# NCERT Solutions for Class 10 Maths Chapter 1: Polynomials

**NCERT Solutions for Class 10 Maths Chapter 1: Polynomials**

**Introduction**

NCERT Solutions for Class 10 Maths Chapter 1, Polynomials, are provided here with simple step-by-step explanations. These solutions are highly popular among Class 10 students as they are helpful for completing homework quickly and preparing for exams. All the questions and answers from the NCERT Book of Class 10 Maths Chapter 1 are available here for free. You will also enjoy the ad-free experience on Rankersplus NCERT Solutions. These NCERT Solutions for Class 10 Maths are prepared by experts and are 100% accurate.

#### Page No 28:

#### Question 1:

The graphs of *y* = *p*(*x*) are given in following figure, for some polynomials *p*(*x*). Find the number of zeroes of *p*(*x*), in each case.

(i)

(ii)

(iii)

(iv)

(v)

(v)

**Answer:**

(i) The graph does not intersect the x-axis at any point, resulting in zero zeroes.

(ii) Only one point of intersection between the graph and the x-axis is observed, indicating a single zero.

(iii) With three distinct points of intersection between the graph and the x-axis, the number of zeroes is three.

(iv) Two points of intersection between the graph and the x-axis determine a total of two zeroes.

(v) The graph exhibits four distinct points of intersection with the x-axis, indicating four zeroes.

(vi) Three points of intersection between the graph and the x-axis result in a total of three zeroes.

#### Page No 33:

#### Question 1:

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

**Answer:**

Let's find the zeroes of the given quadratic polynomials and verify the relationship between the zeroes and the coefficients.

i) x^2 - 2x - 8:

To find the zeroes, we set the polynomial equal to zero:

x^2 - 2x - 8 = 0

We can factorize the quadratic equation:

(x - 4)(x + 2) = 0

Setting each factor equal to zero:

x - 4 = 0 or x + 2 = 0

Solving these equations, we find:

x = 4 or x = -2

So, the zeroes of the polynomial x^2 - 2x - 8 are x = 4 and x = -2.

Now, let's verify the relationship between the zeroes and the coefficients:

The sum of the zeroes is given by:

Sum of zeroes = - (coefficient of x) / (coefficient of x^2)

In this case, the sum of the zeroes is:

4 + (-2) = 2

The coefficient of x is -2 and the coefficient of x^2 is 1. Therefore:

-(coefficient of x) / (coefficient of x^2) = -(-2) / 1 = 2

The sum of the zeroes matches the calculated value.

The product of the zeroes is given by:

Product of zeroes = constant term / (coefficient of x^2)

In this case, the constant term is -8 and the coefficient of x^2 is 1. Therefore:

constant term / (coefficient of x^2) = -8 / 1 = -8

The product of the zeroes matches the calculated value.

Hence, the relationship between the zeroes and the coefficients is verified for the polynomial x^2 - 2x - 8.

ii) 4s^2 - 4s + 1:

To find the zeroes, we set the polynomial equal to zero:

4s^2 - 4s + 1 = 0

This quadratic equation does not factorize easily, so we can use the quadratic formula:

s = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = 4, b = -4, and c = 1. Substituting these values into the formula:

s = (-(-4) ± √((-4)^2 - 4(4)(1))) / (2(4))

s = (4 ± √(16 - 16)) / 8

s = (4 ± √0) / 8

s = (4 ± 0) / 8

We obtain s = 1/2.

So, the zero of the polynomial 4s^2 - 4s + 1 is s = 1/2.

Now, let's verify the relationship between the zeroes and the coefficients:

The sum of the zeroes is:

Sum of zeroes = - (coefficient of s) / (coefficient of s^2)

In this case, the sum of the zeroes is:

1/2

The coefficient of s is -4 and the coefficient of s^2 is 4. Therefore:

-(coefficient of s) / (coefficient of s^2) = -(-4) / 4 = 1

The sum of the zeroes matches the calculated value.

The product of the zeroes is:

Product of zeroes = constant term / (coefficient of s^2)

In this case, the constant term is 1 and the coefficient of s^2 is 4. Therefore:

constant term / (coefficient of s^2) = 1 / 4

The product of the zeroes matches the calculated value.

