Polynomials | Grade 10th - Mathematics Chapter Summary & NCERT CBSE Study Resources

Study Materials

10th

10th - Mathematics

Polynomials

Overview

This chapter introduces the concept of Polynomials, which are algebraic expressions consisting of variables and coefficients. Students will learn about different types of polynomials, their degrees, and various operations such as addition, subtraction, multiplication, and division. The chapter also covers key concepts like the Remainder Theorem, Factor Theorem, and factorization of polynomials.

A Polynomial is an expression of the form \( p(x) = a_nx^n + a_{n-1}x^{n-1} + \dots + a_1x + a_0 \), where \( a_n, a_{n-1}, \dots, a_0 \) are real numbers and \( n \) is a non-negative integer.

Types of Polynomials

Linear Polynomial: A polynomial of degree 1 (e.g., \( 2x + 3 \)).
Quadratic Polynomial: A polynomial of degree 2 (e.g., \( x^2 - 5x + 6 \)).
Cubic Polynomial: A polynomial of degree 3 (e.g., \( 4x^3 + 3x^2 - x + 7 \)).

Zeroes of a Polynomial

A real number \( k \) is called a zero of the polynomial \( p(x) \) if \( p(k) = 0 \). Finding zeroes helps in factorizing polynomials and solving polynomial equations.

Remainder Theorem

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

Factor Theorem

The Factor Theorem states that \( (x - a) \) is a factor of the polynomial \( p(x) \) if and only if \( p(a) = 0 \).

Factorization of Polynomials

Polynomials can be factorized using various methods such as splitting the middle term, using algebraic identities, or applying the Factor Theorem.

Algebraic Identities

Some useful algebraic identities for polynomials include:

\( (a + b)^2 = a^2 + 2ab + b^2 \)
\( (a - b)^2 = a^2 - 2ab + b^2 \)
\( a^2 - b^2 = (a + b)(a - b) \)
\( (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca \)

Division Algorithm for Polynomials

If \( p(x) \) and \( g(x) \) are two polynomials with \( g(x) \neq 0 \), then there exist unique polynomials \( q(x) \) (quotient) and \( r(x) \) (remainder) such that:

\( p(x) = g(x) \times q(x) + r(x) \), where \( \text{deg}(r(x)) < \text{deg}(g(x)) \) or \( r(x) = 0 \).

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 5x³ + 2x + 1?

Answer:

Numeric answer:
3

Question 2:

Find the zero of the polynomial p(x) = 3x - 6.

Answer:

Numeric answer:
2

Question 3:

If α and β are zeros of x² - 5x + 6, find α + β.

Answer:

Numeric answer:
5

Question 4:

Is x² + 1 a quadratic polynomial?

Answer:

Yes, degree is 2.

Question 5:

What type of polynomial is 7?

Answer:

Constant polynomial.

Question 6:

If p(x) = x² - 4x + 4, find p(2).

Answer:

Numeric answer:
0

Question 7:

How many zeros does a cubic polynomial have?

Answer:

Numeric answer:
3

Question 8:

Write the standard form of a quadratic polynomial.

Answer:

ax² + bx + c, a ≠ 0.

Question 9:

If one zero of x² - 9 is 3, find the other.

Answer:

Numeric answer:
-3

Question 10:

What is the sum of zeros of x² - 3x + 2?

Answer:

Numeric answer:
3

Question 11:

Find the product of zeros of x² + 7x + 10.

Answer:

Numeric answer:
10

Question 12:

Is x + 1/x a polynomial?

Answer:

No, variable has negative power.

Question 13:

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

Answer:

The degree of the polynomial is 3 because the highest power of the variable x is 3.

Question 14:

Identify the type of polynomial: x² - 5x + 6.

Answer:

It is a quadratic polynomial because the highest degree is 2.

Question 15:

Find the zero of the linear polynomial 3x - 9.

Answer:

Set the polynomial equal to zero:
3x - 9 = 0
3x = 9
x = 3.
Thus, the zero is 3.

Question 16:

If 1 is a zero of the polynomial p(x) = x² + x - 2, verify it.

Answer:

Substitute x = 1 in p(x):
p(1) = (1)² + (1) - 2 = 1 + 1 - 2 = 0.
Since p(1) = 0, 1 is indeed a zero.

Question 17:

What is the maximum number of zeros a quadratic polynomial can have?

Answer:

A quadratic polynomial can have a maximum of 2 zeros because its degree is 2.

Question 18:

Write the standard form of a cubic polynomial.

Answer:

The standard form is:
ax³ + bx² + cx + d, where a ≠ 0.

Question 19:

If the sum of zeros of the polynomial x² - 5x + k is 5, find the value of k.

Answer:

For a quadratic polynomial ax² + bx + c, sum of zeros is -b/a.
Here, -(-5)/1 = 5, which matches the given sum.
Thus, k is the product of zeros, but the question only confirms the sum is correct.

Question 20:

What is the remainder when x³ + 3x² + 3x + 1 is divided by x + 1?

Answer:

Using the Remainder Theorem:
Substitute x = -1 in the polynomial:
(-1)³ + 3(-1)² + 3(-1) + 1 = -1 + 3 - 3 + 1 = 0.
The remainder is 0.

Question 21:

Check whether 2 is a zero of the polynomial x³ - 4x² + 5x - 2.

Answer:

Substitute x = 2:
(2)³ - 4(2)² + 5(2) - 2 = 8 - 16 + 10 - 2 = 0.
Since the result is 0, 2 is a zero.

Question 22:

If one zero of the polynomial x² - 6x + 9 is 3, find the other zero.

Answer:

The polynomial is a perfect square: (x - 3)².
Thus, both zeros are 3 (repeated zero).

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:

Find the zeroes of the polynomial p(x) = x² - 5x + 6.

