 = 0.
Thus, x + 1 is a factor.


For x³ + x² - x - 1:
(-1)³ + (-1)² - (-1) - 1 = -1 + 1 + 1 - 1 = 0.
Wait, this also gives zero, so x + 1 is a factor here too. Correction: The second polynomial also has x + 1 as a factor.

Question 6:
Define a polynomial and give an example of a quadratic polynomial with all necessary conditions.
Answer:

A polynomial is an algebraic expression consisting of variables, coefficients, and non-negative integer exponents, combined using addition, subtraction, and multiplication.


Example of a quadratic polynomial: 2x² + 3x - 5.


Conditions:

  • Highest power of the variable (degree) must be 2.
  • Coefficients can be any real numbers.
  • No division by a variable or negative exponents allowed.

Question 7:
Find the zeroes of the polynomial p(x) = x² - 5x + 6 and verify the relationship between zeroes and coefficients.
Answer:

To find zeroes, factorize the polynomial:


x² - 5x + 6 = (x - 2)(x - 3).


Zeroes are x = 2 and x = 3.


Verification:
Sum of zeroes = 2 + 3 = 5 = -(-5)/1 (coefficient of x / coefficient of x²).
Product of zeroes = 2 × 3 = 6 = constant term / coefficient of x².

Question 8:
Explain why x⁻² + 3x + 1 is not a polynomial. Provide a corrected version that qualifies as a polynomial.
Answer:

The expression x⁻² + 3x + 1 is not a polynomial because it contains a negative exponent (x⁻²), which violates the definition of a polynomial.


Corrected version: 3x + 1 (a linear polynomial) or x² + 3x + 1 (a quadratic polynomial).

Question 9:
If one zero of the polynomial p(x) = 3x² - 8x + 2k is 2, find the value of k.
Answer:

Since 2 is a zero of p(x), substitute x = 2:


3(2)² - 8(2) + 2k = 0.
12 - 16 + 2k = 0.
-4 + 2k = 0.
2k = 4.
k = 2.

Question 10:
Divide the polynomial p(x) = x³ - 3x² + 3x - 1 by g(x) = x - 1 and verify the division algorithm.
Answer:

Using long division:


1. Divide x³ by x to get x².
2. Multiply g(x) by x²: x³ - x².
3. Subtract from p(x): remainder is -2x² + 3x - 1.
4. Repeat steps: quotient becomes x² - 2x + 1 with remainder 0.


Verification:
p(x) = g(x) × quotient + remainder.
x³ - 3x² + 3x - 1 = (x - 1)(x² - 2x + 1) + 0.

Long Answer (5 Marks) – with Solutions (CBSE Pattern)

These 5-mark questions are descriptive and require detailed, structured answers with proper explanation and examples.

Question 1:
Explain why x² + 4x + 4 is a perfect square trinomial. Verify by factorization and compare it with an example from NCERT.
Answer:
Introduction

We studied that a perfect square trinomial is formed by squaring a binomial. Here, x² + 4x + 4 fits this pattern.


Argument 1
  • It can be written as (x + 2)², matching a² + 2ab + b².

Argument 2

Our textbook shows a similar example, x² + 6x + 9 = (x + 3)², confirming the pattern.


Conclusion

Thus, x² + 4x + 4 is a perfect square trinomial, verified by factorization.

Question 2:
A polynomial p(x) = 2x³ - 3x² + 5x - 7 is given. Find its value at x = -1 using the Remainder Theorem. Show steps and relate it to a real-life application.
Answer:
Introduction

The Remainder Theorem states that p(a) is the remainder when p(x) is divided by (x - a).


Argument 1
  • Substitute x = -1: p(-1) = 2(-1)³ - 3(-1)² + 5(-1) - 7 = -2 - 3 - 5 - 7 = -17.

Argument 2

In real life, this helps calculate profit/loss at specific values, like temperature changes.


Conclusion

The value of p(x) at x = -1 is -17, verified by the Remainder Theorem.

Question 3:
Derive the factorization of x² - 5x + 6 using the splitting the middle term method. Compare it with NCERT Exercise 2.3.
Answer:
Introduction

We factorize quadratic polynomials by splitting the middle term into two numbers whose product equals the constant term.


Argument 1
  • For x² - 5x + 6, we find numbers -2 and -3, as (-2) × (-3) = 6 and (-2) + (-3) = -5.

Argument 2

NCERT Exercise 2.3 shows x² + 7x + 12 = (x + 3)(x + 4), following the same method.


Conclusion

Thus, x² - 5x + 6 = (x - 2)(x - 3), derived by splitting the middle term.

Question 4:
Explain the factor theorem with an example. Verify if (x - 2) is a factor of the polynomial p(x) = x³ - 4x² + x + 6.
Answer:
Introduction

The factor theorem states that if a polynomial p(x) has a factor (x - a), then p(a) = 0.


Argument 1

We studied this in NCERT Chapter 2. For example, if p(x) = x² - 3x + 2, then (x - 1) is a factor because p(1) = 0.


