**KSEEB SSLC Class 10 Maths Solutions Chapter 9 Polynomials Ex 9.3** are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 9 Polynomials Exercise 9.3.

## Karnataka SSLC Class 10 Maths Solutions Chapter 9 Polynomials Exercise 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) = x^{3} – 3x^{2} + 5x – 3; g(x) = x^{2} – 2

(ii) p(x) = x^{4} – 3x^{2} + 4x + 5; g(x) = x^{2} + 1 – x

(iii) p(x) = x^{4} – 5x + 6; g(x) = 2 – x^{2}.

Solution:

(i) p(x) = x^{3} – 3x^{2} + 5x – 3

g(x) =x^{2} – 2

q(x) = ?

r(x) = ?

∴ Quotient, q(x) = (x – 3)

Remainder, r(x) = (7x – 9)

(ii) p(x) = x^{4} – 3x^{2} + 4x + 5

g(x) = x^{2} + 1 – x

q(x) = ?

r(x) = ?

∴ Quotient, q(x) = x^{2} + x – 3

Remainder, r(x) = 8

(iii) p(x) = x^{4} – 5x + 6

g(x) = 2 – x^{2}

q(x) = ?

r(x) = ?

∴ Quotient, q(x) = -x^{2} – 2

Remainder, r(x) = -5x + 10

Polynomial Division Calculator is an online tool that helps to divide two given polynomials.

Question 2.

Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial :

(i) t^{2} – 3; 2t^{4} + 3t^{3} – 2t^{2} – 9t – 12

(ii) x^{2} + 3x + 1; 3x^{4} + 5x^{3} – 7x^{2} + 2x + 2

(iii) x^{3} – 3x + 1; x^{5} – 4x^{3} + x^{2} + 3x + 1

Solution:

(i) t^{2} – 3; 2t^{4} + 3t^{3} – 2t^{2} – 9t – 12

∴ t^{2} – 3 is the factor of 2t^{4} + 3t^{3} – 2t^{2} – 9t – 12 because remainder is zeroes

(ii) x^{2} + 3x + 1; 3x^{4} + 5x^{3} – 7x^{2} + 2x + 2

Here remainder is zero

∴ x^{2} + 3x + 1 is the factor of 3x^{4} + 5x^{3} – 7x^{2} + 2x + 2

(iii) x^{3} – 3x + 1; x^{5} – 4x^{3} + x^{2} + 3x + 1

Here remainder r(x) ≠ 0 ∵ 2 ≠ 0

∴ x^{3} – 3x + 1 is not the factor of x^{5} – 4x^{3} + x^{2} + 3x + 1

Question 3.

Obtain all other zeroes of 3x^{4} + 6x^{3} – 2x^{2} – 10x – 5, if two of its zeroes are \(\sqrt{\frac{5}{3}}\) and \(-\sqrt{\frac{5}{3}}\)

Solution:

Roots of expressions are \(\sqrt{\frac{5}{3}}\) and \(-\sqrt{\frac{5}{3}}\)

∴ In q(x) = x^{2} + 2x + 1, there are two more roots.

x^{2} + 2x + 1 = (x + 1)^{2}

other roots are: x = -1, -1

Question 4.

On dividing x^{3} – 3x^{2} + x + 2 by a polynomial g(x), the quotient and remainder were x – 2 and -2x + 4, respectively. Find g(x).

Solution:

x^{3} – 3x^{2} + x + 2 = g(x) × (x – 2) + (-2x + 4)

g(x) × (x – 2) = x^{3} – 3x^{2} + x + 2 – (-2x + 4)

= x^{3} – 3x^{2} + 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) = 2x^{2} – 2x + 14, g(x) = 2,

q(x) = x^{2} – x + 7 and r(x) = 0

Verification

P(x) = g(x) q(x) + r(x)

2x^{2} – 2x + 14 = 2(x^{2} – x + 7) + 0

= 2x^{2} – 2x + 14

∴ LHS = RHS

ii) P(x) = x^{3} + x^{2} + x + 1, g(x) = x^{2} – 1, q(x) = x + 1 and r(x) = 2x + 2

Verification

P(x) = q(x) g(x) + r(x)

x^{3} + x^{2} + x + 1 =(x+ 1) (x^{2} – 1) + 2x + 2

= x^{3} – x + x^{2} – 1 + 2x + 2

= x^{3} + x^{2} + x + 1

LHS = RHS

iii) P(x) = x^{3} + 2x^{2} – x + 2, g(x) = x^{2} – 1, q(x) = x + 2 and r(x) = 4

p(x) = g(x) q(x) + r(x)

x^{3} + 2x^{2} – x + 2 = (x^{2} – 1) (x + 2) + 4

= x^{3} + 2x^{2} – x – 2 + 4

= x^{3} + 2x^{2} – x + 2

LHS = RHS

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