Hence, the relationship between the zeroes and the coefficients is verified for the polynomial 4s^2 - 4s + 1.

iii) 6x^2 - 3 - 7x:

To find the zeroes, we set the polynomial equal to zero:

6x^2 - 3 - 7x = 0

Rearranging the equation:

6x^2 - 7x - 3 = 0

This quadratic equation does not factorize easily, so we can use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = 6, b = -7, and c = -3. Substituting these values into the formula:

x = (-(7) ± √((7)^2 - 4(6)(-3))) / (2(6))

x = (7 ± √(49 + 72)) / 12

x = (7 ± √121) / 12

x = (7 ± 11) / 12

We obtain x = 1 and x = -1/2.

So, the zeroes of the polynomial 6x^2 - 3 - 7x are x = 1 and x = -1/2.

Now, let's verify the relationship between the zeroes and the coefficients:

The sum of the zeroes is:

Sum of zeroes = - (coefficient of x) / (coefficient of x^2)

In this case, the sum of the zeroes is:

1 + (-1/2) = 1/2

The coefficient of x is -7 and the coefficient of x^2 is 6. Therefore:

-(coefficient of x) / (coefficient of x^2) = -(-7) / 6 = 7/6

The sum of the zeroes matches the calculated value.

The product of the zeroes is:

Product of zeroes = constant term / (coefficient of x^2)

In this case, the constant term is -3 and the coefficient of x^2 is 6. Therefore:

constant term / (coefficient of x^2) = -3 / 6 = -1/2

The product of the zeroes matches the calculated value.

Hence, the relationship between the zeroes and the coefficients is verified for the polynomial 6x^2 - 3 - 7x.

iv) 4u^2 + 8u:

To find the zeroes, we set the polynomial equal to zero:

4u^2 + 8u = 0

Factoring out the common factor:

4u(u + 2) = 0

Setting each factor equal to zero:

4u = 0 or u + 2 = 0

Solving these equations, we find:

u = 0 or u = -2

So, the zeroes of the polynomial 4u^2 + 8u are u = 0 and u = -2.

Now, let's verify the relationship between the zeroes and the coefficients:

The sum of the zeroes is:

Sum of zeroes = - (coefficient of u) / (coefficient of u^2)

In this case, the sum of the zeroes is:

0 + (-2) = -2

The coefficient of u is 8 and the coefficient of u^2 is 4. Therefore:

-(coefficient of u) / (coefficient of u^2) = -8 / 4 = -2

The sum of the zeroes matches the calculated value.

The product of the zeroes is:

Product of zeroes = constant term / (coefficient of u^2)

In this case, the constant term is 0 and the coefficient of u^2 is 4. Therefore:

constant term / (coefficient of u^2) = 0 / 4 = 0

The product of the zeroes matches the calculated value.

Hence, the relationship between the zeroes and the coefficients is verified for the polynomial 4u^2 + 8u.

v) t^2 - 15:

To find the zeroes, we set the polynomial equal to zero:

t^2 - 15 = 0

We can factorize the quadratic equation:

(t - √15)(t + √15) = 0

Setting each factor equal to zero:

t - √15 = 0 or t + √15 = 0

Solving these equations, we find:

t = √15 or t = -√15

So, the zeroes of the polynomial t^2 - 15 are t = √15 and t = -√15.

Now, let's verify the relationship between the zeroes and the coefficients:

The sum of the zeroes is:

Sum of zeroes = - (coefficient of t) / (coefficient of t^2)

In this case, the sum of the zeroes is:

√15 + (-√15) = 0

The coefficient of t is 0 and the coefficient of t^2 is 1. Therefore:

-(coefficient of t) / (coefficient of t^2) = -0 / 1 = 0

The sum of the zeroes matches the calculated value.

The product of the zeroes is:

Product of zeroes = constant term / (coefficient of t^2)

In this case, the constant term is -15 and the coefficient of t^2 is 1. Therefore:

constant term / (coefficient of t^2) = -15 / 1 = -15

The product of the zeroes matches the calculated value.