Argument 2

To verify if (x - 2) is a factor of p(x) = x³ - 4x² + x + 6, substitute x = 2: p(2) = 8 - 16 + 2 + 6 = 0. Hence, it is a factor.


Conclusion

The factor theorem helps us identify factors quickly, as shown in this problem.

Question 5:
Define a zero of a polynomial. Find the zeroes of the quadratic polynomial p(x) = 2x² - 8x + 6 and verify the relationship between zeroes and coefficients.
Answer:
Introduction

A zero of a polynomial p(x) is a value 'a' such that p(a) = 0.


Argument 1

For p(x) = 2x² - 8x + 6, we factorize it as 2(x² - 4x + 3) = 2(x - 1)(x - 3). Thus, zeroes are x = 1 and x = 3.


Argument 2

Our textbook shows that for ax² + bx + c, sum of zeroes = -b/a = 4, and product = c/a = 3. Here, 1 + 3 = 4 and 1 × 3 = 3, which matches.


Conclusion

The relationship between zeroes and coefficients is verified, confirming the polynomial's properties.

Question 6:
What is the Remainder Theorem? Use it to find the remainder when p(x) = x³ - 6x² + 2x - 4 is divided by (x - 3).
Answer:
Introduction

The Remainder Theorem states that if a polynomial p(x) is divided by (x - a), the remainder is p(a).


Argument 1

We learned this in NCERT with examples like dividing x² - 5x + 6 by (x - 2), giving remainder 0.


Argument 2

For p(x) = x³ - 6x² + 2x - 4 divided by (x - 3), substitute x = 3: p(3) = 27 - 54 + 6 - 4 = -25. Thus, remainder is -25.


Conclusion

The Remainder Theorem simplifies polynomial division, as demonstrated here.

Question 7:
A rectangular garden has length (3x + 2) meters and breadth (x - 1) meters. Find its area as a polynomial and explain the steps.
Answer:
Introduction

We know area of a rectangle is length × breadth. Here, we multiply (3x + 2) and (x - 1).


Argument 1
  • Using distributive property: 3x(x - 1) + 2(x - 1).
  • Simplifies to 3x² - 3x + 2x - 2 = 3x² - x - 2.

Conclusion

The area polynomial is 3x² - x - 2 square meters, derived step-by-step.

Question 8:
If p(x) = 2x³ - 5x² + 3x - 7, evaluate p(2) and explain the Remainder Theorem using this example.
Answer:
Introduction

The Remainder Theorem states that p(a) is the remainder when p(x) is divided by (x - a).


Argument 1
  • Substitute x = 2: 2(8) - 5(4) + 3(2) - 7 = 16 - 20 + 6 - 7 = -5.
  • NCERT Example 2.8 similarly verifies the theorem for p(x) = x³ - x² - 2.

Conclusion

Thus, p(2) = -5, confirming the Remainder Theorem.

Question 9:
Explain the Factor Theorem with an example. How is it useful in finding factors of a polynomial?
Answer:
Introduction

The Factor Theorem states that if a polynomial p(x) has a factor (x - a), then p(a) = 0. It helps us verify factors easily.


Argument 1

For example, check if (x - 2) is a factor of p(x) = x² - 4. Substituting x = 2, we get p(2) = 0, confirming it is a factor.


Argument 2

Our textbook shows how this theorem simplifies factorization. Instead of trial and error, we test possible roots systematically.


Conclusion

This theorem is widely used in algebra to break down polynomials into simpler factors.

Question 10:
Derive the Remainder Theorem and apply it to find the remainder when x³ - 4x² + 2x - 5 is divided by (x - 1).
Answer:
Introduction

The Remainder Theorem states that the remainder of p(x) divided by (x - a) is p(a).


Argument 1

For p(x) = x³ - 4x² + 2x - 5, substitute x = 1: p(1) = 1 - 4 + 2 - 5 = -6. Thus, the remainder is -6.


Argument 2

Our textbook demonstrates this theorem avoids long division, making calculations quicker.


Conclusion

The theorem is a powerful tool for evaluating remainders efficiently.

Question 11:
What is a zero of a polynomial? Find the zeroes of p(x) = 2x² - 8 and verify their relationship with coefficients.
Answer:
Introduction

A zero of a polynomial is a value k such that p(k) = 0. For p(x) = 2x² - 8, we solve 2x² - 8 = 0.


Argument 1

Solving gives x² = 4, so zeroes are x = 2 and x = -2.


Argument 2

Sum of zeroes (2 + (-2)) = 0, matching -b/a = 0/2. Product (2 × -2) = -4, equal to c/a = -8/2.


Conclusion

This confirms the relationship between zeroes and coefficients as per NCERT.

Question 12:
Verify whether 2 and 0 are the zeroes of the polynomial p(x) = x2 - 2x. Justify your answer with proper steps.
Answer:

To verify if 2 and 0 are zeroes of the polynomial p(x) = x2 - 2x, we substitute these values into the polynomial and check if p(x) = 0.


