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## Karnataka State Syllabus Class 10 Maths Chapter 9 Polynomials Additional Questions

I. Multiple Choice Questions:

Question 1.

The degree of polynomial is x + 2

a. 2

b. 1

c. 3

d. 4

Answer:

b. 1

Question 2.

The degree of a quadratic polynomial is

a. 0

b. 1

c. 2

d. 3

Answer:

c. 2

Question 3.

The degree polynomial of a bi-quadratic

a. 0

b. 2

c. 4

d. 1

Answer:

c. 4

Question 4.

The standard polynomial. form of a linear

a. ax + c

b. ax^{2} + c

c. ax^{2} + bx + c = 0

d. ax^{3} + bx^{2} + cx + d = 0

Answer:

a. ax + c

Question 5.

The standard form of a quadratic equation.

a. ax^{2} + bx + c = 0

b. ax + c = 0

c. ax^{2} + c = 0

d. ax^{3} + bx + c = 0

Answer:

a. ax^{2} + bx + c = 0

Question 6.

A polynomial of degree 3 is called

a. a linear polynomial

b. a quadratic polynomial

c. a cubic polynomial

d. a biquadratic polynomial

Answer:

c. a cubic polynomial

Question 7.

The value of px. = x^{2} – 3x – 4, x = -1

a. 1

b. – 4

c. 0

d. – 3

Answer:

c. 0

Question 8.

The degree of the polynomial x^{4} + x^{3} is

a. 2

b. 3

c. 5

d. 4

Answer:

d. 4

Question 9.

The number of zeroes of linear polynomial at most is

a. 0

b. 1

c. 2

d. 3

Answer:

b. 1

Question 10.

The degree of polynomial x + 2. x + 1. is

a. 1

b. 3

c. 4

d. 2

Answer:

d. 2

Question 11.

The degree of a zero polynomial is

a. not defined

b. 1

c. 2

d. 3

Answer:

a. not defined

Question 12.

The zeroes of the polynomial x^{3} -4x are

a. 0, ± 2

b. 0, ± 1

c. 0, ± 3

d. 0, 0

Answer:

a. 0, ± 2

Question 13.

The zeroes of the polynomials, t^{2} – 15 are.

a. ± \(\sqrt{15}\)

b. ± \(\sqrt{5}\)

c. ± \(\sqrt{3}\)

d. ± 3

Answer:

a. ± \(\sqrt{15}\)

Question 14.

Dividend is equal to.

a. divisor × quotient + remainder

b. divisor × quotient

c. divisor × quotient – remainder

d. divisor × quotient × remainder

Answer:

a. divisor × quotient + remainder

Question 15.

If the divisor is x^{2} and quotient is x while the number remainder is 1, then the dividend is.

a. x^{2}

b. x

c. x^{3}

d. x^{3} + 1

Answer:

d. x^{3} + 1

Question 16.

What is the co-efficient of the first term of the quotient when 2x^{2} + 2x + 1 is divided by x + 2?

a. 1

b. 2

c. 3

d. – 2

Answer:

b. 2

Question 17.

The zeroes of the polynomial x^{2} – 3x – 4 are

a. 4, – 1

b. 4, 1

c. – 4, 1

d.- 4, – 1

Answer:

a. 4, – 1

Question 18.

If f(x) = x^{2} – 1 the value of f(2) is

a. 1

b. 3

c. 4

d. 0

Answer:

b. 3

Question 19.

If f(x) = 8 f(x) is called

a. Constant polynomials

b. linear polynomials

c. quadratic polynomials

d. Cubic polynomials

Answer:

a. Constant polynomials

Question 20.

A cubic polynomial has at most

a. 1 zeroes

b. 2 zeroes

c. 3 zeroes

d. 4 zeroes.

Answer:

c. 3 zeroes

II. Short Answer Questions:

Question 1.

Find the value of P(x) = x^{2} + 2x – 5 at x = 1

Answer:

P(x) = x^{2} + 2x – 5

P (1) = (1)^{2} + 2(1) – 5 = 1 + 2 – 5

= 3 – 5

P(1) = – 2

Question 2.

