# NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Exercise 2.3

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 Book Name Class 10 Mathematics NCERT Textbook Chapter Chapter 2 Polynomials Exercise Ex 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.