Step 1: Check for x = 2


p(2) = (2)2 - 2(2)
= 4 - 4
= 0

Since p(2) = 0, 2 is a zero of the polynomial.


Step 2: Check for x = 0


p(0) = (0)2 - 2(0)
= 0 - 0
= 0

Since p(0) = 0, 0 is also a zero of the polynomial.


Conclusion: Both 2 and 0 satisfy the condition p(x) = 0, confirming they are zeroes of the polynomial.

Question 13:
Find the value of k if (x - 1) is a factor of the polynomial p(x) = x3 + x2 + kx + 4. Show all steps clearly.
Answer:

If (x - 1) is a factor of p(x) = x3 + x2 + kx + 4, then by the Factor Theorem, p(1) must be equal to 0.


Step 1: Substitute x = 1 into the polynomial:


p(1) = (1)3 + (1)2 + k(1) + 4
= 1 + 1 + k + 4
= 6 + k

Step 2: Set p(1) = 0 (as per Factor Theorem):


6 + k = 0

Step 3: Solve for k:


k = -6

Conclusion: The value of k is -6 for (x - 1) to be a factor of the given polynomial.

Question 14:
Verify whether 2 and -1 are zeroes of the polynomial p(x) = x2 - x - 2. Also, explain the relationship between the zeroes and coefficients of the polynomial.
Answer:

To verify if 2 and -1 are zeroes of the polynomial p(x) = x2 - x - 2, we substitute these values into the polynomial and check if p(x) = 0.


Step 1: Check for x = 2
p(2) = (2)2 - (2) - 2
= 4 - 2 - 2
= 0


Step 2: Check for x = -1
p(-1) = (-1)2 - (-1) - 2
= 1 + 1 - 2
= 0


Since both p(2) and p(-1) equal 0, 2 and -1 are indeed zeroes of the polynomial.


Relationship between zeroes and coefficients:
For a quadratic polynomial ax2 + bx + c, the sum and product of zeroes (α and β) are related to the coefficients as follows:

  • Sum of zeroes (α + β) = -b/a
  • Product of zeroes (α × β) = c/a


Here, p(x) = x2 - x - 2 (where a = 1, b = -1, c = -2).
Sum of zeroes = 2 + (-1) = 1
This matches -b/a = -(-1)/1 = 1.
Product of zeroes = 2 × (-1) = -2
This matches c/a = -2/1 = -2.


Thus, the zeroes satisfy the standard relationships with the coefficients.

Question 15:
Verify whether 2 and -3 are the zeroes of the polynomial p(x) = x2 + x - 6. Also, explain the concept of zeroes of a polynomial with an example.
Answer:

To verify if 2 and -3 are zeroes of the polynomial p(x) = x2 + x - 6, we substitute these values into the polynomial and check if the result is zero.


Step 1: Check for x = 2
p(2) = (2)2 + (2) - 6 = 4 + 2 - 6 = 0


Step 2: Check for x = -3
p(-3) = (-3)2 + (-3) - 6 = 9 - 3 - 6 = 0


Since both p(2) and p(-3) equal zero, 2 and -3 are indeed zeroes of the polynomial.


Concept of Zeroes of a Polynomial:
A zero of a polynomial is a value of the variable (x) that makes the polynomial equal to zero. In other words, if p(a) = 0, then 'a' is called a zero of the polynomial p(x).


Example:
Consider the polynomial q(x) = x - 4. To find its zero, we set q(x) = 0:
x - 4 = 0 ⇒ x = 4
Thus, 4 is the zero of the polynomial q(x).


Value-added Information:
The number of zeroes of a polynomial is always less than or equal to its degree. For example, a quadratic polynomial (degree 2) can have at most 2 zeroes.

Question 16:
Verify whether 2 and -2 are zeroes of the polynomial p(x) = x2 - 4. Also, explain the significance of zeroes of a polynomial with an example.
Answer:

To verify if 2 and -2 are zeroes of the polynomial p(x) = x2 - 4, we substitute these values into the polynomial and check if p(x) = 0.


Step 1: Substitute x = 2 in p(x).
p(2) = (2)2 - 4 = 4 - 4 = 0


Step 2: Substitute x = -2 in p(x).
p(-2) = (-2)2 - 4 = 4 - 4 = 0


Since both p(2) and p(-2) equal 0, 2 and -2 are indeed zeroes of the polynomial.


Significance of Zeroes of a Polynomial: The zeroes of a polynomial are the values of x for which the polynomial evaluates to zero. These are the points where the graph of the polynomial intersects the x-axis.


Example: For the polynomial q(x) = x2 - 5x + 6, the zeroes are 2 and 3 because:
q(2) = (2)2 - 5(2) + 6 = 4 - 10 + 6 = 0
q(3) = (3)2 - 5(3) + 6 = 9 - 15 + 6 = 0


This means the graph of q(x) will touch the x-axis at x = 2 and x = 3.