If x = 1 is a zero of the Polynomial f(x) = x^{3} – 2x^{2} + 4x + K, write the value of K.

Answer:

f(x) = x^{3} – 2x^{2} + 4x + K x = 1

f(1) = (1)^{3} – 2(1)^{2} + 4(1) + K

f(1) = 0 zero of the polynomial

0 = 1 – 2 + 4 + K

0 = 3 + K

K = – 3

Question 3.

If α and β are zeroes of Polynomials P(x) = x^{2} – 5x + 6, then find the value of α + β – 3αβ.

Answer:

α + β = \(\frac{-b}{a}=\frac{-(-5)}{1}\) = 5 and

αβ = \(\frac{c}{a}=\frac{6}{1}\) = 6

α + β – 3αβ

= 5 – 3(6)

= 5 – 18 = -13

α + β – 3αβ = – 13

Question 4.

Find the zeroes of the Polynomial P(x) = 4x^{2} – 12x + 9

Answer:

P(x) = 4x^{2} – 12x + 9

4x^{2} – 6x – 6x + 9 = 0

2x(2x – 3) – 3(2x – 3) = 0

(2x – 3) (2x – 3) = 0

x = \(\frac{3}{2}\), \(\frac{3}{2}\)

Question 5.

If 1 is a zero of the Polynomial P(x) = ax^{2} – 3(a – 1) x – 1, then find the value of a.

Answer:

P(x) = ax^{2} – 3(a – 1) x – 1

∴ x = 1

P(1) = a(1)^{2} – 3(a – 1) 1 – 1

P(1) = a – 3a + 3 – 1

0 = – 2a + 2

2a = 2

a = \(\frac{2}{2}\) = 1

a = 1

Question 6.

Write the degree of the Polynomial 3x^{3} – x^{4} + 5x + 3

Answer:

3x^{3} – x^{4} + 5x + 3

– x^{4} + 3x^{3} + 5x + 3

∴ degree of the Polynomial is 4

Question 7.

Write the standard form of a cubic Polynomial.

Answer:

ax^{3} + bx^{3} + cx + d.

Question 8.

Find the zero of the Polynomial , P(x) = a^{2}x, a ≠ 0

Answer:

P(x) = a^{2}x

a^{2}x = 0

x = \(\frac{0}{a^{2}}\)

x = 0

Question 9.

If α and β are the zeroes of the polynomial 4x^{2} + 3x + 7, then find the value of \(\frac{1}{\alpha}+\frac{1}{\beta}\)

Answer:

III. Long Answer Questions:

Question 1.

Find the zeroes of the following quadratic polynomials x^{2} + 4x + 4.

Answer:

f(x) = x^{2} + 4x + 4

0 = x^{2} + 2x + 2x + 4

0 = x(x + 2) + 2(x + 2)

(x + 2) (or) (x + 2) = 0

x + 2 = 0 (or) x + 2 = 0

x = – 2 x = – 2.

Zeroes of the Polynomials x^{2} + 4x + 4 are – 2 and – 2

Question 2.

If x = 1 is a zero of the polynomial f(x) = x^{3} – 2x^{2} + 4x + K, find the value of K.

Answer:

f(x) = x^{3} – 2x^{2} + 4x + K

∴ x = 1

f(x) = (1)^{3} – 2(1)^{2} + 4(1) + K

0 = 1 – 2(1) + 4 + K [f(1) = 0]

0 = 1 – 2 + 4 + K.

K + 3 = 0

K = – 3

Question 3.

For what value of K, – 4 is a zero of the polynomial x^{2} – x – (2K + 2)?

Answer:

f(x) = x^{2} – x – (2K + 2) and f(x) = 0

x = – 4

f(- 4) = (- 4)^{2} (- 4) – (2K + 2)

0 = 16 + 4 – 2K – 2

20 – 2 – 2K = 0 ⇒ 18 = 2K

K = \(\frac{18}{2}\) ⇒ K = 9

Question 4.