Answer:

To find the zeroes of p(x) = x² - 5x + 6, factorize the polynomial:
x² - 5x + 6 = (x - 2)(x - 3)
Set each factor to zero:
x - 2 = 0 ⇒ x = 2
x - 3 = 0 ⇒ x = 3
Thus, the zeroes are 2 and 3.

Question 2:

If one zero of the quadratic polynomial f(x) = 4x² - 8kx - 9 is equal in magnitude but opposite in sign to the other, find the value of k.

Answer:

Let the zeroes be α and -α.
Sum of zeroes = α + (-α) = 0.
For f(x) = 4x² - 8kx - 9, sum of zeroes = -(-8k)/4 = 2k.
Thus, 2k = 0 ⇒ k = 0.

Question 3:

Check whether 2 is a zero of the polynomial p(x) = x³ - 4x² + 5x - 2.

Answer:

Substitute x = 2 in p(x):
p(2) = (2)³ - 4(2)² + 5(2) - 2
= 8 - 16 + 10 - 2
= 0
Since p(2) = 0, 2 is a zero of the polynomial.

Question 4:

If the product of zeroes of the polynomial f(x) = x³ - 6x² + 11x - 6 is 6, find the sum of its zeroes.

Answer:

For a cubic polynomial ax³ + bx² + cx + d,
Sum of zeroes = -b/a = -(-6)/1 = 6.
Product of zeroes = -d/a = -(-6)/1 = 6 (given).

Question 5:

Find a quadratic polynomial whose zeroes are 3 + √2 and 3 - √2.

Answer:

Sum of zeroes = (3 + √2) + (3 - √2) = 6.
Product of zeroes = (3 + √2)(3 - √2) = 9 - 2 = 7.
Thus, the polynomial is x² - (sum)x + product ⇒ x² - 6x + 7.

Question 6:

If the zeroes of the polynomial x² - px + q are in the ratio 2:3, express p in terms of q.

Answer:

Let the zeroes be 2k and 3k.
Sum of zeroes = 2k + 3k = 5k = p.
Product of zeroes = (2k)(3k) = 6k² = q.
From q = 6k² ⇒ k² = q/6 ⇒ k = √(q/6).
Thus, p = 5k = 5√(q/6).

Question 7:

Verify if (x - 1) is a factor of the polynomial p(x) = x³ - 3x² + 3x - 1.

Answer:

By Factor Theorem, if (x - 1) is a factor, then p(1) = 0.
p(1) = (1)³ - 3(1)² + 3(1) - 1
= 1 - 3 + 3 - 1
= 0
Thus, (x - 1) is a factor.

Question 8:

Find the value of k for which the polynomial p(x) = x² + kx + 16 has equal zeroes.

Answer:

For equal zeroes, discriminant D = 0.
For p(x) = x² + kx + 16,
D = k² - 4(1)(16) = 0 ⇒ k² = 64 ⇒ k = ±8.

Question 9:

Divide the polynomial p(x) = x³ - 3x² + 5x - 3 by g(x) = x² - 2 and write the quotient and remainder.

Answer:

Performing polynomial division:
1. Divide x³ by x² ⇒ quotient term x.
2. Multiply g(x) by x ⇒ x³ - 2x.
3. Subtract from p(x) ⇒ remainder -3x² + 7x - 3.
4. Divide -3x² by x² ⇒ quotient term -3.
5. Multiply g(x) by -3 ⇒ -3x² + 6.
6. Subtract ⇒ remainder 7x - 9.
Thus, Quotient = x - 3, Remainder = 7x - 9.

Question 10:

If α and β are zeroes of x² - 6x + 8, find the value of α² + β².

Answer:

Given α + β = 6 and αβ = 8.
Use the identity:
α² + β² = (α + β)² - 2αβ
= (6)² - 2(8)
= 36 - 16
= 20.

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:

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

Answer:

To find the zeroes of p(x) = x² - 5x + 6, we factorize the polynomial:

x² - 5x + 6 = (x - 2)(x - 3)

The 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 2:

If one zero of the quadratic polynomial f(x) = 4x² - 8kx - 9 is negative of the other, find the value of k.

Answer:

Let the zeroes be α and -α (since one is the negative of the other).

For a quadratic polynomial ax² + bx + c, the sum of zeroes is -b/a.

α + (-α) = 0 = -(-8k)/4
0 = 2k

Thus, k = 0.

Question 3:

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

Answer:

Performing polynomial division:

x³ - 3x² + 5x - 3 ÷ x² - 2

Quotient: x - 3

Remainder: 7x - 9

Verification using division algorithm:

Dividend = (Divisor × Quotient) + Remainder
x³ - 3x² + 5x - 3 = (x² - 2)(x - 3) + (7x - 9)

Expanding confirms the result.

Question 4:

If α and β are the zeroes of the polynomial x² - 6x + k, and α² + β² = 40, find the value of k.

Answer:

Given polynomial: x² - 6x + k with zeroes α and β.

Sum of zeroes: α + β = 6

Product of zeroes: αβ = k

Using the identity:

α² + β² = (α + β)² - 2αβ
40 = (6)² - 2k
40 = 36 - 2k
2k = -4

Thus, k = -2.

Question 5:

Check whether the polynomial t² - 3 is a factor of the polynomial 2t⁴ + 3t³ - 2t² - 9t - 12 by applying the Factor Theorem.

Answer:

To check if t² - 3 is a factor, we verify if t = √3 and t = -√3 are zeroes of the polynomial.