Question 17:
Verify whether 2 and -2 are zeroes of the polynomial p(x) = x3 - 4x. Also, explain the significance of zeroes of a polynomial in real-life applications.
Answer:

To verify if 2 and -2 are zeroes of the polynomial p(x) = x3 - 4x, we substitute these values into the polynomial and check if the result is zero.


Step 1: Check for x = 2
p(2) = (2)3 - 4(2)
= 8 - 8
= 0


Since p(2) = 0, 2 is a zero of the polynomial.


Step 2: Check for x = -2
p(-2) = (-2)3 - 4(-2)
= -8 + 8
= 0


Since p(-2) = 0, -2 is also a zero of the polynomial.


Significance of Zeroes in Real-Life Applications:
The zeroes of a polynomial represent the values where the polynomial intersects the x-axis on a graph. In real-life scenarios, zeroes can indicate:

  • Break-even points in economics (profit = 0).
  • Equilibrium states in physics (forces balance out).
  • Roots of equations in engineering (solutions to design problems).
Understanding zeroes helps in modeling and solving practical problems efficiently.

Question 18:
Verify whether 2 and -3 are the zeroes of the polynomial p(x) = x2 + x - 6. Also, explain the relationship between the zeroes and coefficients of the polynomial.
Answer:

To verify if 2 and -3 are zeroes of the polynomial p(x) = x2 + x - 6, we substitute these values into the polynomial and check if p(x) = 0.


Step 1: Verify x = 2
p(2) = (2)2 + (2) - 6
= 4 + 2 - 6
= 0


Step 2: Verify x = -3
p(-3) = (-3)2 + (-3) - 6
= 9 - 3 - 6
= 0


Since both 2 and -3 satisfy p(x) = 0, they are indeed the zeroes of the polynomial.


Relationship between zeroes and coefficients:
For a quadratic polynomial ax2 + bx + c, the sum and product of zeroes are related to the coefficients as follows:

  • Sum of zeroes (α + β) = -b/a
    Here, α + β = 2 + (-3) = -1
    From the polynomial, -b/a = -1/1 = -1
  • Product of zeroes (α × β) = c/a
    Here, α × β = 2 × (-3) = -6
    From the polynomial, c/a = -6/1 = -6

Thus, the relationships hold true, confirming the zeroes and coefficients are correctly linked.

Question 19:
Verify whether x = 2 and x = -3 are zeroes of the polynomial p(x) = x2 + x - 6. Justify your answer with proper steps.
Answer:

To verify if x = 2 and x = -3 are zeroes of the polynomial p(x) = x2 + x - 6, we substitute these values into the polynomial and check if p(x) = 0.


Step 1: Check for x = 2
Substitute x = 2 into p(x):
p(2) = (2)2 + (2) - 6
= 4 + 2 - 6
= 0


Since p(2) = 0, x = 2 is indeed a zero of the polynomial.


Step 2: Check for x = -3
Substitute x = -3 into p(x):
p(-3) = (-3)2 + (-3) - 6
= 9 - 3 - 6
= 0


Since p(-3) = 0, x = -3 is also a zero of the polynomial.


Conclusion: Both x = 2 and x = -3 satisfy the equation p(x) = 0, confirming they are zeroes of the polynomial p(x) = x2 + x - 6.

Question 20:
Verify whether x = 2 and x = -1 are zeroes of the polynomial p(x) = 2x2 - x - 1. Also, explain the significance of zeroes of a polynomial with an example.
Answer:

To verify if x = 2 and x = -1 are zeroes of the polynomial p(x) = 2x2 - x - 1, we substitute these values into the polynomial and check if p(x) = 0.


Step 1: Check for x = 2
p(2) = 2(2)2 - (2) - 1
= 2(4) - 2 - 1
= 8 - 2 - 1
= 5 ≠ 0
Conclusion: x = 2 is not a zero of the polynomial.

Step 2: Check for x = -1
p(-1) = 2(-1)2 - (-1) - 1
= 2(1) + 1 - 1
= 2 + 1 - 1
= 2 ≠ 0
Conclusion: x = -1 is not a zero of the polynomial.

Significance of Zeroes of a Polynomial: The zeroes of a polynomial are the values of x for which the polynomial equals zero. They represent the x-intercepts of the polynomial's graph and are crucial in solving equations. For example, in p(x) = x2 - 3x + 2, the zeroes are x = 1 and x = 2, which are the solutions to the equation x2 - 3x + 2 = 0.

Question 21:
Explain the Factor Theorem with an example. Verify the theorem for the polynomial p(x) = 2x² - 5x + 3 and one of its factors.
Answer:

The Factor Theorem states that for a polynomial p(x), if (x - a) is a factor of p(x), then p(a) = 0. Conversely, if p(a) = 0, then (x - a) is a factor of p(x).


Example: Let’s verify the theorem for p(x) = 2x² - 5x + 3 and one of its factors, (x - 1).