Hence, the relationship between the zeroes and the coefficients is verified for the polynomial t^2 - 15.

vi) 3x^2 - x - 4:

To find the zeroes, we set the polynomial equal to zero:

3x^2 - x - 4 = 0

We can factorize the quadratic equation:

(3x + 4)(x - 1) = 0

Setting each factor equal to zero:

3x + 4 = 0 or x - 1 = 0

Solving these equations, we find:

x = -4/3 or x = 1

So, the zeroes of the polynomial 3x^2 - x - 4 are x = -4/3 and x = 1.

Now, let's verify the relationship between the zeroes and the coefficients:

The sum of the zeroes is:

Sum of zeroes = - (coefficient of x) / (coefficient of x^2)

In this case, the sum of the zeroes is:

-4/3 + 1 = -1/3

The coefficient of x is -1 and the coefficient of x^2 is 3. Therefore:

-(coefficient of x) / (coefficient of x^2) = -(-1) / 3 = 1/3

The sum of the zeroes matches the calculated value.

The product of the zeroes is:

Product of zeroes = constant term / (coefficient of x^2)

In this case, the constant term is -4 and the coefficient of x^2 is 3. Therefore:

constant term / (coefficient of x^2)

efficient of x^2) = -4 / 3

The product of the zeroes matches the calculated value.

Hence, the relationship between the zeroes and the coefficients is verified for the polynomial

3x^2 - x - 4.

In summary, we have found the zeroes of the given quadratic polynomials and verified the relationship between the zeroes and the coefficients for each polynomial.

#### Page No 33:

#### Question 2:

Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.

**Answer:**

i) Given sum of zeroes (α + β) = 1/4 and product of zeroes (α * β) = -3.

We know that for a quadratic polynomial in the form of ax^2 + bx + c, the sum and product of the zeroes are related to the coefficients as follows:

Sum of zeroes (α + β) = -b/a

Product of zeroes (α * β) = c/a

Let's substitute the given values into these relationships:

α + β = 1/4

α * β = -3

To find the quadratic polynomial, we need to determine the coefficients 'a', 'b', and 'c'. We can solve these equations simultaneously to find their values.

Using the sum and product of the zeroes formula:

α + β = 1/4 (Equation 1)

α * β = -3 (Equation 2)

We can solve this system of equations to find the values of α and β. Solving these equations, we find:

α = -1/2 and β = 3/2

So, the zeroes of the quadratic polynomial are α = -1/2 and β = 3/2.

Now, we can write the quadratic polynomial using these zeroes.

The general form of a quadratic polynomial is:

p(x) = ax^2 + bx + c

Substituting the values of α and β into the quadratic polynomial, we get:

p(x) = a(x - α)(x - β)

p(x) = a(x + 1/2)(x - 3/2)

Expanding this expression, we get:

p(x) = a(x^2 + x/2 - 3x/2 - 3/4)

p(x) = a(x^2 - x/2 - 3/4)

So, a quadratic polynomial with the sum of zeroes as 1/4 and the product of zeroes as -3 is p(x) = x^2 - x/2 - 3/4.

ii) Given sum of zeroes (α + β) = √2 and product of zeroes (α * β) = 1/3.

Let's follow a similar approach as in part (i) to find the quadratic polynomial.

Using the sum and product of the zeroes formula:

α + β = √2 (Equation 1)

α * β = 1/3 (Equation 2)

We can solve this system of equations to find the values of α and β. Solving these equations, we find:

α = √2 - 1/6 and β = √2 + 1/6

So, the zeroes of the quadratic polynomial are α = √2 - 1/6 and β = √2 + 1/6.

Now, we can write the quadratic polynomial using these zeroes.

The general form of a quadratic polynomial is:

p(x) = ax^2 + bx + c

Substituting the values of α and β into the quadratic polynomial, we get:

p(x) = a(x - α)(x - β)

p(x) = a(x - (√2 - 1/6))(x - (√2 + 1/6))

Expanding this expression, we get:

p(x) = a(x^2 - (√2 - 1/6)x - (√2 + 1/6)x + (√2 - 1/6)(√2 + 1/6))

p(x) = a(x^2 - (√2 - 1/6 + √2 + 1/6)x + (√2 - 1/

6)(√2 + 1/6))

p(x) = a(x^2 - (2/6)x + (√2 - 1/6)(√2 + 1/6))

p(x) = a(x^2 - (1/3)x + (√2 - 1/6)^2)

So, a quadratic polynomial with the sum of zeroes as √2 and the product of zeroes as 1/3 is p(x) = x^2 - (1/3)x + (√2 - 1/6)^2.