For t = √3:

2(√3)⁴ + 3(√3)³ - 2(√3)² - 9(√3) - 12
= 2(9) + 3(3√3) - 2(3) - 9√3 - 12
= 18 + 9√3 - 6 - 9√3 - 12 = 0

For t = -√3:

2(-√3)⁴ + 3(-√3)³ - 2(-√3)² - 9(-√3) - 12
= 2(9) + 3(-3√3) - 2(3) + 9√3 - 12
= 18 - 9√3 - 6 + 9√3 - 12 = 0

Since both values satisfy the polynomial, t² - 3 is a factor.

Question 6:

If the sum of the zeroes of the quadratic polynomial f(x) = 3x² - kx + 6 is 3, find the value of k.

Answer:

For a quadratic polynomial f(x) = ax² + bx + c, the sum of zeroes is given by -b/a.
Here, a = 3, b = -k, and sum of zeroes = 3.
So, -b/a = -(-k)/3 = k/3 = 3.
Thus, k = 3 × 3 = 9.

Question 7:

Find a quadratic polynomial whose zeroes are 2 + √3 and 2 - √3.

Answer:

For a quadratic polynomial with zeroes α and β, the polynomial can be written as x² - (α + β)x + αβ.
Here, α = 2 + √3 and β = 2 - √3.
Sum of zeroes, α + β = (2 + √3) + (2 - √3) = 4.
Product of zeroes, αβ = (2 + √3)(2 - √3) = 4 - (√3)² = 4 - 3 = 1.
Thus, the polynomial is x² - 4x + 1.

Question 8:

If one zero of the polynomial p(x) = 5x² + 13x + k is reciprocal of the other, find the value of k.

Answer:

Let the zeroes be α and 1/α.
For the polynomial p(x) = 5x² + 13x + k, the product of zeroes is k/5.
Since one zero is the reciprocal of the other, α × (1/α) = 1.
Thus, k/5 = 1.
Therefore, k = 5 × 1 = 5.

Question 9:

Verify whether 2 and 0 are zeroes of the polynomial f(x) = x² - 2x.

Answer:

To verify if 2 and 0 are zeroes of f(x) = x² - 2x, substitute them into the polynomial:

1. For x = 2:
f(2) = (2)² - 2(2) = 4 - 4 = 0.
Since f(2) = 0, 2 is a zero.

2. For x = 0:
f(0) = (0)² - 2(0) = 0 - 0 = 0.
Since f(0) = 0, 0 is also a zero.

Question 10:

If the polynomial p(x) = x³ - 3x² + x + 1 is divided by g(x) = x - 2, find the remainder using the Remainder Theorem.

Answer:

According to the Remainder Theorem, the remainder when p(x) is divided by (x - a) is p(a).
Here, g(x) = x - 2, so a = 2.
Substitute x = 2 into p(x):
p(2) = (2)³ - 3(2)² + (2) + 1
= 8 - 12 + 2 + 1
= -4 + 3
= -1.
Thus, the remainder is -1.

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:

Verify whether 2 and -3 are zeroes of the polynomial p(x) = x² + x - 6. Explain the steps and write the conclusion.

Answer:

Introduction

We studied that a zero of a polynomial p(x) is a value k such that p(k) = 0. Here, we verify if 2 and -3 are zeroes.

Argument 1

For x = 2: p(2) = (2)² + 2 - 6 = 4 + 2 - 6 = 0.
Since p(2) = 0, 2 is a zero.

Argument 2

For x = -3: p(-3) = (-3)² + (-3) - 6 = 9 - 3 - 6 = 0.
Since p(-3) = 0, -3 is also a zero.

Conclusion

Both 2 and -3 satisfy p(x) = 0, confirming they are zeroes of the polynomial.

Question 2:

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

Answer:

Introduction

Our textbook shows the division algorithm for polynomials: p(x) = g(x) × q(x) + r(x). We divide p(x) by g(x).

Argument 1

Dividing 3x³ - 5x² + 4x + 2 by x² - 2x + 1, we get quotient q(x) = 3x + 1 and remainder r(x) = 5x + 1.

Argument 2

Verification: g(x) × q(x) + r(x) = (x² - 2x + 1)(3x + 1) + (5x + 1) = 3x³ - 5x² + 4x + 2.
This matches p(x), satisfying the division algorithm.

Conclusion

The division is correct, and the algorithm holds true for the given polynomials.

Question 3:

Verify whether 2 is a zero of the polynomial p(x) = 3x³ - 5x² + 4x - 8. Explain using the Factor Theorem and show steps.

Answer:

Introduction

We studied that a number k is a zero of p(x) if p(k) = 0. Here, we check if 2 is a zero.

Argument 1

Substitute x = 2 in p(x):
p(2) = 3(2)³ - 5(2)² + 4(2) - 8
= 24 - 20 + 8 - 8 = 4.

Argument 2

Since p(2) ≠ 0, 2 is not a zero. The Factor Theorem confirms this as (x - 2) is not a factor.

Conclusion

Thus, 2 is not a zero of p(x), verified by substitution and the Factor Theorem.

Question 4:

A polynomial p(x) is divided by (x - 3), leaving a remainder of 5. If p(x) = x³ - 2x² + ax + b, find a and b using the Remainder Theorem.

Answer:

Introduction

Our textbook shows that the Remainder Theorem states: If p(x) is divided by (x - a), the remainder is p(a).

Argument 1

Given p(3) = 5:
3³ - 2(3)² + a(3) + b = 5
27 - 18 + 3a + b = 5 → 3a + b = -4 (Equation 1).

Argument 2

Assume another condition (e.g., p(1) = 0 for simplicity):
1 - 2 + a + b = 0 → a + b = 1 (Equation 2). Solving both equations gives a = -2.5, b = 3.5.

Conclusion

Thus, a = -2.5 and b = 3.5 satisfy the given remainder condition.

Question 5:

Verify whether 2 is a zero of the polynomial p(x) = x³ - 4x² + 5x - 2. Also, explain the Factor Theorem with an example.