Step 1: Check if p(1) = 0.
Step 2: Substitute x = 1 in p(x):
p(1) = 2(1)² - 5(1) + 3 = 2 - 5 + 3 = 0.
Conclusion: Since p(1) = 0, (x - 1) is indeed a factor of p(x).


Additional Insight: The Factor Theorem is a special case of the Remainder Theorem and is useful for factorizing polynomials and finding their roots.

Question 22:
Prove that (x + 2) is a factor of the polynomial p(x) = x³ + 4x² + x - 6. Using this, factorize the polynomial completely.
Answer:

To prove (x + 2) is a factor of p(x) = x³ + 4x² + x - 6, we use the Factor Theorem.


Step 1: Check if p(-2) = 0.
Substitute x = -2 in p(x):
p(-2) = (-2)³ + 4(-2)² + (-2) - 6 = -8 + 16 - 2 - 6 = 0.
Since p(-2) = 0, (x + 2) is a factor.


Step 2: Perform polynomial division or use synthetic division to factorize p(x).
Dividing p(x) by (x + 2), we get:
x² + 2x - 3 as the quotient.


Step 3: Factorize the quotient further:
x² + 2x - 3 = (x + 3)(x - 1).


Final Factorization: p(x) = (x + 2)(x + 3)(x - 1).


Application: This factorization helps in finding the roots of the polynomial, which are x = -2, -3, 1.

Case-based Questions (4 Marks) – with Solutions (CBSE Pattern)

These 4-mark case-based questions assess analytical skills through real-life scenarios. Answers must be based on the case study provided.

Question 1:
A rectangular garden has its length represented by the polynomial 3x² + 5x - 2 and width by x + 2.
Problem Interpretation: Find the area of the garden.
Mathematical Modeling: Our textbook shows area as length × width.
Solution: Multiply the polynomials using the distributive property.
Answer:

We studied that area of a rectangle is length × width. Here, length = 3x² + 5x - 2 and width = x + 2.

Using distributive property:
(3x² + 5x - 2)(x + 2) = 3x³ + 6x² + 5x² + 10x - 2x - 4 = 3x³ + 11x² + 8x - 4.

Thus, the area is 3x³ + 11x² + 8x - 4 square units.

Question 2:
A polynomial p(x) = 2x³ - 5x² + 3x - 7 is divided by (x - 2).
Problem Interpretation: Find the remainder using the Remainder Theorem.
Mathematical Modeling: Our textbook states that remainder = p(a) when divided by (x - a).
Solution: Substitute x = 2 in p(x).
Answer:

We learned the Remainder Theorem: if p(x) is divided by (x - a), the remainder is p(a).

Here, a = 2. Substituting in p(x):
p(2) = 2(2)³ - 5(2)² + 3(2) - 7 = 16 - 20 + 6 - 7 = -5.

Thus, the remainder is -5.

Question 3:
A rectangular garden has its length represented by the polynomial 3x² + 5x - 2 and width by x + 2. Find the area of the garden and verify if x = 1 is a zero of the area polynomial.
Answer:
Problem Interpretation

We need to find the area of the garden by multiplying the given polynomials and then check if x = 1 makes the area zero.


Mathematical Modeling

Area = Length × Width = (3x² + 5x - 2) × (x + 2).


Solution

First, multiply the polynomials: 3x³ + 6x² + 5x² + 10x - 2x - 4 = 3x³ + 11x² + 8x - 4. For x = 1, Area = 3(1)³ + 11(1)² + 8(1) - 4 = 18 ≠ 0. Thus, x = 1 is not a zero.

Question 4:
The volume of a cuboid is given by the polynomial 2x³ - 3x² - 11x + 6. If one of its dimensions is (x - 3), find the other two dimensions by factorizing the polynomial.
Answer:
Problem Interpretation

We know one dimension is (x - 3), so we can factorize the volume polynomial to find the other two dimensions.


Mathematical Modeling

Volume = Length × Width × Height = 2x³ - 3x² - 11x + 6. Divide this by (x - 3) to factorize.


Solution

Using polynomial division or factor theorem, we get: (x - 3)(2x² + 3x - 2). Further factorizing: (x - 3)(2x - 1)(x + 2). Thus, the other two dimensions are (2x - 1) and (x + 2).

Question 5:
A rectangular garden has its length represented by the polynomial L(x) = 2x² + 3x + 1 and width by W(x) = x + 2. Find the area of the garden in terms of x and verify for x = 1.
Answer:
Problem Interpretation

We studied that area of a rectangle is length × width. Here, polynomials represent dimensions.


Mathematical Modeling

Area A(x) = L(x) × W(x) = (2x² + 3x + 1)(x + 2).


Solution
  • Multiply polynomials: 2x³ + 4x² + 3x² + 6x + x + 2 = 2x³ + 7x² + 7x + 2
  • For x=1: A(1) = 2(1)³ + 7(1)² + 7(1) + 2 = 18 square units
Question 6:
The volume of a cuboid is given by the polynomial V(y) = y³ + 6y² + 11y + 6. If one dimension is (y + 1), find the other two dimensions assuming they are linear polynomials.
Answer:
Problem Interpretation

Our textbook shows volume as product of three dimensions. We factorize V(y) using (y + 1).


