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Karnataka State Syllabus Class 10 Maths Chapter 9 Polynomials Ex 9.3
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 – 3x3 + 5x – 3, g(x) = x2 – 2
Answer:
p(x) = x3 – 3x3 + 5x – 3, g(x) = x2 + 0x – 2
∴ quotient q(x) = x – 3
Remainder r(x) = – 7x – 9
(ii) p(x) = x4 – 3×2 + 4x + 5, g(x) = x2 + 1 – x
Answer:
p(x) = x4 + 0x3 – 3x3 – 3x2 + 4x + 5 & g(x) = x2 – x + 1
∴ Quotient q(x) = x2 + x – 3
Remainder r (x) = 8
(iii) p(x) = x4 – 5x + 6, g(x) = 2 – x2
Answer:
P(x) = x4 + 0x3 + 0x2 – 5x + 6 & g(x) = – x2 + 0x + 2
∴ Quotient q(x) = – x2 – 2
Remainder r (x) = – 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
Answer:
P(t) = 2t4 + 3t3 – 2t2 – 9t – 12 and
g(t) = t2 – 3 (or) g(t) = t2 + 0t – 3
Since Remainder is zero, there fore, the first polynomial is a factor of second polynomial.
(ii) x2 + 3x + 1, 3x4 + 5x3 – 7x2 + 2x + 2
Answer:
P(x) = 3x4 + 5x3 – 7x2 + 2x + 2 & g(x) = x2 + 3x + 1
Since, remainder is zero, there fore first polynomial is factor of second polynomial
(iii) x3 – 3x + 1, x5 – 4x3 + x2 + 3x + 1
Answer:
p(x) = x5 + 0x4 – 4x3 + x2 + 3x + 1 , and g(x) = x3 + 0x2 – 3x +1
Since Remainder not equal to zero, first polynomial is not a factor of second polynomial.
Question 3.
Obtain all other zeroes of 3x4 + 6x3 – 2x2 – 10x – 5, if two of its zeroes are \(\sqrt{\frac{5}{3}}\) and – \(\sqrt{\frac{5}{3}}\).
Answer:
Since, two zeros are \(\sqrt{\frac{5}{3}}\) and – \(\sqrt{\frac{5}{3}}\) there fore, By factor theorem,
x2 – \(\frac{5}{3}\) is a factor of
3x4 + 6x3 – 2x2 – 10x – 5
Apply the division algorithm
p(x) = 3x4 + 6x3 – 2x2 – 10x – 5 and
g(x) = x2 + 0x – \(\frac{5}{3}\)
Remainder is zero,
∴ P(x) = g (x) q (x)
3x4 + 6x3 – 2x2 – 10x – 5
= (x2 – \(\frac{5}{3}\))(3x2 + 6x + 3)
Other zeroes are given by the second Polynomial factors 3x2 + 6x + 3
Put 3x2 + 6x + 3 = 0
3x2 + 3x + 3x + 3 = 0
3x (x + 1) + 3(x + 1) = 0
(x + 1)(3x + 3) = 0
x + 1 = 0 (or) 3x + 3 = 0
x = – 1 (or) 3x = – 3
x = \(\frac{- 3}{3}\) = – 1
∴ x = – 1, – 1
Therefore other zeroes of the Polynomial , 3x4 + 6x3 – 2x2 – 10x – 5 are – 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:
p(x) = x3 – 3x2 + x + 2,
q(x) = x – 2 and r(x) = – 2x + 4
By division algorithm for Polynomials.
p(x) = g(x) q(x) + r(x)
x3 – 3x2 + x + 2 = g(x) (x – 2) + (- 2x + 4)
g(x) (x – 2) = x3 – 3x2 + x + 2 + 2x – 4
g (x) (x – 2) = x3 – 3x2 + 3x – 2
Therefore g(x) = x2 – x + 1
Question 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)
(ii) deg q(x) = deg r(x)
(iii) deg r(x) = 0
Answer:
(i) Degree p(x) = Degree q(x)
p(x) = 2x2 + 2x + 8
q(x) = 2x2 = 2
q(x) = x2 + x + 4
r(x) = 0
(ii) Degree q(x) = Degree r(x)
p(x) = 2x2 + 2x + 8
q(x) = x2 + x + 9
q(x) = 2
r(x) = -10
(iii) Degree r(x) = 0
p(x) = x3 + x + 5
q(x) = x2 + 1
q(x) = x
r(x) = 5.