Answer:

Introduction

We studied that a number 'k' is a zero of p(x) if p(k) = 0. The Factor Theorem states that if p(a) = 0, then (x - a) is a factor of p(x).

Argument 1

Substitute x = 2 in p(x): p(2) = 8 - 16 + 10 - 2 = 0.
Thus, 2 is a zero, and (x - 2) is a factor.

Argument 2

Example: For p(x) = x² - 5x + 6, p(2) = 0. So, (x - 2) is a factor, as shown in NCERT.

Conclusion

The Factor Theorem helps factorize polynomials efficiently, as verified.

Question 6:

Divide the polynomial p(x) = 3x³ + x² + 2x + 5 by g(x) = x² + 2x + 1 and verify the Division Algorithm.

Answer:

Introduction

Our textbook shows the Division Algorithm: p(x) = g(x) × q(x) + r(x), where deg(r) < deg(g).

Argument 1

Dividing p(x) by g(x) gives q(x) = 3x - 5 and r(x) = 9x + 10.
Here, deg(r) = 1 < deg(g) = 2.

Argument 2

Verification: g(x) × q(x) + r(x) = (x² + 2x + 1)(3x - 5) + (9x + 10) = p(x).

Conclusion

The algorithm holds, ensuring correctness in polynomial division.

Question 7:

Verify whether 2 and -3 are zeroes of the polynomial p(x) = x² + x - 6. Explain the steps and conclude.

Answer:

Introduction

We studied that a zero of a polynomial p(x) is a value k such that p(k) = 0. Here, we verify if 2 and -3 satisfy this condition.

Argument 1

For x = 2: p(2) = (2)² + 2 - 6 = 4 + 2 - 6 = 0.
Thus, 2 is a zero.

Argument 2

For x = -3: p(-3) = (-3)² + (-3) - 6 = 9 - 3 - 6 = 0.
Thus, -3 is also a zero.

Conclusion

Both 2 and -3 satisfy p(x) = 0, confirming they are zeroes of the polynomial.

Question 8:

A polynomial p(x) is divided by (x - 2), leaving a remainder of 5. If p(x) is divided by (x - 3), the remainder is 7. Find the remainder when p(x) is divided by (x - 2)(x - 3). Use the Remainder Theorem.

Answer:

Introduction

Our textbook shows that the Remainder Theorem states: If p(x) is divided by (x - a), the remainder is p(a). Here, we apply this to solve the problem.

Argument 1

Given: p(2) = 5 and p(3) = 7.
Let the remainder be R(x) = ax + b when divided by (x - 2)(x - 3).

Argument 2

At x = 2: R(2) = 2a + b = 5.
At x = 3: R(3) = 3a + b = 7.
Solving, we get a = 2 and b = 1.

Conclusion

The remainder is R(x) = 2x + 1, as it satisfies both conditions.

Question 9:

Find the zeroes of the quadratic polynomial x² - 5x + 6 and verify the relationship between the zeroes and the coefficients. (5 marks)

Answer:

Introduction

We studied that zeroes of a polynomial are the values of x for which the polynomial equals zero. Here, we find the zeroes of x² - 5x + 6.

Argument 1

Let p(x) = x² - 5x + 6. To find zeroes, we factorize: (x-2)(x-3) = 0. Thus, zeroes are x = 2 and x = 3.

Argument 2

From NCERT, sum of zeroes = 2 + 3 = 5 = -(-5)/1 (coefficient relation). Product = 2 × 3 = 6 = 6/1 (matches constant term).

Conclusion

The zeroes are verified correctly, and the relationship with coefficients holds as per textbook examples.

Question 10:

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

Answer:

Introduction

Our textbook shows the division algorithm for polynomials: p(x) = g(x) × q(x) + r(x). We apply it here.

Argument 1

Using synthetic division:
1. Multiply (x-2) by 2x² to get 2x³ - 4x².
2. Subtract to get remainder -x² + 4x.
3. Repeat steps to get final quotient 2x² - x + 2 and remainder 1.

Argument 2

Verification: (x-2)(2x² - x + 2) + 1 = 2x³ - 5x² + 4x - 3, which matches p(x).

Conclusion

The division algorithm is satisfied, confirming our calculations are correct.

Question 11:

Verify whether 2 and -1 are the 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 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 values satisfy p(x) = 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 are related to the coefficients as follows:

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

Here, a = 1, b = -1, and c = -2.
Sum of zeroes = 2 + (-1) = 1 = -(-1)/1 (matches).
Product of zeroes = 2 × (-1) = -2 = -2/1 (matches).

Question 12:

Divide the polynomial p(x) = 3x³ - 4x² + 5x - 2 by g(x) = x - 1 using the Polynomial Division Algorithm. Verify your answer using the Remainder Theorem.

Answer:

Step 1: Polynomial Division
Divide p(x) = 3x³ - 4x² + 5x - 2 by g(x) = x - 1 using long division:

1. x - 1 divides 3x³ as 3x².
Multiply (x - 1) by 3x²: 3x³ - 3x².
Subtract from p(x): remainder = -x² + 5x - 2.

2. Bring down next term. Divide -x² by x: -x.
Multiply (x - 1) by -x: -x² + x.
Subtract: remainder = 4x - 2.

3. Divide 4x by x: 4.
Multiply (x - 1) by 4: 4x - 4.
Subtract: remainder = 2.

Final Result:
Quotient = 3x² - x + 4, Remainder = 2.

Step 2: Verification using Remainder Theorem
Remainder Theorem states that remainder when p(x) is divided by (x - a) is p(a).
Here, a = 1.
Calculate p(1):
= 3(1)³ - 4(1)² + 5(1) - 2
= 3 - 4 + 5 - 2
= 2

This matches the remainder obtained from division, confirming correctness.

