KSEEB SSLC Class 10 Maths Solutions Chapter 9 Polynomials Ex 9.3

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Karnataka State Syllabus Class 10 Maths Solutions 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 – 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
q(x) = ?
r(x) = ?

∴ Quotient, q(x) = (x – 3)
Remainder, r(x) = (7x – 9)

(ii) p(x) = x⁴ – 3x² + 4x + 5
g(x) = x² + 1 – x
q(x) = ?
r(x) = ?

∴ Quotient, q(x) = x² + x – 3
Remainder, r(x) = 8

(iii) p(x) = x⁴ – 5x + 6
g(x) = 2 – x²
q(x) = ?
r(x) = ?

∴ 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
(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

∴ t² – 3 is the factor of 2t⁴ + 3t³ – 2t² – 9t – 12 because remainder is zeroes

(ii) x² + 3x + 1; 3x⁴ + 5x³ – 7x² + 2x + 2

Here remainder is zero
∴ x² + 3x + 1 is the factor of 3x⁴ + 5x³ – 7x² + 2x + 2

(iii) x³ – 3x + 1; x⁵ – 4x³ + x² + 3x + 1

Here remainder r(x) ≠ 0 ∵ 2 ≠ 0
∴ x³ – 3x + 1 is not the 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 53−−√ and −53−−√
Solution:
Roots of expressions are 53−−√ and −53−−√

∴ In q(x) = x² + 2x + 1, there are two more roots.
x² + 2x + 1 = (x + 1)²
other roots are: x = -1, -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).
Solution:
x³ – 3x² + x + 2 = g(x) × (x – 2) + (-2x + 4)
g(x) × (x – 2) = x³ – 3x² + x + 2 – (-2x + 4)
= x³ – 3x² + x + 2

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
Solution:
i) P(x) = 2x² – 2x + 14, g(x) = 2,
q(x) = x² – x + 7 and r(x) = 0
Verification
P(x) = g(x) q(x) + r(x)
2x² – 2x + 14 = 2(x² – x + 7) + 0
= 2x² – 2x + 14
∴ LHS = RHS

ii) P(x) = x³ + x² + x + 1, g(x) = x² – 1, q(x) = x + 1 and r(x) = 2x + 2
Verification
P(x) = q(x) g(x) + r(x)
x³ + x² + x + 1 =(x+ 1) (x² – 1) + 2x + 2
= x³ – x + x² – 1 + 2x + 2
= x³ + x² + x + 1
LHS = RHS

iii) P(x) = x³ + 2x² – x + 2, g(x) = x² – 1, q(x) = x + 2 and r(x) = 4
p(x) = g(x) q(x) + r(x)
x³ + 2x² – x + 2 = (x² – 1) (x + 2) + 4
= x³ + 2x² – x – 2 + 4
= x³ + 2x² – x + 2
LHS = RHS

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