**KSEEB Solutions for Class 9 Maths Chapter 4 Polynomials Ex 4.4** are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 4 Polynomials Exercise 4.4.

## Karnataka Board Class 9 Maths Chapter 4 Polynomials Ex 4.4.

**Exercise 4.4 Class 9 Polynomials Question 1.**

Determine which of the following polynomials has (x+1) a factor :

i) x^{3} + x^{2} + x + 1

ii) x^{4} + x^{3} + x^{2} + x + 1

iii) x^{4} + 3x^{3} + 3x^{2} + x + 1

iv) x^{3} – x^{2} – (2 + \(\sqrt{2}\) )x + \(\sqrt{2}\)

Answer:

i) x- 1 is a factor of p(x)

x + 1 = x – a

a = -1

For the value of p(a),

value of r(x) = 0.

∴ p(x) = x^{3} + x^{2} + x + 1

p(-1) = (-1)^{3} + (-1)^{2} + (-1) + 1

= -1 + 1 -1 + 1

p(-1) = 0

Here, p(a) = r(x) = 0

∴ x + 1 is a factor.

ii) If x + 1 = x – a, then a = -1

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

p(-1)= (-1)^{4} + (-1)^{3} +(-1)^{2} + (-1)+ 1

= 1 – 1 + 1 – 1 + 1

= 3 – 2 p(-1)= 1

Here, r(x) = p(a)= 1 Reminder is not zero.

∴ x+1 is not a factor.

iii) If x + 1 = x – a then

a = -1

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

p(-1) = (-1)^{4} + 3(-1 )^{3} +3(-1 )^{2} + (-1) + 1

= 1 + 3(-1) + 3(1) + 1(-1) + 1

= 1- 3 + 3 – 1 + 1

= 5 – 4

P(-1)= 1

Here, r(x) = p(a)=l Remainder is not zero

∴ x+1 is not a factor.

iv) If x + 1 = x – a then,

a = -1

p(x) = x^{3} – x^{2} – (2 + \(\sqrt{2}\))x+ \(\sqrt{2}\)

p(-1) = (-1)^{3} – (-1)^{2} -(2 + \(\sqrt{2}\))(-1) + \(\sqrt{2}\)

= -1 -(+1) – (2 – \(\sqrt{2}\))+ \(\sqrt{2}\)

= -1 – 1 + 2 + \(\sqrt{2}\) + \(\sqrt{2}\)

= -2 + 2 + \(2 \sqrt{2}\)

= = + \(2 \sqrt{2}\)

p(-1) = \(2 \sqrt{2}\)

Here, r(x) = p(a) = \(2 \sqrt{2}\) Value of remainder r(x) is not zero.

∴ x + 1 is not a factor.

**KSEEB Solutions For Class 9 Maths Polynomials Question 2.**

Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases :

i) p(x) = 2x^{3} + x^{2} – 2x – 1, g(x) = x+1

ii) p(x) = x^{3} + 3x^{2} + 3x + 1, g(x) = x + 2

iii) p(x) = x^{3} – 4x^{2} + x + 6, g(x) = x – 3

Answer:

i) p(x) = 2x^{3} + x^{2} – 2x – 1

g(x) = x + 1

If x + 1 = 0, then x = -1

p(x) = 2x^{3} + x^{2} – 2x – 1

p(-1)= 2(-1)^{3} + (-1)^{2} -2(- 1) – 1

= 2(-1) + (1) + 2 – 1

= -2 + 1 + 2 – 1

P(-1)= 0

Here, r(x) = p(a) = 0,

∴ g(x) is the a factor f(x)

ii) p(x) = x^{3} + 3x^{2} + 3x + 1

g(x) = x + 2

If x + 2 = 0, then

x = -2

∴ p(x) = x^{3} + 3x^{2} + 3x + 1

p(-2) = (-2)^{3} + 3(-2)^{2} + 3 (-2) + 1

= -8 + 3(4) + 3(-2) + 1

= -8 + 12 – 6 + 1

= 13 – 24

p(-2)= -11

Here we have r(x) = p(a) =-11.

Value of r(x) is not equal to zero.

∴ g(x) is not a factor of f(x).

iii) p(x) = x^{3} – 4x^{2} + x + 6

g(x) = x – 3

Let, x – 3 = 0, then

x = 3

p(x) = x^{3} – 4x^{2} + x + 6

p(3) = (3)^{2} – 4(3)^{2} + 3 + 6

= 27 – 4(9) + 3 + 6

= 27 – 36 + 3 + 6

= 36 – 36

p(3) = 0

Here, we have r(x) = p(a) = 0

∴ (x – 3) is the factor of p(x).

**9th Maths Polynomials Exercise 4.4 Question 3.**

Find the value of k, if x – 1 is a factor of p(x) in each of the following cases :

i) p(x) = x^{2} + x + k

ii) p(x) = 2x^{2} + kx + \(\sqrt{2}\)

iii) p(x) = kx^{2} – \(\sqrt{2} \mathrm{x}\) + 1

iv) p(x) = kx^{2} – 3x + k

Answer:

i) p(x) = x^{2} + x + k

g(x) = x – 1

k = ?