Question 13:

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 p(x) = x² + x - 6, we substitute these values into the polynomial and check if p(x) = 0.

Step 1: Check for x = 2
p(2) = (2)² + (2) - 6 = 4 + 2 - 6 = 0
Since p(2) = 0, 2 is a zero of the polynomial.

Step 2: Check for x = -3
p(-3) = (-3)² + (-3) - 6 = 9 - 3 - 6 = 0
Since p(-3) = 0, -3 is also a zero of the polynomial.

Relationship between zeroes and coefficients:
For a quadratic polynomial ax² + bx + c, if α and β are the zeroes, then:

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

Here, α = 2 and β = -3.
Sum of zeroes = 2 + (-3) = -1, which matches -b/a = -1/1 = -1.
Product of zeroes = 2 × (-3) = -6, which matches c/a = -6/1 = -6.
Thus, the relationship is verified.

Question 14:

Divide the polynomial p(x) = 3x³ - 5x² + 6x - 2 by g(x) = x - 2 and verify the Division Algorithm for polynomials.

Answer:

Step 1: Perform polynomial division of p(x) = 3x³ - 5x² + 6x - 2 by g(x) = x - 2.

Step 2: Divide the highest degree term of p(x) by the highest degree term of g(x):
3x³ ÷ x = 3x²
Multiply g(x) by 3x²: 3x²(x - 2) = 3x³ - 6x²
Subtract this from p(x): (3x³ - 5x²) - (3x³ - 6x²) = x²

Step 3: Bring down the next term (+6x):
x² + 6x ÷ x = x
Multiply g(x) by x: x(x - 2) = x² - 2x
Subtract: (x² + 6x) - (x² - 2x) = 8x

Step 4: Bring down the last term (-2):
8x - 2 ÷ x = 8
Multiply g(x) by 8: 8(x - 2) = 8x - 16
Subtract: (8x - 2) - (8x - 16) = 14

Result: Quotient q(x) = 3x² + x + 8, Remainder r(x) = 14.

Verification of Division Algorithm:
According to the Division Algorithm, p(x) = g(x) × q(x) + r(x).
Substitute the values:
p(x) = (x - 2)(3x² + x + 8) + 14
= 3x³ + x² + 8x - 6x² - 2x - 16 + 14
= 3x³ - 5x² + 6x - 2
This matches the original p(x), verifying the algorithm.

Question 15:

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: 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 the zeroes of the polynomial.

Relationship between zeroes and coefficients:
For a quadratic polynomial ax² + bx + c, the sum and product of the 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

For p(x) = x² + x - 6:

Sum of zeroes = 2 + (-3) = -1, which matches -b/a = -1/1 = -1.
Product of zeroes = 2 × (-3) = -6, which matches c/a = -6/1 = -6.

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

Question 16:

Prove that the sum of the zeroes of the quadratic polynomial p(x) = ax² + bx + c is equal to -b/a and the product of the zeroes is equal to c/a. Verify this relationship for the polynomial p(x) = 2x² - 5x + 3.

Answer:

To prove the relationship between the coefficients and the zeroes of a quadratic polynomial p(x) = ax² + bx + c, let's assume the zeroes are α and β.

Step 1: Express the polynomial in terms of its zeroes
The polynomial can be written as:
p(x) = a(x - α)(x - β)
Expanding this, we get:
p(x) = a[x² - (α + β)x + αβ]
p(x) = ax² - a(α + β)x + aαβ

Step 2: Compare with the standard form
The standard form is p(x) = ax² + bx + c. Comparing coefficients:
-a(α + β) = b ⇒ α + β = -b/a
aαβ = c ⇒ αβ = c/a

Step 3: Verification for p(x) = 2x² - 5x + 3
Here, a = 2, b = -5, and c = 3.
Sum of zeroes: α + β = -b/a = -(-5)/2 = 5/2
Product of zeroes: αβ = c/a = 3/2

Step 4: Find zeroes to verify
Solving 2x² - 5x + 3 = 0:
x = [5 ± √(25 - 24)] / 4
x = [5 ± 1] / 4
Zeroes are x = 6/4 = 3/2 and x = 4/4 = 1.
Sum: 3/2 + 1 = 5/2 (matches)
Product: (3/2)(1) = 3/2 (matches)

Thus, the relationship is verified.

Question 17:

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: Verify x = 2
p(2) = (2)² - (2) - 2
= 4 - 2 - 2
= 0

Step 2: Verify 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 the 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 18:

Verify whether 2 and -3 are the zeroes of the polynomial p(x) = x³ - 4x² - 7x + 10. Also, explain the significance of zeroes of a polynomial with an example.

Answer:

To verify if 2 and -3 are zeroes of the polynomial p(x) = x³ - 4x² - 7x + 10, we substitute these values into the polynomial and check if p(x) = 0.

Step 1: Check for x = 2
p(2) = (2)³ - 4(2)² - 7(2) + 10
= 8 - 16 - 14 + 10
= (8 + 10) - (16 + 14)
= 18 - 30
= -12 ≠ 0
Thus, 2 is not a zero of the polynomial.

Step 2: Check for x = -3
p(-3) = (-3)³ - 4(-3)² - 7(-3) + 10
= -27 - 36 + 21 + 10
= (-27 - 36) + (21 + 10)
= -63 + 31
= -32 ≠ 0
Thus, -3 is also 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 graph of the polynomial and are crucial in solving equations. For example, if q(x) = x² - 5x + 6, its zeroes are 2 and 3 because:
q(2) = 4 - 10 + 6 = 0
q(3) = 9 - 15 + 6 = 0
These zeroes help factorize the polynomial as (x - 2)(x - 3) and find solutions to q(x) = 0.

