Class10 NCRT Polynomials Exercise – 2.3 pdf || UP Board
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) = x3-3x2+5x-3, g(x) = x2-2
Solution:
Quotient = x – 2 and remainder = 7x – 9
(ii) p(x) = x4-3x2+4x+5, g(x) = x2+1-x
Solution:
Quotient = x2 +
x - 3 and
remainder = 8
(iii) p(x) = x4-5x+6, g(x) = 2 – x2
Solution:
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:
(i) t2 – 3, 2t4 + 3t3 – 2t2 – 9t – 12
Solution:
Since, remainder is 0, therefore the polynomial
t2 – 3 is a
factor of the polynomial 2t4 + 3t3 – 2t2 – 9t
– 12.
(ii) x2 + 3x + 1, 3x4 + 5x3 – 7x2 +2x +2
Solution:
Since, remainder is 0, therefore the polynomial x2
+ 3x + 1 is a factor of the polynomial 3x4 + 5x3 –
7x2 +2x +2.
(iii) x3 -3x + 1, x5 -4x3 +x2 +3x +1
Solution:
The
polynomial x3 -3x + 1 is not a factor of the polynomial x5
-4x3 +x2 +3x +1 because the remainder is 0 not 2.
3. Obtain
all other zeroes of 3x4 +6x3 -2x2 -10x -5,
if two of its zeroes are √(5/3) and -√(5/3).
Solution:
Let p(x) = 3x4 +6x3 -2x2 -10x -5
Therefore, x =√(5/3),
x =-√(5/3)
Or
3x2
– 5 = 0
3x2 – 5 is a factor of p(x),
by division method,
p(x)
= (3x2
-5) (x2 + 2x +1)
Now, on factoring x2 + 2x +1,
x2 + 2x +1 = 0
x2 + x + x +1 = 0
x (x+1) +1(x+1) = 0
( x+1) (x+1) = 0
x = -1, -1
Hence two other zeroes are -1, -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).
Solution:
Given that,
Dividend p(x) = x3 – 3x2 + x +2
Quotient q(x) = x – 2
Remainder r(x) = -2x +4
Divisor g(x) = ?
We know that,
p(x) = g(x)×
q(x) + r(x)
x3 – 3x2 + x +2 = g(x)×(x – 2) + (-2x +4)
by division method,
Hence, g(x) = x2
- x +1
5. 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)
Solution:
The degree of the dividend and the quotient can be equal only if the
divisor is a constant (to the degree 0) number.
Let p(x)
= 3x2 -6x +5
Let g(x)
= 3
Therefore, q(x) = x2 -2x +1 and r(x) = 2
(ii) deg
q(x)
= deg r(x)
Solution:
Let p(x)
= 2x2 -4x +3
Let g(x)
= x2 -2x +1
Therefore, q(x)
= 2 and r(x)
= 1
(iii) deg
r(x)
= 0
Solution:
Let p(x)
= 2x2 -4x +3
Let g(x)
= x2 -2x +1
Therefore, q(x) = 2 and r(x) = 1
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