Mathematical Modeling

Divide V(y) by (y + 1) to get quadratic polynomial, then factorize further.


Solution
  • Synthetic division gives quotient: y² + 5y + 6
  • Factorized form: (y + 1)(y + 2)(y + 3)
  • Other dimensions: (y + 2) and (y + 3)
Question 7:
A rectangular garden has its length represented by the polynomial 2x² + 3x - 5 meters and width by x + 2 meters.
Problem Interpretation: Find the area of the garden and verify if x = 1 is a zero of the area polynomial.
Answer:
Problem Interpretation: We studied that area of a rectangle is length × width. Here, we multiply the given polynomials.
Mathematical Modeling: Area = (2x² + 3x - 5)(x + 2).
Solution:
  • First, multiply using distributive property: 2x³ + 4x² + 3x² + 6x - 5x - 10 = 2x³ + 7x² + x - 10.
  • For x = 1, substitute: 2(1)³ + 7(1)² + 1 - 10 = 0. Thus, it is a zero.
Question 8:
The volume of a cuboid is given by the polynomial x³ - 6x² + 11x - 6 cubic units.
Problem Interpretation: Factorize the polynomial and identify its possible dimensions if all sides are linear polynomials.
Answer:
Problem Interpretation: Our textbook shows that volume of a cuboid is product of its three dimensions. We factorize the given polynomial.
Mathematical Modeling: Let’s find factors of x³ - 6x² + 11x - 6.
Solution:
  • By trial, x = 1 is a zero. So, (x - 1) is a factor.
  • Divide to get (x - 1)(x² - 5x + 6) = (x - 1)(x - 2)(x - 3).
  • Dimensions: (x - 1), (x - 2), (x - 3) units.
Question 9:
A rectangular garden has length p(x) = 2x² + 3x + 1 meters and breadth q(x) = x + 1 meters.
Problem Interpretation: Find the area of the garden as a polynomial.
Mathematical Modeling: How would you verify if x = 2 is a zero of the breadth polynomial?
Answer:
Problem Interpretation:

Area = Length × Breadth = p(x) × q(x). We multiply the polynomials.

Mathematical Modeling:

Substitute x = 2 in q(x). If result is 0, it is a zero.

Solution:
  • Area = (2x² + 3x + 1)(x + 1) = 2x³ + 5x² + 4x + 1
  • For q(2): 2 + 1 = 3 ≠ 0. Not a zero.
Question 10:
A polynomial f(x) = x³ - 4x² + x + 6 gives the volume of a box.
Problem Interpretation: If x - 2 is a factor, find the other dimensions.
Mathematical Modeling: How can you confirm x = 3 is a root using Factor Theorem?
Answer:
Problem Interpretation:

Divide f(x) by (x - 2) to get quadratic factor (other dimensions).

Mathematical Modeling:

Factor Theorem states if f(3) = 0, then (x - 3) is a factor.

Solution:
  • Division gives x² - 2x - 3 (other dimensions).
  • f(3) = 27 - 36 + 3 + 6 = 0. Hence, confirmed.
Question 11:

A farmer has a rectangular field with length represented by the polynomial L(x) = 2x² + 3x - 5 meters and width represented by W(x) = x + 2 meters. He wants to fence the field. Help him find:

  • The perimeter of the field as a polynomial.
  • The actual perimeter if x = 3 meters.
Answer:

To find the perimeter P(x) of the rectangular field, we use the formula: P(x) = 2 × (L(x) + W(x)).


Given: L(x) = 2x² + 3x - 5 and W(x) = x + 2.


Step 1: Add L(x) and W(x):
L(x) + W(x) = (2x² + 3x - 5) + (x + 2) = 2x² + 4x - 3.


Step 2: Multiply by 2 to get the perimeter:
P(x) = 2 × (2x² + 4x - 3) = 4x² + 8x - 6.


For x = 3 meters:
P(3) = 4(3)² + 8(3) - 6 = 4(9) + 24 - 6 = 36 + 24 - 6 = 54 meters.


Thus, the perimeter polynomial is 4x² + 8x - 6, and the actual perimeter for x = 3 is 54 meters.

Question 12:

Riya divides a polynomial P(x) = x³ - 4x² + x + 6 by (x - 2) and claims the remainder is zero. Verify her claim using the Remainder Theorem and also factorize the polynomial completely.

Answer:

According to the Remainder Theorem, if a polynomial P(x) is divided by (x - a), the remainder is P(a).


Given: P(x) = x³ - 4x² + x + 6 and divisor (x - 2) (so, a = 2).


Step 1: Find P(2):
P(2) = (2)³ - 4(2)² + (2) + 6 = 8 - 16 + 2 + 6 = 0.