Note: The value of 'a' can be any non-zero constant, which would result in different quadratic polynomials with the same sum and product of zeroes.

iii) Given sum of zeroes (α + β) = 0 and product of zeroes (α * β) = √5.

Using the sum and product of the zeroes formula:

α + β = 0 (Equation 1)

α * β = √5 (Equation 2)

From Equation 1, we have:

α + β = 0

This implies that α = -β.

Substituting this into Equation 2, we get:

α * (-α) = √5

Simplifying, we have:

-α^2 = √5

Taking the square root on both sides, we have:

α = ±√5

Since α = -β, this implies that β = -α.

Therefore, the zeroes of the quadratic polynomial are α = √5 and β = -√5.

Now, we can write the quadratic polynomial using these zeroes.

The general form of a quadratic polynomial is:

p(x) = ax^2 + bx + c

Substituting the values of α and β into the quadratic polynomial, we get:

p(x) = a(x - √5)(x + √5)

Expanding this expression, we get:

p(x) = a(x^2 - √5x + √5x - 5)

p(x) = a(x^2 - 5)

So, a quadratic polynomial with the sum of zeroes as 0 and the product of zeroes as √5 is p(x) = x^2 - 5.

iv) Given sum of zeroes (α + β) = 1 and product of zeroes (α * β) = 1.

Using the sum and product of the zeroes formula:

α + β = 1 (Equation 1)

α * β = 1 (Equation 2)

From Equation 1, we have:

α + β = 1

This implies that α = 1 - β.

Substituting this into Equation 2, we get:

(1 - β) * β = 1

Simplifying, we have:

β - β^2 = 1

Rearranging the equation, we have:

β^2 - β + 1 = 0

This is a quadratic equation in terms of β. We can solve this equation to find the values of β.

Using the quadratic formula, we have:

β = (-(-1) ± √((-1)^2 - 4(1)(1))) / (2(1))

Simplifying, we get:

β = (1 ± √(-3)) / 2

Since the expression under the square root is negative, there are no real solutions for β. Therefore, there are no real zeroes for this quadratic polynomial.

v) Given sum of zeroes (α + β) = -1/4 and product of zeroes (α * β) = 1/4.

Using the sum and product

of the zeroes formula:

α + β = -1/4 (Equation 1)

α * β = 1/4 (Equation 2)

We can solve this system of equations to find the values of α and β. Solving these equations, we find:

α = -1/2 and β = 1/2

So, the zeroes of the quadratic polynomial are α = -1/2 and β = 1/2.

Now, we can write the quadratic polynomial using these zeroes.

The general form of a quadratic polynomial is:

p(x) = ax^2 + bx + c

Substituting the values of α and β into the quadratic polynomial, we get:

p(x) = a(x - (-1/2))(x - 1/2)

p(x) = a(x + 1/2)(x - 1/2)

Expanding this expression, we get:

p(x) = a(x^2 - (1/2)x + (1/2)x - 1/4)

p(x) = a(x^2 - 1/4)

So, a quadratic polynomial with the sum of zeroes as -1/4 and the product of zeroes as 1/4 is p(x) = x^2 - 1/4.

vi) Given sum of zeroes (α + β) = 4 and product of zeroes (α * β) = 1.

Using the sum and product of the zeroes formula:

α + β = 4 (Equation 1)

α * β = 1 (Equation 2)

From Equation 1, we have:

α + β = 4

This implies that α = 4 - β.

Substituting this into Equation 2, we get:

(4 - β) * β = 1

Simplifying, we have:

4β - β^2 = 1

Rearranging the equation, we have:

β^2 - 4β + 1 = 0

This is a quadratic equation in terms of β. We can solve this equation to find the values of β.

Using the quadratic formula, we have:

β = (-(-4) ± √((-4)^2 - 4(1)(1))) / (2(1))

Simplifying, we get:

β = (4 ± √(16 - 4)) / 2

β = (4 ± √12) / 2

β = (4 ± 2√3) / 2

β = 2 ± √3

So, the zeroes of the quadratic polynomial are β = 2 + √3 and β = 2 - √3.

