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 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) = x^{3} – 3x^{2} + 5x – 3, g(x) = x^{2} – 2
(ii) p(x) = x^{4} – 3x^{2} + 4x + 5, g(x) = x^{2} + 1 – x
(iii) p(x) = x^{4} – 5x + 6, g(x) = 2 – x^{2}
Solution
(i) p(x) = x^{3} – 3x^{2} + 5x – 3, g(x) = x^{2} – 2
Quotient = x-3 and remainder 7x - 9
(ii) p(x) = x^{4} – 3x^{2} + 4x + 5, g(x) = x^{2} + 1 – x
Quotient = x^{2} + x - 3 and remainder 8
(iii) p(x) = x^{4} – 5x + 6, g(x) = 2 – x^{2}
^{}
Quotient = -x^{2} -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) t^{2} – 3, 2t^{4} + 3t^{3} – 2t^{2} – 9t – 12
(ii) x^{2} + 3x + 1, 3x^{4} + 5x^{3} – 7x^{2} + 2x + 2
(iii) x^{3} – 3x + 1, x^{5} – 4x^{3} + x^{2} + 3x + 1
Solution
(i) t^{2} – 3, 2t^{4} + 3t^{3} – 2t^{2} – 9t – 12
t^{2} – 3 exactly divides 2t^{4} + 3t^{3} – 2t^{2} – 9t – 12 leaving no remainder. Hence, it is a factor of 2t^{4} + 3t^{3} – 2t^{2} – 9t – 12.
(ii) x^{2} + 3x + 1, 3x^{4} + 5x^{3} – 7x^{2} + 2x + 2
x^{2} + 3x + 1 exactly divides 3x^{4} + 5x^{3} – 7x^{2} + 2x + 2 leaving no remainder. Hence, it is factor of 3x^{4} + 5x^{3} – 7x^{2} + 2x + 2.
(iii) x^{3} – 3x + 1, x^{5} – 4x^{3} + x^{2} + 3x + 1
x^{3} – 3x + 1 didn't divides exactly x^{5} – 4x^{3} + x^{2} + 3x + 1 and leaves 2 as remainder. Hence, it not a factor of x^{5} – 4x^{3} + x^{2} + 3x + 1.
3. Obtain all other zeroes of 3x^{4} + 6x^{3} – 2x^{2} – 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 x^{2} - 5/3, we have,
So, 3x^{4} + 6x^{3} - 2x^{2} - 10x -5 = (3x^{2} + 6x + ) (x^{2} - 5/3) [By division algorithm]
= 3 (x^{2} + 2x + 1) (3x^{2} -5)/3
= (x + 1)2 (3x^{2} - 5) = (x + 1) (x + 1) (3x^{2} - 5)
Thus, remaining zeroes are -1, -1.
4. On dividing x^{3} - 3x^{2} + x + 2 by a polynomial g(x), the quotient and remainder were x - 2 and -2x + 4, respectively. Find g(x).
Solution
x^{3} - 3x^{2} + x + 2 = g(x) × (x -2) + (-2x + 4) [by division algorithm]
⇒ g (x) × (x - 2) = (x^{3} - 3x^{2} + x +2) - (-2x + 4)
⇒ g(x) × (x-2) = x^{3} - 3x^{2} + 3x -2
⇒ g(x) = x^{3} - 3x^{2} + 3x -2/ x-2 ....(i)
Thus, x^{3} - 3x^{2} + 3x - 2 = (x -2) × (x2 - x + 1) ...(ii)
From (i) and (ii), g(x) = (x^{2} - x + 1) × (x × 2)/(x-2)
⇒ g(x) = x^{2} - 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 6x^{2} + 2x + 2 by 2
Here, p(x) = 6x^{2} + 2x + 2
g(x) = 2
q(x) = 3x^{2} + 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, 6x^{2} + 2x + 2 = 2x (3x^{2} + x + 1)
Hence, division algorithm is satisfied.
(ii) Let us assume the division of x^{3}+ x by x^{2},
Here, p(x) = x^{3} + x
g(x) = x^{2}
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)
x^{3} + x = (x^{2} ) × x + x
x^{3} + x = x^{3} + x
Thus, the division algorithm is satisfied.
(iii) Let us assume the division of x^{3}+ 1 by x^{2}.
Here, p(x) = x^{3} + 1
g(x) = x^{2}
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)
x^{3} + 1 = (x^{2} ) × x + 1
x^{3} + 1 = x^{3} + 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.