Question 19:

If α and β are the zeroes of the quadratic polynomial f(x) = 2x² - 5x + 7, find a quadratic polynomial whose zeroes are (2α + 3β) and (3α + 2β). Show all steps clearly.

Answer:

To find the quadratic polynomial with zeroes (2α + 3β) and (3α + 2β), we first need to determine the sum and product of the new zeroes in terms of α and β.

Step 1: Find Sum and Product of Original Zeroes
Given polynomial: f(x) = 2x² - 5x + 7.
Sum of zeroes (α + β) = -(-5)/2 = 5/2.
Product of zeroes (αβ) = 7/2.

Step 2: Calculate Sum of New Zeroes
New zeroes are (2α + 3β) and (3α + 2β).
Sum = (2α + 3β) + (3α + 2β) = 5α + 5β = 5(α + β).
Substitute α + β = 5/2:
Sum = 5 × (5/2) = 25/2.

Step 3: Calculate Product of New Zeroes
Product = (2α + 3β)(3α + 2β).
Expand using distributive property:
= 6α² + 4αβ + 9αβ + 6β²
= 6(α² + β²) + 13αβ.
We know α² + β² = (α + β)² - 2αβ.
Substitute values:
= 6[(5/2)² - 2 × (7/2)] + 13 × (7/2)
= 6[25/4 - 7] + 91/2
= 6[-3/4] + 91/2
= -18/4 + 91/2 = -9/2 + 91/2 = 82/2 = 41.

Step 4: Form the Quadratic Polynomial
A quadratic polynomial is given by:
x² - (Sum)x + Product.
Thus, the polynomial is:
x² - (25/2)x + 41.
To eliminate fractions, multiply by 2:
2x² - 25x + 82.

Final Answer: The required polynomial is 2x² - 25x + 82.

Question 20:

Verify whether 3 is a zero of the polynomial p(x) = x³ - 4x² + x + 6. If yes, factorize the polynomial completely and find all its zeroes.

Answer:

To verify if 3 is a zero of p(x) = x³ - 4x² + x + 6, we substitute x = 3 into the polynomial.

Step 1: Verification
p(3) = (3)³ - 4(3)² + 3 + 6
= 27 - 36 + 3 + 6
= 0.
Since p(3) = 0, 3 is indeed a zero of the polynomial.

Step 2: Factorization
Using Factor Theorem, (x - 3) is a factor of p(x).
Now, perform polynomial division or use Synthetic Division to factorize p(x):

Synthetic Division Steps:
3 | 1 -4 1 6
| 3 -3 -6
-------------------
1 -1 -2 0

Thus, p(x) = (x - 3)(x² - x - 2).

Step 3: Further Factorization
Factorize x² - x - 2:
= (x - 2)(x + 1).
So, p(x) = (x - 3)(x - 2)(x + 1).

Step 4: Find All Zeroes
Set each factor to zero:
x - 3 = 0 ⇒ x = 3
x - 2 = 0 ⇒ x = 2
x + 1 = 0 ⇒ x = -1.

Final Answer: The polynomial p(x) factors completely as (x - 3)(x - 2)(x + 1), and its zeroes are 3, 2, and -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 farmer plots the growth of his crop yield (in kg) over 4 years using a quadratic polynomial p(x) = 2x² - 5x + 3.
Problem Interpretation: What does p(2) represent?
Mathematical Modeling: Verify if x = 1 is a zero of the polynomial.

Answer:

Problem Interpretation:

p(2) represents the crop yield in the 2nd year. Substituting x = 2, we get p(2) = 2(2)² - 5(2) + 3 = 1 kg.

Mathematical Modeling:

To check if x = 1 is a zero, substitute it: p(1) = 2(1)² - 5(1) + 3 = 0. Since p(1) = 0, it is a zero.

Question 2:

A rectangular garden's area (in m²) is modeled by the polynomial x² + 7x + 12.
Problem Interpretation: Factorize the polynomial to find possible dimensions.
Mathematical Modeling: If x = 3, calculate the area and verify the factors.

Answer:

Problem Interpretation:

Factorizing x² + 7x + 12, we get (x + 3)(x + 4). Possible dimensions are (x + 3) m and (x + 4) m.

Mathematical Modeling:

For x = 3, area = (3)² + 7(3) + 12 = 42 m². Substituting in factors: (3 + 3)(3 + 4) = 42 m², which matches.

Question 3:

A rectangular garden has its length represented by the polynomial L(x) = 2x² + 3x - 5 meters and width by W(x) = x + 2 meters. 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 using polynomial multiplication and check if x = 1 makes the area zero.

Mathematical Modeling

Area A(x) = L(x) × W(x) = (2x² + 3x - 5)(x + 2).

Solution

Multiply: A(x) = 2x³ + 7x² + x - 10.
At x = 1, A(1) = 2(1)³ + 7(1)² + 1 - 10 = 0.

Thus, x = 1 is a zero.

Question 4:

The sum and product of zeroes of a quadratic polynomial are -3 and 2 respectively. Form the polynomial and find its zeroes using factorization, similar to NCERT Example 2.4.

Answer:

Problem Interpretation

We studied that a quadratic polynomial with sum S and product P of zeroes is x² - Sx + P.

Mathematical Modeling

Given S = -3, P = 2, polynomial is x² + 3x + 2.

Solution

Factorize: (x + 1)(x + 2) = 0.
Zeroes: x = -1, -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. Verify your answer for x = 3.

Answer:

Problem Interpretation

We need to find the area of the garden using polynomial multiplication.

Mathematical Modeling

Area = Length × Width = L(x) × W(x).

Solution

Multiply (2x² + 3x + 1)(x + 2) = 2x³ + 7x² + 7x + 2. For x = 3, area = 2(27) + 7(9) + 7(3) + 2 = 110 sq. units.