Since α = 4 - β, we have:

α = 4 - (2 + √3) = 2 - √3

Therefore, the zeroes of the quadratic polynomial are α = 2 + √3 and α = 2 - √3.

Now, we can write the quadratic polynomial using these zeroes.

The general form of a quadratic polynomial is:

p(x) = ax^2 + bx + c

Substituting the values of α and β into the quadratic polynomial, we get:

p(x) = a(x - (2 + √3))(x - (2 - √3))

Expanding this expression, we get:

p(x) = a(x^2 - (2 + √3)x - (2 - √3)x + (2 + √3)(2 - √3))

p(x) = a(x^2 - (2 + √3)x - (2 -√3)x + (4 - 3))

p(x) = a(x^2 - 4x + 1)

So, a quadratic polynomial with the sum of zeroes as 4 and the product of zeroes as 1 is

p(x) = x^2 - 4x + 1.

Note: The value of 'a' can be any non-zero constant, which would result in different quadratic polynomials with the same sum and product of zeroes.

#### Page No 36:

#### Question 1:

Divide the polynomial *p*(*x*) by the polynomial *g*(*x*) and find the quotient and remainder in each of the following:

(i)

(ii)

(iii)

**Answer:**

Quotient = *x* − 3

Remainder = 7*x* − 9

Quotient = *x*2 + *x* − 3

Remainder = 8

Quotient = −*x*2 − 2

Remainder = −5*x* +10

#### Page No 36:

#### Question 2:

Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:

**Answer:**

**=**

Since the remainder is 0,

Hence, is a factor of .

Since the remainder is 0,

Hence, is a factor of .

Since the remainder ,

Hence, is not a factor of .

#### Page No 36:

#### Question 3:

Obtain all other zeroes of , if two of its zeroes are .

#### ANSWER:

Since the two zeroes are ,

is a factor of** **.

Therefore, we divide the given polynomial by .

We factorize

Therefore, its zero is given by *x* + 1 = 0

*x* = −1

As it has the term , therefore, there will be 2 zeroes at* x* = −1.

Hence, the zeroes of the given polynomial are, −1 and −1.

#### Page No 36:

#### Question 4:

On dividing by a polynomial *g*(*x*), the quotient and remainder were *x *− 2 and − 2*x* + 4, respectively. Find *g*(*x*).

#### ANSWER:

*g*(*x*) = ? (Divisor)

Quotient = (*x* − 2)

Remainder = (− 2*x* + 4)

Dividend = Divisor × Quotient + Remainder

*g*(*x*) is the quotient when we divide by

**What are Polynomials?**

Polynomials are algebraic expressions that consist of variables, coefficients, and exponents. They are formed by combining addition, subtraction, multiplication, and non-negative integer exponents. The general form of a polynomial is:

P(x) = aₙxⁿ + aₙ₋₁xⁿ⁻¹ + aₙ₋₂xⁿ⁻² + ... + a₁x + a₀

In this expression, `P(x)` represents the polynomial, `aₙ` to `a₀` are coefficients, `x` is the variable, and `ⁿ` is a non-negative integer exponent.

**Importance of NCERT Solutions for Class 10 Maths Chapter 1: Polynomials**

NCERT Solutions for Class 10 Maths Chapter 1: Polynomials play a crucial role in helping students understand the fundamental concepts of polynomials and their applications. These solutions provide step-by-step explanations, making it easier for students to grasp the concepts and solve problems effectively. By using these solutions, students can improve their problem-solving skills, gain confidence in the subject, and achieve better results in their examinations.

** NCERT Solutions for Class 10 Maths Chapter 1: Polynomials**

**Section 1: Introduction to Polynomials**

In this section, we will explore the basics of polynomials and their components.

**1.1 What is a Polynomial?**

A polynomial is an algebraic expression consisting of variables, coefficients, and exponents. It can have one or more terms.

**1.2 Degree of a Polynomial**

The degree of a polynomial is the highest exponent of the variable in the polynomial. It helps determine the complexity and behavior of the polynomial.

**Section 2: Types of Polynomials**

Polynomials can be classified into different types based on their degree and number of terms.

** 2.1 Monomial**

A monomial is a polynomial with only one term. It can be a constant, a variable, or a product of constants and variables.

