# KSEEB Solutions for Class 9 Maths Chapter 4 Polynomials Ex 4.4

In this chapter, we provide KSEEB SSLC Class 9 Maths Chapter 4 Polynomials Ex 4.4 for English medium students, Which will very helpful for every student in their exams. Students can download the latest KSEEB SSLC Class 9 Maths Chapter 4 Polynomials Ex 4.4 pdf, free KSEEB SSLC Class 9 Maths Chapter 4 Polynomials Ex 4.4 pdf download. Now you will get step by step solution to each question.

## 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) x3 + x2 + x + 1
ii) x4 + x3 + x2 + x + 1
iii) x4 + 3x3 + 3x2 + x + 1
iv) x3 – x2 – (2 + 2–√ )x + 2–√
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) = x3 + x2 + 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) = x4 + x3 + x2 + 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) = x4 + 3x3 + 3x2 + 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) = x3 – x2 – (2 + 2–√)x+ 2–√
p(-1) = (-1)3 – (-1)2 -(2 + 2–√)(-1) + 2–√
= -1 -(+1) – (2 – 2–√)+ 2–√
= -1 – 1 + 2 + 2–√ + 2–√
= -2 + 2 + 22–√
= = + 22–√
p(-1) = 22–√
Here, r(x) = p(a) = 22–√ 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) = 2x3 + x2 – 2x – 1, g(x) = x+1
ii) p(x) = x3 + 3x2 + 3x + 1, g(x) = x + 2
iii) p(x) = x3 – 4x2 + x + 6, g(x) = x – 3
i) p(x) = 2x3 + x2 – 2x – 1
g(x) = x + 1
If x + 1 = 0, then x = -1
p(x) = 2x3 + x2 – 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) = x3 + 3x2 + 3x + 1
g(x) = x + 2
If x + 2 = 0, then
x = -2
∴ p(x) = x3 + 3x2 + 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) = x3 – 4x2 + x + 6
g(x) = x – 3
Let, x – 3 = 0, then
x = 3
p(x) = x3 – 4x2 + 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) = x2 + x + k
ii) p(x) = 2x2 + kx + 2–√
iii) p(x) = kx2 – 2–√x + 1
iv) p(x) = kx2 – 3x + k
i) p(x) = x2 + x + k
g(x) = x – 1
k = ?
If x – 1 = 0, then
x = 1
p(x) = x2 + 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) = 2x2 + kx + 2–√
g(x) = x – 1 k = ?
If x – 1 = 0, then x = 1
p(x) = 2x2 + kx + 2–√
p(1) = 2(1)2 + k(1) + 2–√
= 2(1) + k(l) + 2–√
p(1) = 2 + k + 2–√
If (x – 1) is the factor of p(x), then we have p(1) = 0.
∴ 2 + k + 2–√ = 0
k = -2 – 2–√

iii) p(x) = kx2 – 2–√x + 1
g(x) = x – 1 k = ?
If x – 1 = 0, then
x = 1
p(x) = kx2 – 2–√x + 1
p(1) = k(1)2– 2–√(1) + 1
= k( 1) – 2–√ + 1
p(1) = k- 2–√ + 1
If (x – 1) is the factor of p(x) then we have p(1) = 0.
∴ p(1) = k – 2–√ + 1 = 0
∴ k= 2–√ – 1

iv) p(x) = kx2 – 3x + k
g(x) = x – 1 k = ?
If x – 1 = 0, then x – 1
p(x) = kx2 – 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) 12x2 – 7x + 1
ii) 2x2 + 7x + 3
iii) 6x2 + 5x – 6
iv) 3x2 – x – 4
i) 12x2 – 7x + 1 = 12x2 – 4x – 3x + 1
= 4x (3x – 1) – 1(3x – 1)
= (3x – 1) (4x – 1)

ii) 2x2 + 7x + 3 = 2x2 + 6x + x + 3
= 2x(x + 3) + 1 (x + 3)
= (x + 3) (2x + 1)

iii) 6x2 + 5x – 6 = 6x2 + 9x – 4x – 6
= 3x(2x + 3) – 2(2x + 3)
= (2x + 3) (3x – 2)

iv) 3x2 – x – 4 = 3x2 – 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) x2 – 2x2 – x + 2
ii) x2 – 3x2 – 9x – 5
iii) x3 + 13x2 + 32x + 20
iv) 2y3 + y2 – 2y – 1
i) x3 – 2x2 – x + 2
= x2(x – 2) – 1 (x – 2)
= (x – 2) (x2 – 1)
= (x – 2) (x + 1) (x- 1)

ii) x3 – 3x2 – 9x – 5
= x3 – 5x2 + 2x2 – 10x + x – 5
= x2(x – 5) + 2x(x – 5) + 1 (x – 5)
= (x – 5) (x2 + 2x + 1)
= (x – 5) {x2 + x + x + 1}
= (x – 5) (x(x + 1) + 1(x + 1)}
= (x- 5)(x + 1) (x + 1)

iii) x3 + 13x2 + 32x + 20
= x3 + 10x2 + 3x2 + 30x + 2x + 26
= x2(x + 10) + 3x(x + 10) + 2(x + 10)
= (x + 10) (x2 + 3x + 2)
= (x + 10) {x2 + 2x + x + 2)
= (x + 10) {x(x + 2) + 1 (x + 2))
= (x + 10) (x + 2) (x + 1)

iv) 2y3 + y2 – 2y – 1
= y2(2y + 1) – 1(2y + 1)
= (2y + 1) (y2– 1)
= (2y + 1) {(y)2 – (1)2}
= (2y+ 1) (y + 1) (y- 1)

All Chapter KSEEB Solutions For Class 9 Maths

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All Subject KSEEB Solutions For Class 9

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