NCERT Solutions Class 10th 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

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

Therefore, upon division we get,
Quotient = x – 3
Remainder = 7x – 9

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

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

Therefore, upon division we get,
Quotient = x² + x – 3
Remainder = 8

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

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

Therefore, upon division we get,
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

Solution: Given,
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

Solution: Given,
First polynomial = x² + 3x + 1
Second polynomial = 3x⁴ + 5x³– 7x² + 2x + 2

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

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

Solution: Given,
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. O btain all other zeroes of 3x⁴ + 6x³– 2x² – 10x – 5, if two of its zeroes are √(5/3) and – √(5/3).

Solution: Since this is a polynomial equation of degree 4, hence there will be total 4 roots.
√(5/3) and – √(5/3) are zeroes of polynomial f(x).
∴ (x – √(5/3)) (x + √(5/3) = x²– (5/3) = 0
(3x² − 5) = 0, is a factor of given polynomial f(x).
Now, when we will divide f(x) by (3x² − 5) the quotient obtained will also be a factor of f(x) and the remainder will be 0.

Therefore, 3x⁴ + 6x³ − 2x² − 10x – 5 = (3x² – 5)(x² + 2x + 1)
Now, 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 given polynomial equation are:
√(5/3),- √(5/3) , −1 and −1.
Hence, is the answer.

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: Given,
Dividend, p(x) = x³– 3x² + x + 2
Quotient = x – 2
Remainder = – 2x + 4
We have to find the value of Divisor, g(x) =?
As 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– 3 x² + 3x – 2
Now, for finding g(x) we will divide x³– 3x² + 3 x – 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)

Solution: 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)

Solution: 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

Solution: 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.