PSEB Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.3
PSEB Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.3
PSEB 10th Class Maths Solutions Chapter 2 Polynomials Ex 2.3
Question 1.
Apply the division algorithm to find the quotient and remainder on dividing p (x) by g (x) as given below:
(i) p (x) = x3 – 3x2 + 5x – 3, g (x) = x2 – 2 (Pb. 2018 Set I, II, III)
(ii) p (x) = x4 – 3x2 + 4x + 5, g(x) = x2 + 1 – x
(iii) p (x) = x4 – 5x + 6, g (x) = 2 – x2 [Pb. 2017 Set-B]
Solution:
(i) Given that p (x) = x3 – 3x2 + 5x – 3 and g (x) = x2 – 2,
By division algorithm,
x3 – 3x2 + 5x – 3 = (x – 3) (x2 – 2) + (7x – 9)
Hence, quotient = x – 3 and remainder = 7x – 9
(ii) Given that p (x) = x4 – 3x2 + 4x + 5
or p (x) = x4 + 0x3 – 3x2 + 4x + 5
and g (x) = x2 + 1 – x
or g (x) = x2 – x + 1
By Division Algorithm,
x4 – 3x2 + 4x + 5 = (x2 + x – 3) (x2 – x + 1) + 8
Hence, Quotient = x2 + x – 3 and remainder = 8
(iii) Given that p (x) = x4 – 5x + 6
or p (x) = x4 + 0x3 + 0x2 – 5x + 6
and g (x) = 2 – x2
or g (x) = – x2 + 2
By division algorithtm,
x4 – 5x + 6 = (- x2 – 2) (- x2 + 2) + (- 5x + 10)
Hence, quotient = -x2 – 2.
remainder = – 5x + 10
Question 2.
Check whether the first polynomial is a factor of the second polynomial by applying the division algorithm:
(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:
∵ remainder is zero
∵ By division algorithm,
t2 – 3 is factor of 2t4 + 3t3 – 2t2 – 9t – 12
(ii)
∵ remainder is zero
∴ By division algorithm, x2 + 3x + 1 is a factor of 3x4 + 5x3 – 7x2 + 2x + 2
(iii)
∵ remainder is not zero.
∴ By division algorithm, x3 – 3x + 1 is not a factor of x5 – 4x3 + x2 + 3x + 1.
Question 5.
Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the dhriskm algorithm and
(i) deg p (x) = deg q (x)
(ii) deg r (x) = 0
(iii) deg q (x) = deg r (x)
Solution:
(i) Let p(x) = 5x2 – 5x +10; g(x) = 5 q(x) = x2 – x + 2; r(x) = 0
∴ By division algorithm,
5x2 – 5x + 10 = 5(x2 – x + 2) + 0
or p(x) = g(x) q(x) + r(x)
Also, deg p(x) = deg q(x) = 2
(ii) Let p(x) = 7x3 – 42x + 53
g(x) = x3 – 6x + 7
q(x) = 7; r(x) = 4
∴ By division algorithm,
7x3 -42x + 53 = 7(x3 – 6x+ 7)+ 4
or p(x) = q(x) g(x) + r(x)
Also, deg q(x) = 0 = deg r(x)
(iii) Let p(x) = 4x3 + x2 + 3x + 6;
g(x) = x2 + 3x + 1;
q(x) = 4x – 11;
r(x) = 32x + 17
∴ By division algorithm,
4x3 + x2 + 3x + 6 = (4x – 11) (x2 + 3x + 1) + (32x + 17)
or p(x) = q(x) . g(x) + r(x)
Also, deg q(x) = deg r(x)
Follow on Facebook page – Click Here
Google News join in – Click Here
Read More Asia News – Click Here
Read More Sports News – Click Here
Read More Crypto News – Click Here