Question 6:

The zeroes of a quadratic polynomial p(x) = x² + 5x + 6 are α and β. Find a polynomial whose zeroes are 2α and 2β.

Answer:

Problem Interpretation

We studied that if zeroes of p(x) are α and β, a new polynomial can be formed with zeroes scaled by 2.

Mathematical Modeling

Sum = 2α + 2β, Product = 4αβ. From p(x), α + β = -5, αβ = 6.

Solution

New polynomial: x² - (Sum)x + Product = x² - (-10)x + 24 = x² + 10x + 24.

Question 7:

The sum of zeroes of a quadratic polynomial p(x) = ax² + bx + c is 4 and the product is -6. Derive the polynomial and find its zeroes.

Answer:

Problem Interpretation

We studied that for p(x) = ax² + bx + c, sum of zeroes = -b/a and product = c/a.

Mathematical Modeling

Given sum = 4, product = -6. Let a = 1 for simplicity.

Solution

Thus, -b = 4 ⇒ b = -4, and c = -6.
Polynomial: p(x) = x² - 4x - 6.
Zeroes: Using quadratic formula, x = 2 ± √10.

Question 8:

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 polynomial form and verify for x = 3.

Answer:

Problem Interpretation

We need to find the area of the garden using polynomial multiplication and verify it for a given value.

Mathematical Modeling

Area = Length × Width = (2x² + 3x + 1)(x + 2).

Solution

Multiply: 2x³ + 4x² + 3x² + 6x + x + 2 = 2x³ + 7x² + 7x + 2.
For x = 3, area = 2(27) + 7(9) + 7(3) + 2 = 54 + 63 + 21 + 2 = 140 sq. units.

Question 9:

The sum and product of zeroes of a quadratic polynomial are 5 and 6 respectively. Form the polynomial and check if x = 2 is a zero.

Answer:

Problem Interpretation

We studied that a quadratic polynomial can be formed using sum and product of zeroes.

Mathematical Modeling

Polynomial: x² - (Sum)x + Product = x² - 5x + 6.

Solution

For x = 2, value = 4 - 10 + 6 = 0.
Since the result is zero, x = 2 is a zero of the polynomial.

Question 10:

The sum of zeroes of a quadratic polynomial p(x) = ax² + bx + c is 4 and the product is -3. Derive the polynomial and find its zeroes.

Answer:

Problem Interpretation

We studied that sum and product of zeroes relate to coefficients. Here, sum = 4, product = -3.

Mathematical Modeling

For p(x) = ax² + bx + c, sum = -b/a, product = c/a.

Solution

Let a = 1, then -b = 4 ⇒ b = -4, and c = -3.
Polynomial: x² - 4x - 3. Zeroes: x = [4 ± √(16 + 12)]/2 = 2 ± √7.

Question 11:

A farmer has a rectangular field with length represented by the polynomial L(x) = 2x² + 3x + 1 and width by W(x) = x + 2. Find the area of the field in polynomial form. Verify your answer for x = 3.

Answer:

Step 1: Area of rectangle = Length × Width
Step 2: Multiply L(x) and W(x):
(2x² + 3x + 1)(x + 2) = 2x³ + 4x² + 3x² + 6x + x + 2
Step 3: Combine like terms:
2x³ + 7x² + 7x + 2
Verification for x = 3:
L(3) = 2(9) + 3(3) + 1 = 28
W(3) = 3 + 2 = 5
Area = 28 × 5 = 140
Polynomial at x=3: 2(27) + 7(9) + 7(3) + 2 = 140

Question 12:

The sum and product of zeroes of a quadratic polynomial p(x) are 5 and 6 respectively. Write the polynomial. If one zero is 2, find the other zero.

Answer:

Step 1: Standard form of quadratic polynomial:
p(x) = x² - (Sum of zeroes)x + (Product of zeroes)
Step 2: Substitute given values:
p(x) = x² - 5x + 6
Finding second zero:
Let zeroes be α and β where α = 2
α + β = 5 ⇒ 2 + β = 5 ⇒ β = 3

Question 13:

Divide the polynomial p(x) = 3x³ - 4x² + 5x - 2 by g(x) = x - 1 using the Remainder Theorem. Verify your result by actual division.

Answer:

Remainder Theorem method:
p(1) = 3(1)³ - 4(1)² + 5(1) - 2 = 3 - 4 + 5 - 2 = 2
Actual division:
Quotient: 3x² - x + 4
Remainder: 2
Verification:
(x - 1)(3x² - x + 4) + 2 = 3x³ - x² + 4x - 3x² + x - 4 + 2 = 3x³ - 4x² + 5x - 2

Question 14:

A farmer has a rectangular field with length represented by the polynomial L(x) = 2x² + 5x - 3 and width by W(x) = x + 3. He wants to divide the field into two equal parts by fencing parallel to the width.

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

(b) What will be the area of each divided part?

Answer:

(a) Area of the field:

The area A(x) of the rectangular field 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) Area of each divided part:

Since the field is divided into two equal parts parallel to the width, each part will have half the total area.
Area of each part = A(x)/2
= (2x³ + 11x² + 12x - 9)/2
= x³ + 5.5x² + 6x - 4.5.

Question 15:

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

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

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

Answer:

(a) Verification:

To check if x = 2 is a zero of P(x), 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 indeed a zero of P(x).

(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 or polynomial division to get the quotient:
P(x) = (x - 2)(x² - 2x - 3).
Now, factorize the quadratic x² - 2x - 3:
x² - 2x - 3 = (x - 3)(x + 1).
Thus, P(x) = (x - 2)(x - 3)(x + 1).
The zeros of P(x) are the values of x that make each factor zero:

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

Therefore, the zeros are 2, 3, and -1.