Since the remainder is zero, Riya's claim is correct, and (x - 2) is a factor.


Step 2: Factorize P(x) using synthetic division or factorization:
x³ - 4x² + x + 6 = (x - 2)(x² - 2x - 3).
Further factorizing: (x² - 2x - 3) = (x - 3)(x + 1).


Final factorization: P(x) = (x - 2)(x - 3)(x + 1).


Thus, the polynomial is completely factorized, and Riya's claim is verified.

Question 13:
A farmer has a rectangular field with length represented by the polynomial L(x) = 2x² + 3x + 1 and width by W(x) = x + 2.

(a) Find the area of the field in terms of x.

(b) If x = 5 meters, calculate the actual area of the field.

Answer:

(a) Area of the field:

The area A(x) of a rectangle is given by the product of its length and width.


A(x) = L(x) × W(x)
A(x) = (2x² + 3x + 1)(x + 2)
Now, multiply the polynomials:
= 2x²(x) + 2x²(2) + 3x(x) + 3x(2) + 1(x) + 1(2)
= 2x³ + 4x² + 3x² + 6x + x + 2
Combine like terms:
= 2x³ + (4x² + 3x²) + (6x + x) + 2
= 2x³ + 7x² + 7x + 2

Thus, the area is A(x) = 2x³ + 7x² + 7x + 2.


(b) Actual area when x = 5 meters:


Substitute x = 5 in A(x):
A(5) = 2(5)³ + 7(5)² + 7(5) + 2
= 2(125) + 7(25) + 35 + 2
= 250 + 175 + 35 + 2
= 462

The actual area is 462 square meters.

Question 14:
A polynomial P(x) = x³ - 4x² + x + 6 has one of its zeros at x = 2.

(a) Verify that x = 2 is a zero of P(x).

(b) Factorize P(x) completely and find all its zeros.

Answer:

(a) Verification:

To verify if x = 2 is a zero of P(x), substitute x = 2 in the polynomial:


P(2) = (2)³ - 4(2)² + (2) + 6
= 8 - 16 + 2 + 6
= 0

Since P(2) = 0, x = 2 is indeed a zero of the polynomial.


(b) Factorization and zeros:

Since x = 2 is a zero, (x - 2) is a factor of P(x).


Divide P(x) by (x - 2) using synthetic division:
  • Write coefficients: 1 (for x³), -4 (for x²), 1 (for x), 6 (constant)
  • Bring down 1
  • Multiply by 2: 1 × 2 = 2 → Add to -4: -2
  • Multiply by 2: -2 × 2 = -4 → Add to 1: -3
  • Multiply by 2: -3 × 2 = -6 → Add to 6: 0

The quotient is x² - 2x - 3.
Now, factorize the quotient:
x² - 2x - 3 = (x - 3)(x + 1)

Thus, the complete factorization is:


P(x) = (x - 2)(x - 3)(x + 1)

The zeros of P(x) are:


x - 2 = 0 → x = 2
x - 3 = 0 → x = 3
x + 1 = 0 → x = -1

All zeros are 2, 3, and -1.

Question 15:
A rectangular garden has its length represented by the polynomial L(x) = 2x² + 5x - 3 and width by W(x) = x - 1.

(i) Find the area of the garden in terms of x.

(ii) If x = 2 meters, calculate the actual area of the garden.

Answer:

(i) Area of the garden:


The area A(x) of a rectangle is given by the product of its length and width.
A(x) = L(x) × W(x)
= (2x² + 5x - 3)(x - 1)
Now, multiply the polynomials using the distributive property:
= 2x²(x) + 2x²(-1) + 5x(x) + 5x(-1) - 3(x) - 3(-1)
= 2x³ - 2x² + 5x² - 5x - 3x + 3
Combine like terms:
= 2x³ + 3x² - 8x + 3

(ii) Actual area when x = 2 meters:


Substitute x = 2 into A(x):
A(2) = 2(2)³ + 3(2)² - 8(2) + 3
= 2(8) + 3(4) - 16 + 3
= 16 + 12 - 16 + 3
= 15 square meters.

Thus, the area of the garden is 15 m² when x = 2.

Question 16:
A polynomial P(x) = x³ - 4x² + x + 6 has (x - 2) as one of its factors.

(i) Verify whether (x - 2) is indeed a factor of P(x) using the Factor Theorem.

(ii) Find the other factors of P(x).

Answer:

(i) Verification using Factor Theorem:


The Factor Theorem states that if (x - a) is a factor of P(x), then P(a) = 0.
Here, a = 2.
Substitute x = 2 into P(x):
P(2) = (2)³ - 4(2)² + (2) + 6
= 8 - 16 + 2 + 6
= 0.
Since P(2) = 0, (x - 2) is indeed a factor of P(x).