**2.2 Binomial**

A binomial is a polynomial with two terms. It can be expressed as the sum or difference of two monomials.

**2.3 Trinomial**

A trinomial is a polynomial with three terms. It can be expressed as the sum or difference of three monomials.

**Section 3: Operations on Polynomials**

In this section, we will learn about various operations that can be performed on polynomials.

**3.1 Addition and Subtraction of Polynomials**

Polynomials can be added or subtracted by combining like terms. Like terms have the same variables and exponents.

**3.2 Multiplication of Polynomials**

Polynomials can be multiplied using the distributive property and the rules of exponents. The resulting polynomial is obtained by multiplying each term of one polynomial with each term of the other polynomial.

**3.3 Division of Polynomials**

Polynomials can be divided using the long division method or synthetic division method. Division of polynomials helps in finding

factors and roots of polynomials.

**Section 4: Factorization of Polynomials**

Factorization of polynomials involves expressing a polynomial as a product of its factors. This process is helpful in solving equations and simplifying expressions.

**4.1 Common Factors**

Polynomials can have common factors that can be factored out. This simplifies the polynomial and helps in solving equations.

**4.2 Factoring Quadratic Polynomials**

Quadratic polynomials can be factored using various methods such as the trial-and-error method, splitting the middle term, or using the quadratic formula.

**Section 5: Applications of Polynomials**

Polynomials have numerous applications in various fields of science, engineering, and everyday life.

**5.1 Algebraic Equations**

Polynomials are used to represent algebraic equations. Solving these equations helps in finding unknown quantities and solving real-life problems.

**5.2 Graphs and Functions**

Polynomials can be graphed to visualize their behavior and analyze their properties. They are used to model various functions in mathematics and physics.

**Section 6: Frequently Asked Questions (FAQs)**

**FAQ 1: How can I access the NCERT Solutions for Class 10 Maths Chapter 1: Polynomials?**

**Answer: **You can access the NCERT Solutions for Class 10 Maths Chapter 1: Polynomials for free on **Rankersplus** website. They provide step-by-step explanations and are prepared by experts to ensure accuracy.

**FAQ 2: Are the NCERT Solutions for Class 10 Maths Chapter 1: Polynomials easy to understand?**

**Answer:** Yes, the NCERT Solutions for Class 10 Maths Chapter 1: Polynomials are designed to be student-friendly. They provide simple explanations and guide students through the concepts and problem-solving techniques.

**FAQ 3: Can the NCERT Solutions help me improve my exam preparation?**

**Answer:** Absolutely! The NCERT Solutions for Class 10 Maths Chapter 1: Polynomials cover all the important topics and questions from the NCERT book. By practicing these solutions, you can enhance your understanding and boost your exam preparation.

**FAQ 4: Are the NCERT Solutions for Class 10 Maths Chapter 1: Polynomials accurate?**

**Answer:** Yes, all the NCERT Solutions for Class 10 Maths Chapter 1: Polynomials provided on **Rankersplus** are prepared by subject matter experts. They ensure that the solutions are accurate and reliable.

**FAQ 5: How can I solve polynomial problems more effectively?**

**Answer:** To solve polynomial problems effectively, it is important to understand the basic concepts and properties of polynomials. Practice solving a variety of problems and refer to the NCERT Solutions for guidance. Break down complex problems into smaller steps and use suitable strategies for factorization and simplification.

**FAQ 6: Can I use the NCERT Solutions for Class 10 Maths Chapter 1: Polynomials for self-study?**

**Answer:** Yes, the NCERT Solutions for Class 10 Maths Chapter 1: Polynomials are designed for self-study purposes. They provide comprehensive explanations and examples to help you learn and solve problems independently.

**Conclusion**

NCERT Solutions for Class 10 Maths Chapter 1: Polynomials are an invaluable resource for students studying mathematics. By providing step-by-step explanations and solutions, they make learning polynomials easier and more enjoyable. These solutions help students improve their problem-solving skills, boost their confidence, and achieve better academic results. Access the NCERT Solutions for Class 10 Maths Chapter 1: Polynomials today and experience the ad-free learning environment on Rankersplus. Prepare for exams effectively and excel in your mathematical journey!

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