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NCERT Solutions for Class 10 Maths Chapter 2:Polynomial Exercise 3<




In this page we have NCERT Solutions for Class 10 Maths Chapter 2:Polynomial for EXERCISE 3 . 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
NCERT Solutions  for Class 10 Maths Chapter 2:Polynomial
Quotient = x2 + - 3 and remainder 8
(iii) p(x) = x4 – 5x + 6, g(x) = 2 – x2
NCERT Solutions  for Class 10 Maths Chapter 2:Polynomial
Quotient = -x2 -2 and remainder -5x +10 
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
NCERT Solutions  for Class 10 Maths Chapter 2:Polynomial
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
NCERT Solutions  for Class 10 Maths Chapter 2:Polynomial
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
NCERT Solutions  for Class 10 Maths Chapter 2:Polynomial
So
P(x)= 3x4 + 6x3 – 2x2 – 10x – 5
   =(3x2 +6x+3)(x2 -5/3)
    =3(x2 +2x+1)( x2 -5/3)
We factorize x2 + 2+ 1
= (+ 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)
NCERT Solutions  for Class 10 Maths Chapter 2:Polynomial
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 + 1)
Hence, division algorithm is satisfied.

(ii)
Now degree q(x) = degree r(x)
 Let us assume the division of x3x 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 ) × + 5
x3 + 5 = x3 + 5
Thus, the division algorithm is satisfied
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