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 State Syllabus Class 9 Maths Chapter 4 Polynomials Ex 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) x3 + x2 + x + 1
ii) x4 + x3 + x2 + x + 1
iii) x4 + 3x3 + 3x2 + x + 1
iv) x3 – x2 – (2 + 2–√ )x + 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) = 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
Answer:
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
Answer:
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
Answer:
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
Answer:
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
—————————————————————————–
All Subject KSEEB Solutions For Class 9
*************************************************
I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.
If these solutions have helped you, you can also share kseebsolutionsfor.com to your friends.
Best of Luck!!