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Chapter 1:Real Numbers

Chapter 2:Polynomials

Chapter 3:Pair of Linear Equations in Two Variables

Chapter 4:Quadratic Equations

Chapter 5: Arithmetic Progression

Chapter 6: Triangles

Chapter 7:Coordinate Geometry

Chapter 8: Introduction to Trigonometry

Chapter 9:Some Applications of Trigonometry

Chapter 10: Circles

Chapter 11:Constructions

Chapter 12:Area Related to Circles

Chapter 13:Surface Areas and Volumes

Chapter 14:Statistics

Chapter 15:Probability

NCERT Solutions for class 10 maths chapter 2

Polynomials

NCERT Solutions for class 10 maths chapter 2 Exercise 2.3

Question 1

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

(i)p(x) = x3-3x2+5x–3 , g(x) = x2–2

(ii) p(x) = x4-3x2+4x+5 , g(x) = x2+1-x

(iii) p(x) =x4–5x+6, g(x) = 2–x2

Answer

(i)p(x) = x3-3x2+5x–3 , g(x) = x2–2

Here

Dividend = p(x) = x3-3x2+5x–3

Divisor = g(x) = x2– 2

Ex-2.3 chapter 2 class 10

we get,

Quotient = x–3

Remainder = 7x–9

(ii) p(x) = x4-3x2+4x+5 , g(x) = x2+1-x

Answer

Given,

Dividend = p(x) = x4 – 3x2 + 4x +5

Divisor = g(x) = x2 +1-x

Ex-2.3 chapter 2 class 10
Ex-2.3 chapter 2 class 10

we get,

Quotient = x2 + x–3

Remainder = 8

(iii) p(x) =x4–5x+6, g(x) = 2–x2

Answer

Given,

Dividend = p(x) =x4 – 5x + 6 = x4 +0x2–5x+6

Divisor = g(x) = 2–x2 = –x2+2

Ex-2.3 chapter 2 class 10

Therefore, upon division we get,

Quotient = -x2-2

Remainder = -5x + 10

Question 2

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

(i) t2-3, 2t4 +3t3-2t2-9t-12

(ii)x2+3x+1 , 3x4+5x3-7x2+2x+2

(iii) x3-3x+1, x5-4x3+x2+3x+1

Answer

(i) t2-3, 2t4 +3t3-2t2-9t-12

Given,

First polynomial = t2-3

Second polynomial = 2t4 +3t3-2t2 -9t-12

Ex-2.3 chapter 2 class 10

As remainder is 0. so we can say that, t2-3 is a factor of 2t2+3t+4.

(ii)x2+3x+1 , 3x4+5x3-7x2+2x+2

Answer

Given,

First polynomial = x2+3x+1

Second polynomial = 3x4+5x3-7x2+2x+2

2.3 class 10 chapter 2

As remainder is 0. so, we can say that, x2 + 3x + 1 is a factor of 3x4+5x3-7x2+2x+2.

(iii) x3-3x+1, x5-4x3+x2+3x+1

Answer

Given,

First polynomial = x3-3x+1

Second polynomial = x5-4x3+x2+3x+1

2.3 class 10 chapter 2

As remainder is not equal to 0. So, we say that, x3-3x+1 is not a factor of x5-4x3+x2+3x+1 .

Question 3

Obtain all other zeroes of 3x4+6x3-2x2-10x-5, if two of its zeroes are √(5/3) and – √(5/3).

Answer

√(5/3) and – √(5/3) are zeroes of f(x).
so

(x –√(5/3) (x+√(5/3) = x2-(5/3) = 0

(3x2−5)=0, is a factor of given polynomial f(x).

Now,

2.3 class 10 chapter 2

Therefore, 3x4 +6x3 −2x2 −10x–5 = (3x2 –5)(x2+2x+1)

On further factorizing (x2+2x+1) we get,

x2+2x+1 = x2+x+x+1 = 0

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

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

So, its zeroes are given by: x= −1 and x = −1.

Therefore, all four zeroes of polynomial equation are:

√(5/3),- √(5/3) , −1 and −1.

Question 4

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).

Answer

Given,

Dividend, p(x) = x3-3x2+x+2

Quotient = x-2

Remainder = –2x+4

we know,

Dividend = Divisor × Quotient + Remainder

∴ x3-3x2+x+2 = g(x)×(x-2) + (-2x+4)

x3-3x2+x+2-(-2x+4) = g(x)×(x-2)

Therefore, g(x) × (x-2) = x3-3x2+x-2

g(x) =
x3-3x2+x - 2 / (x-2)

2.3 class 10 chapter 2

Therefore, g(x) = (x2–x+1)

Question 5

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

Answer

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

We know
Dividend = Divisor × Quotient + Remainder

∴ p(x) = g(x)×q(x)+r(x)

Where r(x) = 0 or degree of r(x)< degree of g(x).

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

Answer

Degree of dividend is equal to degree of quotient, only when the divisor is a constant term.

Let us assume an example, 6x2+6x+6 is a polynomial to be divided by 3.

p(x)= 6x2+6x+6
g(x)= 3

So, (6x2+6x+6)/3 = 2x2+2x+2 = q(x)

so we can see, the degree of p(x) and q(x) is equal.

Hence, division algorithm is satisfied here.

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

Answer

Let us p(x)=x4+x
g(x)=x3.

So, (x4+x)/x3 = x = q(x)

Also, remainder, r(x) = x

we can see, the degree of q(x) and g(x) is equal .

Hence, division algorithm is satisfied here.

(iii) deg r(x) = 0

Answer

Let us take an example, p(x) = x3+1 is a polynomial to be divided by g(x)=x.

So,( x3+1)/x= x=q(x)

And r(x)=2

Clearly, the degree of remainder r(x) is 0.

Hence, division algorithm is satisfied here.