NCERT Solutions for Class 10 Maths Chapter 2 Exercise 2.3
In this page we have NCERT Solutions for Class 10 Maths Chapter 2 Polynomial for
Ex 2.3 on page 36. Hope you like them and do not forget to like , social_share
and comment at the end of the page.
Question 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 Answer
(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 + x - 3 and remainder 8
Question 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 Answer
(i) t2 – 3, 2t4 + 3t3 – 2t2 – 9t – 12 So 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.
Question 3.
Obtain all other zeroes of 3x4 + 6x3 – 2x2 – 10x – 5, if two of its zeroes are √(5/3) and - √(5/3). Answer p(x) = 3x4 + 6x3 – 2x2 – 10x – 5
Since the two zeroes are √(5/3) and - √(5/3).
Therefore
[x-√(5/3)] and [x+√(5/3)] are factors of the polynomial p(x)
So [x-√(5/3)] [x+√(5/3)] =(x2 -5/3) is a factor of the polynomial p(x)
Let’s divide the p(x) by (x2 -5/3) to get remaining factors
So
P(x)= 3x4 + 6x3 – 2x2 – 10x – 5
=(3x2 +6x+3)(x2 -5/3)
=3(x2 +2x+1)( x2 -5/3)
We factorize x2 + 2x + 1
= (x + 1)2
Therefore, its zero is given by x + 1 = 0 x = -1
So the other 2 zeroes are x = - 1.
Hence, the zeroes of the given polynomial are √(5/3) and - √(5/3), - 1 and - 1.
Question 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). Answer
Here in the given question,
Dividend = x3 - 3x2 + x + 2
Quotient = x - 2
Remainder = -2x + 4
Divisor = g(x)
We know that,
Dividend = Quotient × Divisor + Remainder
⇒ x3 - 3x2 + x + 2 = (x - 2) × g(x) + (-2x + 4) x3 - 3x2 + x + 2 - (-2x + 4) = (x - 2) × g(x) x3 - 3x2 + 3x - 2 = (x - 2) × g(x) g(x) = (x3 - 3x2 + 3x - 2)/ (x - 2) g(x) = (x2 - x + 1)
Question 5
Give examples of polynomial p(x), g(x), q(x) and r(x), which satisfy the division algorithm and
(i) degree p(x) = degree q(x)
(ii) degree q(x) = degree r(x)
(iii) degree r(x) = 0 Answer
According to the division algorithm, if p(x) and g(x) are two polynomials
with g(x) ≠ 0, then we can find polynomials q(x) and r(x) such that
p(x) = g(x) × q(x) + r(x), --(A)
where r(x) = 0 or degree of r(x) < degree of g(x)
(i)
degree p(x) = degree q(x)
From equation (A), then r(x)=0 and q(x) =constant term
Here Let us assume the division of 9x2 + 6x + 3 by 3
Here, p(x) = 9x2 + 6x + 3 g(x) = 3 q(x) = 3x2 +2x + 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, 9x2 + 6x + 3= 3 × (3x2 +2 x + 1)
Hence, division algorithm is satisfied.
(ii)
Now degree q(x) = degree r(x)
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)
degree r(x) = 0
Let us assume the division of x3+ 5 by x2.
Here, p(x) = x3 + 5
g(x) = x2 q(x) = x and r(x) = 5
Clearly, the degree of r(x) is 0.
Checking for division algorithm, p(x) = g(x) × q(x) + r(x) x3 + 5 = (x2 ) × x + 5 x3 + 5 = x3 + 5
Thus, the division algorithm is satisfied Download Polynomials Class10 EXERCISE 2.3 as pdf link to this page by copying the following text Also Read
