NCERT Solutions 10th Maths Chapter 1 Polynomials Exercise 1.3

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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:
Answer
(i) p(x) = x3 ñ 3x2 + 5x ñ 3, g(x) = x2 ñ 2
Quotient = x-3 and remainder 7x - 9
(ii) p(x) = x4 ñ 3x2 + 4x + 5, g(x) = x2 + 1 ñ x
Quotient = x2 + x - 3 and remainder 8
(iii) p(x) = x4 ñ 5x + 6, g(x) = 2 ñ 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:
Answer
(i) t2 ñ 3, 2t4 + 3t3 ñ 2t2 ñ 9t ñ 12
t2 ñ 3 exactly divides 2t4 + 3t3 ñ 2t2 ñ 9t ñ 12 leaving no remainder. Hence, it is a factor of 2t4 + 3t3 ñ 2t2 ñ 9t ñ 12.
(ii) x2 + 3x + 1, 3x4 + 5x3 ñ 7x2 + 2x + 2
x2 + 3x + 1 exactly divides 3x4 + 5x3 ñ 7x2 + 2x + 2 leaving no remainder. Hence, it is factor of 3x4 + 5x3 ñ 7x2 + 2x + 2.
(iii) x3 ñ 3x + 1, x5 ñ 4x3 + x2 + 3x + 1
x3 ñ 3x + 1 didn't divides exactly x5 ñ 4x3 + x2 + 3x + 1 and leaves 2 as remainder. Hence, it not a factor of x5 ñ 4x3 + x2 + 3x + 1.
3. Obtain all other zeroes of 3x4 + 6x3 ñ 2x2 ñ 10x ñ 5, if two of its zeroes are v(5/3)
and - v(5/3).
Answer
p(x) = 3x4 + 6x3 ñ 2x2 ñ 10x ñ 5
Since the two zeroes are v(5/3) and - v(5/3).
We factorize x2 + 2x + 1
= (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 2 zeroes at x = - 1.
Hence, the zeroes of the given polynomial are v(5/3) and - v(5/3), - 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).
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)

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
Answer
(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.