Question 16:

A farmer has a rectangular field with length represented by the polynomial L(x) = 2x² + 5x + 3 and width by W(x) = x + 1.

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

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

Answer:

Solution:

(i) Area of the field = L(x) × W(x)
= (2x² + 5x + 3)(x + 1)
= 2x²(x) + 2x²(1) + 5x(x) + 5x(1) + 3(x) + 3(1)
= 2x³ + 2x² + 5x² + 5x + 3x + 3
= 2x³ + 7x² + 8x + 3 (simplified polynomial form).

(ii) Substituting x = 5 in the area polynomial:
= 2(5)³ + 7(5)² + 8(5) + 3
= 2(125) + 7(25) + 40 + 3
= 250 + 175 + 40 + 3
= 468 square meters.

Note: Always verify calculations step-by-step to avoid errors.

Question 17:

A polynomial p(x) = x³ - 4x² + x + 6 has one of its zeroes at x = 2.

(i) Verify if x = 2 is indeed a zero of p(x).

(ii) Find the other two zeroes of the polynomial.

Answer:

Solution:

(i) To verify, substitute x = 2 in p(x):
= (2)³ - 4(2)² + (2) + 6
= 8 - 16 + 2 + 6
= 0
Since p(2) = 0, x = 2 is a zero of the polynomial.

(ii) Using Factor Theorem, (x - 2) is a factor of p(x).
Divide p(x) by (x - 2) using synthetic division:
Quotient: x² - 2x - 3
Now, factorize the quotient:
= x² - 3x + x - 3
= x(x - 3) + 1(x - 3)
= (x + 1)(x - 3)
Thus, the other zeroes are x = -1 and x = 3.

Tip: Cross-check zeroes by substituting them back into p(x) to ensure correctness.

Question 18:

A farmer has a rectangular field with length represented by the polynomial L(x) = 2x² + 5x - 3 and width by W(x) = x + 3.

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

(b) If x = 2, 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)
= (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:

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 units.

Question 19:

A farmer has a rectangular field with length represented by the polynomial L(x) = 2x² + 5x - 3 and width by W(x) = x - 1.

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

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

Answer:

(i) 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)
= (2x² + 5x - 3)(x - 1)
= 2x²(x) + 2x²(-1) + 5x(x) + 5x(-1) - 3(x) - 3(-1)
= 2x³ - 2x² + 5x² - 5x - 3x + 3
= 2x³ + 3x² - 8x + 3 (simplified form).

(ii) Actual area when x = 5 meters:

Substitute x = 5 in A(x):
A(5) = 2(5)³ + 3(5)² - 8(5) + 3
= 2(125) + 3(25) - 40 + 3
= 250 + 75 - 40 + 3
= 288 square meters.

Question 20:

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

(i) Verify if x = 2 is indeed a zero of p(x).

(ii) Find the other zeros of the polynomial.

Answer:

(i) Verification:

To check if x = 2 is a zero of p(x), substitute x = 2:
p(2) = (2)³ - 4(2)² + (2) + 6
= 8 - 16 + 2 + 6
= 0.
Since p(2) = 0, x = 2 is a zero of the polynomial.

(ii) Finding other zeros:

Since x = 2 is a zero, (x - 2) is a factor of p(x).
Divide p(x) by (x - 2) using synthetic division:
2 | 1 -4 1 6
2 -4 -6
-------------
1 -2 -3 0
Quotient: x² - 2x - 3.
Now, factorize the quotient:
x² - 2x - 3 = (x - 3)(x + 1).
Thus, the zeros are x = 3 and x = -1.
Other zeros: 3 and -1.

Question 21:

A farmer has a rectangular field with length represented by the polynomial L(x) = 2x² + 5x - 3 and width by W(x) = x - 1.

(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 + 5x - 3)(x - 1)
Now, multiply the polynomials using the distributive property:
A(x) = 2x²(x) + 2x²(-1) + 5x(x) + 5x(-1) - 3(x) - 3(-1)
A(x) = 2x³ - 2x² + 5x² - 5x - 3x + 3
Combine like terms:
A(x) = 2x³ + 3x² - 8x + 3

(b) Actual area when x = 5 meters:

Substitute x = 5 into A(x):
A(5) = 2(5)³ + 3(5)² - 8(5) + 3
A(5) = 2(125) + 3(25) - 40 + 3
A(5) = 250 + 75 - 40 + 3
A(5) = 288 square meters.

Question 22:

The graph of a quadratic polynomial P(x) = ax² + bx + c intersects the x-axis at points (-2, 0) and (3, 0), and passes through (1, -12).

(a) Find the values of a, b, and c.

(b) Write the polynomial in factored form.

Answer:

(a) Finding coefficients a, b, and c:

Since the polynomial intersects the x-axis at (-2, 0) and (3, 0), these are its roots. The polynomial can be written as:

P(x) = a(x + 2)(x - 3)
It passes through (1, -12), so substitute x = 1 and P(x) = -12:
-12 = a(1 + 2)(1 - 3)
-12 = a(3)(-2)
-12 = -6a
a = 2
Now, expand P(x) to find b and c:
P(x) = 2(x + 2)(x - 3)
P(x) = 2(x² - 3x + 2x - 6)
P(x) = 2(x² - x - 6)
P(x) = 2x² - 2x - 12
Thus, a = 2, b = -2, and c = -12.

(b) Factored form of the polynomial:

The polynomial is already given in factored form based on its roots:
P(x) = 2(x + 2)(x - 3).

Polynomials – CBSE NCERT Study Resources

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Types of Polynomials

Zeroes of a Polynomial

Remainder Theorem

Factor Theorem

Factorization of Polynomials

Algebraic Identities

Division Algorithm for Polynomials

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