If x – 1 = 0, then

x = 1

p(x) = x^{2} + x + k

p(1) = (1)^{2} + 1 + k

p( 1) = 1 + 1 + k

p( 1) = 2 + k

If g(x) is a factor, then we have r(x) = 0

∴ p(1) = 0

2 + k= 0

∴ k = 0 – 2

k = -2

ii) p(x) = 2x^{2} + kx + \(\sqrt{2}\)

g(x) = x – 1 k = ?

If x – 1 = 0, then x = 1

p(x) = 2x^{2} + kx + \(\sqrt{2}\)

p(1) = 2(1)^{2} + k(1) + \(\sqrt{2}\)

= 2(1) + k(l) + \(\sqrt{2}\)

p(1) = 2 + k + \(\sqrt{2}\)

If (x – 1) is the factor of p(x), then we have p(1) = 0.

∴ 2 + k + \(\sqrt{2}\) = 0

k = -2 – \(\sqrt{2}\)

iii) p(x) = kx^{2} – \(\sqrt{2} x\) + 1

g(x) = x – 1 k = ?

If x – 1 = 0, then

x = 1

p(x) = kx^{2} – \(\sqrt{2} x\) + 1

p(1) = k(1)^{2}– \(\sqrt{2}\)(1) + 1

= k( 1) – \(\sqrt{2}\) + 1

p(1) = k- \(\sqrt{2}\) + 1

If (x – 1) is the factor of p(x) then we have p(1) = 0.

∴ p(1) = k – \(\sqrt{2}\) + 1 = 0

∴ k= \(\sqrt{2}\) – 1

iv) p(x) = kx^{2} – 3x + k

g(x) = x – 1 k = ?

If x – 1 = 0, then x – 1

p(x) = kx^{2} – 3x + k

p(1) = k(1)^{2} – 3(1) + k

= k(1) – 3(1) + k

= k – 3 + k

p(1) = 2k – 3

If (x – 1) is the factor of p(x), then we have p(1) = 0.

**Polynomials Class 9 Exercise 4.4 Question 4.**

Factorise :

i) 12x^{2} – 7x + 1

ii) 2x^{2} + 7x + 3

iii) 6x^{2} + 5x – 6

iv) 3x^{2} – x – 4

Answer:

i) 12x^{2} – 7x + 1

= 12x^{2} – 4x – 3x + 1

= 4x (3x – 1) – 1(3x – 1)

= (3x – 1) (4x – 1)

ii) 2x^{2} + 7x + 3

= 2x^{2} + 6x + x + 3

= 2x(x + 3) + 1 (x + 3)

= (x + 3) (2x + 1)

iii) 6x^{2} + 5x – 6

= 6x^{2} + 9x – 4x – 6

= 3x(2x + 3) – 2(2x + 3)

= (2x + 3) (3x – 2)

iv) 3x^{2} – x – 4

= 3x^{2} – 4x + 3x – 4

=x(3x – 4) + 1 (3x – 4)

= (3x – 4) (x + 1)

**KSEEB Solutions For Class 9 Maths Polynomials Exercise 4.4 Question 5.**

Factorise :

i) x^{2} – 2x^{2} – x + 2

ii) x^{2} – 3x^{2} – 9x – 5

iii) x^{3} + 13x^{2} + 32x + 20

iv) 2y^{3} + y^{2} – 2y – 1

Answer:

i) x^{3} – 2x^{2} – x + 2

= x^{2}(x – 2) – 1 (x – 2)

= (x – 2) (x^{2} – 1)

= (x – 2) (x + 1) (x- 1)

ii) x^{3} – 3x^{2} – 9x – 5

= x^{3} – 5x^{2} + 2x^{2} – 10x + x – 5

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

= (x – 5) (x^{2} + 2x + 1)

= (x – 5) {x^{2} + x + x + 1}

= (x – 5) (x(x + 1) + 1(x + 1)}

= (x- 5)(x + 1) (x + 1)

iii) x^{3} + 13x^{2} + 32x + 20

= x^{3} + 10x^{2} + 3x^{2} + 30x + 2x + 26

= x^{2}(x + 10) + 3x(x + 10) + 2(x + 10)

= (x + 10) (x^{2} + 3x + 2)

= (x + 10) {x^{2} + 2x + x + 2)

= (x + 10) {x(x + 2) + 1 (x + 2))

= (x + 10) (x + 2) (x + 1)

iv) 2y^{3} + y^{2} – 2y – 1

= y^{2}(2y + 1) – 1(2y + 1)

= (2y + 1) (y^{2}– 1)

= (2y + 1) {(y)^{2} – (1)^{2}}

= (2y+ 1) (y + 1) (y- 1)

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