NCERT Solutions Class 10 Maths Chapter 2 Polynomials Ex 2.3

NCERT Solutions Class 10 Maths 
Chapter – 2 (Polynomials) 

The NCERT Solutions in English Language for Class 10 Mathematics Chapter – 2 Polynomials Exercise 2.3 has been provided here to help the students in solving the questions from this exercise. 

Chapter : 2 Polynomials

• NCERT Class 10 Maths Solution Ex – 2.1
• NCERT Class 10 Maths Solution Ex – 2.2
• NCERT Class 10 Maths Solution Ex – 2.4

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
Given,
Dividend = p(x) = x³ – 3x² + 5x – 3
Divisor = g(x) = x² – 2

Quotient = x – 3
Remainder = 7x – 9

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

Quotient = x² + x–3
Remainder = 8

(iii) p(x) =x⁴ – 5x + 6,                g(x) = 2 – x²
Dividend = p(x) =x⁴ – 5x + 6 = x⁴ + 0x² – 5x + 6
Divisor = g(x) = 2 – x² = – x² + 2

Quotient = -x² – 2
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

Solutions –
(i) t² – 3, 2t⁴ + 3t³ – 2t² – 9t – 12
First polynomial = t² – 3
Second polynomial = 2t⁴ + 3t³ – 2t² – 9t – 12

As we can see, the remainder is left as 0. Therefore, we say that, t² – 3 is a factor of 2t⁴  + 3t³ – 2t² – 9t – 12.

(ii) x² + 3x + 1 , 3x⁴ + 5x³ – 7x² + 2x + 2
First polynomial = x² + 3x + 1
Second polynomial = 3x⁴ + 5x³ – 7x² + 2x + 2

As we can see, the remainder is left as 0. Therefore, we say that, x² + 3x + 1 is a factor of 3x⁴ + 5x³ – 7x² + 2x + 2.

(iii) x³ – 3x + 1, x⁵ – 4x³ + x² + 3x + 1
First polynomial = x³ – 3x + 1
Second polynomial = x⁵ – 4x³ + x² + 3x + 1

As we can see, the remainder is not equal to 0. Therefore, we say that, x³ – 3x + 1 is 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 and – .
Solutions –
p (x) = 3x⁴ + 6x³ – 2x² – 10x – 5
Since the two zeroes are and – .
∴ = is a factor of 3x⁴ + 6x³ – 2x² – 10x – 5.

Therefore, 3x⁴ + 6x³ – 2x² – 10x – 5 = (3x² + 6x + 3) + 0
= 3 (x² – ) (x² + 2x + 1)
On factorizing x² + 2x + 1, we get (x + 1)²
Therefore, its zero is given by
x + 1 = 0
x = – 1
As it has the term (x + 1)²,
Therefore, there will be two identical zeroes at x = – 1
Hence the zeroes of the given polynomial are and – , – 1 and – 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 –
Dividend, p(x) = x³ – 3x² + x + 2
Quotient = x – 2
Remainder = –2x + 4
We have to find the value of Divisor, g(x) =?
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 + 3x – 2
Now, for finding g(x) we will divide x³ – 3x² + 3x – 2 with (x – 2)

Therefore, g(x) = (x² – x + 1)

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
Solutions –  According to the division algorithm, dividend p(x) and divisor g(x) are two polynomials, where g(x)≠0. Then we can find the value of quotient q(x) and remainder r(x), with the help of below given formula;
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).
Now let us proof the three given cases as per division algorithm by taking examples for each.

(i) deg p(x) = deg q(x)
Degree of dividend is equal to degree of quotient, only when the divisor is a constant term.
Let us take an example, p(x) = 3x²+3x+3 is a polynomial to be divided by g(x) = 3.
So, (3x²+3x+3)/3 = x²+x+1 = q(x)
Thus, you can see, the degree of quotient q(x) = 2, which also equal to the degree of dividend p(x).
Hence, division algorithm is satisfied here.

(ii) deg q(x) = deg r(x)
Let us take an example, p(x) = x² + 3 is a polynomial to be divided by g(x) = x – 1.
So, x² + 3 = (x – 1)×(x) + (x + 3)
Hence, quotient q(x) = x
Also, remainder r(x) = x + 3
Thus, you can see, the degree of quotient q(x) = 1, which is also equal to the degree of remainder r(x).
Hence, division algorithm is satisfied here.

(iii) deg r(x) = 0
The degree of remainder is 0 only when the remainder left after division algorithm is constant.
Let us take an example, p(x) = x² + 1 is a polynomial to be divided by g(x) = x.
So, x² + 1 = (x)×(x) + 1
Hence, quotient q(x) = x
And, remainder r(x) = 1
Clearly, the degree of remainder here is 0.
Hence, division algorithm is satisfied here.

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