Our ncert solutions for Exercise 2.3 Class 10 maths 2. Polynomials - ncert solutions - Toppers Study is the best material for English Medium students cbse board and other state boards students.

# Exercise 2.3 Class 10 maths 2. Polynomials - ncert solutions - Toppers Study

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## Exercise 2.3 Class 10 maths 2. Polynomials - ncert solutions - Toppers Study

CBSE board students who preparing for **class 10 ncert solutions maths and Mathematics** solved exercise **chapter 2. Polynomials** available and this helps in upcoming exams
2024-2025.

### You can Find Mathematics solution Class 10 Chapter 2. Polynomials

- All Chapter review quick revision notes for chapter 2. Polynomials Class 10
- NCERT Solutions And Textual questions Answers Class 10 Mathematics
- Extra NCERT Book questions Answers Class 10 Mathematics
- Importatnt key points with additional Assignment and questions bank solved.

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### Exercise 2.3 class 10 Mathematics Chapter 2. Polynomials

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- Exercise 2.3 Class 10 Maths 2. Polynomials - Ncert Solutions - Toppers Study
- Class 10 Ncert Solutions
- Solution Chapter 2. Polynomials Class 10
- Solutions Class 10
- Chapter 2. Polynomials Exercise 2.3 Class 10

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## 2. Polynomials

### | Exercise 2.3 |

## Exercise 2.3 Class 10 maths 2. Polynomials - ncert solutions - Toppers Study

### Exercise 2.3 class 10 maths chapter 2. Polynomials

**Q1. **Divide the polynomial *p*(*x*) by the polynomial *g*(*x*) and find the quotient and remainder in each of the following:

(i) p(x) = x^{3} – 3x^{2} + 5x – 3, g(x) = x^{2} – 2

(ii) p(x) = x^{4} – 3x^{2} + 4x + 5, g(x) = x^{2} + 1 – x

(iii) p(x) = x^{4} – 5x + 6, g(x) = 2 – x^{2}

**Solution:** **(i) p(x) = x ^{3} – 3x^{2} + 5x – 3, g(x) = x^{2} – 2**

Quotients q(x) = x **– ** 3 and Remainder = 7x **– ** 9

**Solution:** **(ii) p(x) = x ^{4} – 3x^{2} + 4x + 5, g(x) = x^{2} + 1 – x**

Quotients q(x) = x^{2} + x **– ** 3 and Remainder = 8

**Solution:** **(iii) p(x) = x ^{4} – 5x + 6, g(x) = 2 – x^{2}**

Quotients q(x) = **– **x^{2} **– ** 2 and Remainder = **– ** 5x + 10

**Q2. Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:**

**(i) t ^{2} – 3, 2t^{4} + 3t^{3} – 2t^{2} – 9t – 12**

**(ii) x ^{2} + 3x + 1, 3x^{4} + 5x^{3} – 7x^{2} + 2x + 2**

**(iii) x ^{3} – 3x + 1, x^{5} – 4x^{3} + x^{2} + 3x + 1**

**Solution:** **(i) t ^{2} – 3, 2t^{4} + 3t^{3} – 2t^{2} – 9t – 12**

Hence Remainder r(x) is 0

Therefore, t^{2} – 3 is the factor of 2t^{4} + 3t^{3} – 2t^{2} – 9t – 12

**Solution:** **(ii) x ^{2} + 3x + 1, 3x^{4} + 5x^{3} – 7x^{2} + 2x + 2**

Hence Remainder r(x) is 0

Therefore, x^{2} + 3x + 1 is the factor of 3x^{4} + 5x^{3} – 7x^{2} + 2x + 2

**Solution:** **(iii) x ^{3} – 3x + 1, x^{5} – 4x^{3} + x^{2} + 3x + 1**

Hence Remainder r(x) = 2

Therefore, x3 – 3x + 1, is not a factor of x^{5} – 4x^{3} + x^{2} + 3x + 1

**Q3. Obtain all other zeroes of 3 x^{4} + 6x^{3} – 2x^{2} – 10x – 5, if two of its zeroes are**

**Solution:**

Given that : p(x) = 3*x*^{4} + 6*x*^{3} – 2*x*^{2} – 10*x *– 5

Or 3x^{2} - 5 = 0

Therefore, 3x^{2} - 5 is the factor of p(x)

Now Dividing 3x^{4} + 6x^{3} - 2x^{2} - 10x - 5 by 3x^{2} - 5

Therefore, p(x) = (3x^{2} **– ** 5) (x^{2} + 2x + 1)

Now, factorizing and getting zeroes x^{2} + 2x + 1 -

= x^{2} + x + x + 1 = 0

= x(x + 1) + 1(x + 1) = 0

= (x + 1) (x + 1) = 0

Or x + 1 = 0, x + 1 = 0

Or x = **– **1, x = **– **1

Therefore, two zeroes are **– **1 and **– **1.

**Q4. ****On dividing x3 – 3x2 + x + 2 by a polynomial g(x), the quotient and remainder were x – 2 and –2x + 4, respectively. Find g(x).**

**Solution:**

Given that: Dividend p(x) = x^{3} **– **3x^{2} + x + 2

Quotient q(x) = x **– **2,

Remainder r(x) = **– **2x + 4

Divisor g(x) =?

Dividend = divisor × quotient + remainder

p(x) = g(x) × q(x) + r(x)

x^{3} **–** 3x^{2} + x + 2 = g(x) (x **–** 2) + (**–** 2x + 4)

x^{3} **–** 3x^{2} + x + 2 + 2x **–** 4 = g(x) (x **–** 2)

g(x) (x **–** 2) = x^{3} **–** 3x^{2} + 3x **–** 2

Dividing x^{3} **–** 3x^{2} + 3x **–** 2 by x - 2 we obtain g(x)-

Therefore, Divisor g(x) = x^{2} **–** x + 1

**Q5. Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and**

(i) deg *p*(*x*) = deg *q*(*x*)

(ii) deg *q*(*x*) = deg *r*(*x*)

(iii) deg *r*(*x*) = 0

**Solution:**

Using Euclid’s Division algorithm:

p(x) = g(x) × q(x) + r(x) where q(x) **–** 0

**(i) ****deg**** p(x) = ****deg**** q(x)**

The deg of dividend p(x) and quotient q(x) can be equal when deg of divisor is 0 or any number.

Example : Let p(x) = 2x^{2} - 6x + 3

And let g(x) = 2

On dividing

p(x) = 2x^{2} - 6x + 2 + 1

= 2(x^{2} - 3x + 1) + 1

Now comparing 2(x^{2} - 3x + 1) + 1 by p(x) = g(x) × q(x) + r(x) we get:

So, q(x) = x^{2} - 3x + 1and r(x) = 1

By which we obtain deg p(x) = deg q(x)

**(ii) ****deg**** q(x) = ****deg**** r(x)**

**Solution:** This situation comes when deg p(x) and deg g(x) is equal-

Let p(x) = 2x^{2} + 6x + 7 and g(x) = x^{2} + 3x + 2

On dividing: q(x) = 2 and r(x) = 3

Therefore, deg q(x) = deg r(x)

**(iii) ****deg**** r(x) = 0**

**Solution:**

r(x) = 0 is obtained when p(x) is completely divisible by g(x):

Let p(x) = x^{2} – 1 and g(x) = x + 1

On dividing we obtain:

q(x) = x – 1 and r(x) = 0

##### Other Pages of this Chapter: 2. Polynomials

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