Divide p(x) by g(x) in each of the following cases and verify division algorithm.

P(x) = x^{2} + 4x + 4, g(x) = x + 2

Answer:

x^{2} + 4x + 4 ÷ x + 2

q(x) = x + 2 and r(x) = 0

Question 5.

On dividing the polynomial. P(x) = 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).

Answer:

P(x) = x^{3} – 3x^{2} + x + 2

q(x) = (x – 2); r(x) = (- 2x + 4)

g(x) = ?

g(x) = x^{2} – x + 1; r(x) = 0

∴ The divisor is x^{2} – x + 1

Question 6.

What must be subtracted from x^{3} + 5x^{2} + 5x + 8 so that the resulting polynomial is exactly divisible by x^{2} + 3x – 2?

Answer:

P(x) = x^{3} + 5x^{2} + 5x + 8

g(x) = x^{2} + 3x – 2

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

q(x) = x + 2; r(x) = x + 12

If we subtract x + 12 from x^{3} + 5x^{2} + 5x + 8 it will be exactly divisible by x^{2} + 3x – 2

Question 7.

What should be added to (x^{4} – 1) so that it’ is exactly divisible by (x^{2} + 2x + 1)?

Answer:

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

x^{4} + 0x^{3} + 0x^{3} + 0x – 1

r(x) = – 4x – 4

– {r(x)} = {4x + 4}

Question 8.

If the polynomials 2x^{3} + ax^{2} + 3x – 5 and x^{3} + x^{2} – 4x – a leave the same remainder when divided by x – 1 find the value of a.

Answer:

x – 1 = 0 [∴ g(x) = (0)]

x = 1

P(x) = 2x^{3} + ax^{2} + 3x – 5

P(1) = 2(1)^{3} + a(1)^{3} + 3(1) – 5

P(1) = 2 + a + 3 – 5

P(1) = a ➝ (1)

g(x) = x^{3} + x^{2} – 4x – a

g(1) = (1)^{3} + (1)^{2} – 4(1) – a

= 1 + 1 – 4 – a

g(x) = – 2 – a ➝ (2)

g(1) = g(1) [∴ Polynomials are leave same remainder]

a = – 2 – a

a + a = – 2

2a = – 2

a = \(\frac{-2}{2}\)

a = – 1

Question 9.

If (x^{3} + ax^{2} – bx + 10) is divisible by x^{3} – 3x + 2, find the values of a and b.

Answer:

P(x) = x^{3} + ax^{2} – bx + 10

g(x) = x^{2} – 3x + 2

x^{2} – 2x – x + 2 = 0

x(x – 2) – 1(x – 2) = 0

(x – 2) (x – 1) = 0

x = 2 and x = 1

P(x) = x^{3} + ax^{2} – bx + 10

P(x) = (1)^{3} + a(1)^{2} – b(1) + 10

= 1 + a – b + 10

0 = a – b + 11

a – b = – 11 ➝ (1)

P(2) = (2)^{3} + a(2)^{2} – b(2) + 10

P(2) = 8 + 4a – 2b – 10

0 = 4a – 2b + 18

4a – 2b = – 18 ➝ (2)

Multiply equation (1) by 4 and subtract with equation (2)

b = 13

Put b = 13 in equation (1)

a – b = – 11

a – 13 = – 11

a = – 11 + 13

a = 2

Question 10.

If both x – 2 and x – \(\frac{1}{2}\) are factors of ax^{2} + 5x + b show that a = b.

Answer:

P(x) = ax^{2} + 5x^{2} + b

x – 2 = 0

x = 2

P(x) = ax^{2} + 5x + b

P(2) = a(2)^{2} + 5(2) + b

0 = 4a + 10 + b ➝ (1)

P(x) = ax^{2} + 5x + b

x – \(\frac{1}{2}\) = 0

x = \(\frac{1}{2}\)

P(x) = ax^{2} + 5x + b

a = – 2 ➝ (3)

Put a = – 2 in equation (1)

4a – 10 + b = 0

4(- 2) + 10 + b = 0

– 8 + 10 + b = 0

2 + b = 0

b = -2 ➝ (4)

From (3) and (4)

a = b.

