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I am trying to find NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Exercise 2.3. Give Class 10 Mathematics NCERT Solutions to complete hometask.

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Given Class 10 NCERT Solutions for Chapter 2 Polynomials Exercise 2.3 can be used to complete hometask and solve difficult questions. Through the help of CBSE NCERT Solutions for Class 10, you can able to solve the problems efficiently.

Book NameClass 10 Mathematics NCERT Textbook
ChapterChapter 2 Polynomials
ExerciseEx 2.3

 

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Exercise 2.3

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

Solution

(i) p(x) = x3 – 3x2 + 5x – 3, g(x) = x2 – 2

Quotient = x-3 and remainder 7x - 9
 

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

Quotient = x2 + - 3 and remainder 8
 

(iii) p(x) = x4 – 5x + 6, g(x) = 2 – x2

Quotient = -x2 -2 and remainder -5x +10
 

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

Solution

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

t2 – 3 exactly divides  2t4 + 3t3 – 2t2 – 9t – 12 leaving no remainder. Hence, it is a factor of  2t4 + 3t3 – 2t2 – 9t – 12.

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

x2 + 3x + 1 exactly divides 3x4 + 5x3 – 7x2 + 2x + 2 leaving no remainder. Hence, it is factor of 3x4 + 5x3 – 7x2 + 2x + 2.

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

x3 – 3x + 1 didn't divides exactly x5 – 4x3 + x2 + 3x + 1 and leaves 2 as remainder. Hence, it not a factor of x5 – 4x3 + x2 + 3x + 1.

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

Solution

Since, two zeroes are √(5/3) and - √(5/3)

  is a factor of given polynomial.

Now, dividing given polynomial by x2 - 5/3, we have,

So, 3x4 + 6x3 - 2x2 - 10x -5 = (3x2 + 6x + ) (x2 - 5/3) [By division algorithm]

= 3 (x2 + 2x + 1) (3x2 -5)/3

= (x + 1)2 (3x2 - 5) = (x + 1) (x + 1) (3x2 - 5)

Thus, remaining zeroes are -1, -1.

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).

Solution

x3 - 3x2 + x + 2 = g(x) × (x -2) + (-2x + 4) [by division algorithm]

⇒ g (x) × (x - 2) = (x3 - 3x2 + x +2) - (-2x + 4)

⇒ g(x) × (x-2) = x3 - 3x2 + 3x  -2

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

Thus, x3 - 3x2 + 3x - 2 = (x -2) × (x2 - x + 1)  ...(ii)

From (i) and (ii), g(x) = (x2 - x + 1) × (x × 2)/(x-2)

⇒ g(x) = x2 - x + 1.

5.Give examples of polynomial 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

(i) Let us assume the division of 6x2 + 2x + 2 by 2
Here, p(x) = 6x2 + 2x + 2
g(x) = 2
q(x) = 3x2 + x + 1
r(x) = 0
Degree of p(x) and q(x) is same i.e. 2.
Checking for division algorithm,
p(x) = g(x) × q(x) + r(x)
Or, 6x2 + 2x + 2 = 2x (3x2 + x + 1)
Hence, division algorithm is satisfied.

(ii) Let us assume the division of x3+ x by x2,
Here, p(x) = x3 + x
g(x) = x2
q(x) = x and r(x) = x
Clearly, the degree of q(x) and r(x) is the same i.e., 1.
Checking for division algorithm,
p(x) = g(x) × q(x) + r(x)
x3 + x = (x2 ) × x + x
x3 + x = x3 + x
Thus, the division algorithm is satisfied.

(iii) Let us assume the division of x3+ 1 by x2.
Here, p(x) = x3 + 1
g(x) = x2
q(x) = x and r(x) = 1
Clearly, the degree of r(x) is 0.
Checking for division algorithm,
p(x) = g(x) × q(x) + r(x)
x3 + 1 = (x2 ) × x + 1
x3 + 1 = x3 + 1
Thus, the division algorithm is satisfied.

These NCERT Solutions for Class 10 Maths will be very helpful in scoring good marks in the examinations.

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