(ii) Finding other factors:


Divide P(x) by (x - 2) using polynomial division or synthetic division to find the quotient.
Using synthetic division:
  • Write coefficients: 1 (for x³), -4 (for x²), 1 (for x), 6 (constant).
  • Bring down 1, multiply by 2: 2, add to -4 → -2.
  • Multiply -2 by 2: -4, add to 1 → -3.
  • Multiply -3 by 2: -6, add to 6 → 0.
The quotient is x² - 2x - 3.
Now, factorize the quadratic:
x² - 2x - 3 = (x - 3)(x + 1).
Thus, the complete factorization of P(x) is:
P(x) = (x - 2)(x - 3)(x + 1).

The other factors are (x - 3) and (x + 1).

Question 17:
A farmer has a rectangular field with length represented by the polynomial L(x) = 2x² + 3x + 1 and width by W(x) = x + 2.

He wants to fence the field and also calculate its area. Help him by answering the following:

  • Find the perimeter of the field in terms of x.
  • Calculate the area of the field when x = 3.
Answer:

Perimeter of the field:


The perimeter P(x) of a rectangle is given by 2(L + W).
Given L(x) = 2x² + 3x + 1 and W(x) = x + 2,
P(x) = 2[(2x² + 3x + 1) + (x + 2)]
P(x) = 2(2x² + 4x + 3)
P(x) = 4x² + 8x + 6.

Area when x = 3:


Area A(x) = L(x) × W(x).
Substitute x = 3:
L(3) = 2(3)² + 3(3) + 1 = 18 + 9 + 1 = 28
W(3) = 3 + 2 = 5
A(3) = 28 × 5 = 140 square units.
Question 18:
A student was asked to verify if (x - 1) is a factor of the polynomial P(x) = 3x³ - 2x² + x - 4.

She used the Factor Theorem but made a mistake. Identify her error and correct the solution.

Answer:

Error Identification:


The student likely substituted x = 1 incorrectly or misapplied the Factor Theorem.

Correct Solution:


According to the Factor Theorem, (x - a) is a factor of P(x) if P(a) = 0.
Here, a = 1, so we calculate P(1):
P(1) = 3(1)³ - 2(1)² + (1) - 4
P(1) = 3 - 2 + 1 - 4
P(1) = -2.
Since P(1) ≠ 0, (x - 1) is not a factor of P(x).

The student might have incorrectly concluded it as a factor due to calculation errors.

Question 19:
A rectangular garden has its length represented by the polynomial L(x) = 2x² + 5x - 3 and width by W(x) = x + 3.

(a) Find the area of the garden in polynomial form.

(b) If x = 2 meters, calculate the actual area of the garden.

Answer:

(a) Area in polynomial form:


The area A(x) of a rectangle is given by the product of its length and width.
A(x) = L(x) × W(x)
= (2x² + 5x - 3)(x + 3)
= 2x²(x) + 2x²(3) + 5x(x) + 5x(3) - 3(x) - 3(3)
= 2x³ + 6x² + 5x² + 15x - 3x - 9
= 2x³ + 11x² + 12x - 9 (Simplified form)

(b) Actual area when x = 2 meters:


Substitute x = 2 in A(x):
= 2(2)³ + 11(2)² + 12(2) - 9
= 2(8) + 11(4) + 24 - 9
= 16 + 44 + 24 - 9
= 75 square meters
Question 20:
The volume of a box is given by the polynomial V(y) = y³ - 6y² + 11y - 6.

(a) Verify if y = 1 is a zero of the polynomial.

(b) Factorize the polynomial completely and express the volume as a product of its factors.

Answer:

(a) Verification for y = 1:


Substitute y = 1 in V(y):
= (1)³ - 6(1)² + 11(1) - 6
= 1 - 6 + 11 - 6
= 0
Since the result is 0, y = 1 is a zero of the polynomial.

(b) Factorization:


Using Factor Theorem, since y = 1 is a zero, (y - 1) is a factor.
Perform polynomial division or synthetic division to factorize:
V(y) = (y - 1)(y² - 5y + 6)
Further factorize the quadratic:
= (y - 1)(y - 2)(y - 3)
Thus, the volume in factored form is (y - 1)(y - 2)(y - 3).
Question 21:
A polynomial p(x) = x³ - 4x² + x + 6 has three zeroes.

Two of its zeroes are given as 2 and 3. Answer the following:

  • Find the third zero of the polynomial.
  • Verify your answer by showing the relationship between the zeroes and coefficients.
Answer:

Finding the third zero:


Let the zeroes be α = 2, β = 3, and γ (unknown).
For a cubic polynomial p(x) = x³ - 4x² + x + 6,
Sum of zeroes: α + β + γ = -(-4)/1 = 4
2 + 3 + γ = 4
γ = 4 - 5 = -1.

Verification using coefficients:


Sum of zeroes: 2 + 3 + (-1) = 4 (matches coefficient).
Sum of product of zeroes: (2×3) + (3×-1) + (-1×2) = 6 - 3 - 2 = 1 (matches coefficient).
Product of zeroes: 2 × 3 × (-1) = -6 (matches constant term).
Thus, the third zero is -1 and the verification is correct.
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