Question 11.

If α and β are the zeroes of the quadratic polynomials f(x) = 2x^{2} – 5x + 7, find a polynomial whose zeroes are 2α + 3β and 3α + 2β.

Answer:

Since α and β are the zeroes of the quadratic polynomial f(x) = 2x^{2} – 5x + 7.

Let S and P denote respectively the sum and product of the zeroes of the required polynomial.

Then, S = (2α + 3β) + (3α + 2β)

= 5(α + β) = 5 × \(\frac{5}{2}=\frac{25}{2}\) and

P = (2α + 3β) (3α + 2β)

⇒ P = (6α^{2} + 6β^{2} + 13αβ)

= 6α^{2} + 12αβ + αβ + 6β^{2}

⇒ P = 6(α^{2} + β^{2} + 2αβ + αβ

= 6(α + β)^{2} + αβ

= 6(α^{2} + β^{2} + 2αβ) + αβ

= 6(α + β)^{2} + αβ

Hence the required polynomial g(x) is given by

g(x) = K(x^{2} – 5x + P)

or g(x) = K\(\left(x^{2}-\frac{25}{2} x+41\right)\), where K and any non zero real number.

Question 12.

If α and β are the zeroes of the polynomial. 6y^{2} – 7y + 2, find a quadratic polynomial whose zeroes are \(\frac{1}{\alpha}\) and \(\frac{1}{\beta}\).

Answer:

Let P(y) = 6y^{2} – 7y + 2

Question 13.

If one zero of the polynomial 3x^{2} – 8x + 2k + 1 and seven times the other, find the value of k.

Answer:

Let α and β be the zeroes of the polynomial. Then as per question β = 7α.

Now sum of zeroes = α + β = α + 7α

Question 14.

If one zero of polynomial (a^{2} + 9)x^{2} + 13x + 6a and reciprocal of the other, find the value of a.

Answer:

Let one zero of the given polynomial be α

Then, the other zero and \(\frac{1}{\alpha}\)

∴ Product of zeroes = α × \(\frac{1}{\alpha}\) = 1

But, as per the given polynomial product of Zeroes = \(\frac{6 a}{a^{2}+9}\)

∴ \(\frac{6 a}{a^{2}+9}\) = 1 ⇒ a^{2} + 9 = 6a

⇒a^{2} – 6a + 9 = 0 ⇒ (a – 3)^{2} = 0

⇒ a – 3 = 0 ⇒ a = 3

Hence a = 3.

Question 15.

Find the zeroes of the polynomial f(x) = x^{3} – 5x^{2} – 2x + 24 if it is given that the product of its two zeroes is 12.

Answer:

Let α, β and γ be the zeroes of polynomial f(x) such that αβ = 12

Putting αβ = 12 in αβγ = – 24 we get

12γ = – 24 ⇒ γ = \(\frac{-24}{12}\) = – 2

Now α + β + γ = 5 ⇒ α + β – 2 = 5

⇒ α + β = 7

⇒ α = 7 – β

∴ αβ =12

⇒ (7 – β) β = 12

⇒ 7β = β^{2} = 12

⇒ β^{2} – 7β + 12 = 0

⇒ β2 – 3β – 4β – 12 = 0

⇒β(β -3) – 4(β – 3) = 0

⇒(β -4)(β – 3) = 0

⇒ β = 4 or β = 3

α = 3 or α = 4.

Question 16.

If α, β, γ be zeroes of polynomial 6x^{3} + 3x^{2} – 5x +1, then find the value of α^{– 1} + β^{– 1} + γ^{– 1}

Answer:

P(x) = 6x^{3} + 3x^{2} – 5x + 1 so

a = 6, b = 3, c = – 5, d = 1

∴ α, β and γ are zeroes of the polynomial p(x)