Exercise 2.3 class 10 maths chapter 2. Polynomials
Q1. 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
Solution: (i) p(x) = x3 – 3x2 + 5x – 3, g(x) = x2 – 2
Quotients q(x) = x – 3 and Remainder = 7x – 9
Solution: (ii) p(x) = x4 – 3x2 + 4x + 5, g(x) = x2 + 1 – x
Quotients q(x) = x2 + x – 3 and Remainder = 8
Solution: (iii) p(x) = x4 – 5x + 6, g(x) = 2 – x2
Quotients q(x) = – x2 – 2 and Remainder = – 5x + 10
Q2. 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
Solution: (i) t2 – 3, 2t4 + 3t3 – 2t2 – 9t – 12
Hence Remainder r(x) is 0
Therefore, t2 – 3 is the factor of 2t4 + 3t3 – 2t2 – 9t – 12
Solution: (ii) x2 + 3x + 1, 3x4 + 5x3 – 7x2 + 2x + 2
Hence Remainder r(x) is 0
Therefore, x2 + 3x + 1 is the factor of 3x4 + 5x3 – 7x2 + 2x + 2
Solution: (iii) x3 – 3x + 1, x5 – 4x3 + x2 + 3x + 1
Hence Remainder r(x) = 2
Therefore, x3 – 3x + 1, is not a factor of x5 – 4x3 + x2 + 3x + 1
Q3. Obtain all other zeroes of 3x4 + 6x3 – 2x2 – 10x – 5, if two of its zeroes are
Solution:
Given that : p(x) = 3x4 + 6x3 – 2x2 – 10x – 5
Or 3x2 - 5 = 0
Therefore, 3x2 - 5 is the factor of p(x)
Now Dividing 3x4 + 6x3 - 2x2 - 10x - 5 by 3x2 - 5
Therefore, p(x) = (3x2 – 5) (x2 + 2x + 1)
Now, factorizing and getting zeroes x2 + 2x + 1 -
= x2 + x + x + 1 = 0
= x(x + 1) + 1(x + 1) = 0
= (x + 1) (x + 1) = 0
Or x + 1 = 0, x + 1 = 0
Or x = – 1, x = – 1
Therefore, two zeroes are – 1 and – 1.
Q4. 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 that: Dividend p(x) = x3 – 3x2 + x + 2
Quotient q(x) = x – 2,
Remainder r(x) = – 2x + 4
Divisor g(x) =?
Dividend = divisor × quotient + remainder
p(x) = g(x) × q(x) + r(x)
x3 – 3x2 + x + 2 = g(x) (x – 2) + (– 2x + 4)
x3 – 3x2 + x + 2 + 2x – 4 = g(x) (x – 2)
g(x) (x – 2) = x3 – 3x2 + 3x – 2
Dividing x3 – 3x2 + 3x – 2 by x - 2 we obtain g(x)-
Therefore, Divisor g(x) = x2 – x + 1
Q5. 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)
(ii) deg q(x) = deg r(x)
(iii) deg r(x) = 0
Solution:
Using Euclid’s Division algorithm:
p(x) = g(x) × q(x) + r(x) where q(x) – 0
(i) deg p(x) = deg q(x)
The deg of dividend p(x) and quotient q(x) can be equal when deg of divisor is 0 or any number.
Example : Let p(x) = 2x2 - 6x + 3
And let g(x) = 2
On dividing
p(x) = 2x2 - 6x + 2 + 1
= 2(x2 - 3x + 1) + 1
Now comparing 2(x2 - 3x + 1) + 1 by p(x) = g(x) × q(x) + r(x) we get:
So, q(x) = x2 - 3x + 1and r(x) = 1
By which we obtain deg p(x) = deg q(x)
(ii) deg q(x) = deg r(x)
Solution: This situation comes when deg p(x) and deg g(x) is equal-
Let p(x) = 2x2 + 6x + 7 and g(x) = x2 + 3x + 2
On dividing: q(x) = 2 and r(x) = 3
Therefore, deg q(x) = deg r(x)
(iii) deg r(x) = 0
Solution:
r(x) = 0 is obtained when p(x) is completely divisible by g(x):
Let p(x) = x2 – 1 and g(x) = x + 1
On dividing we obtain:
q(x) = x – 1 and r(x) = 0
