AP Board Class 9 Maths Chapter 2 Polynomials and Factorisation Ex 2.4 Textbook Solutions PDF: Download Andhra Pradesh Board STD 9th Maths Chapter 2 Polynomials and Factorisation Ex 2.4 Book Answers ~ HSSlive: Plus One & Plus Two Notes & Solutions for Kerala State Board

AP Board Class 9 Maths Chapter 2 Polynomials and Factorisation Ex 2.4 Textbook Solutions PDF: Download Andhra Pradesh Board STD 9th Maths Chapter 2 Polynomials and Factorisation Ex 2.4 Book Answers

AP Board Class 9th Maths Chapter 2 Polynomials and Factorisation Ex 2.4 Textbooks Solutions and answers for students are now available in pdf format. Andhra Pradesh Board Class 9th Maths Chapter 2 Polynomials and Factorisation Ex 2.4 Book answers and solutions are one of the most important study materials for any student. The Andhra Pradesh State Board Class 9th Maths Chapter 2 Polynomials and Factorisation Ex 2.4 books are published by the Andhra Pradesh Board Publishers. These Andhra Pradesh Board Class 9th Maths Chapter 2 Polynomials and Factorisation Ex 2.4 textbooks are prepared by a group of expert faculty members. Students can download these AP Board STD 9th Maths Chapter 2 Polynomials and Factorisation Ex 2.4 book solutions pdf online from this page.

Andhra Pradesh Board Class 9th Maths Chapter 2 Polynomials and Factorisation Ex 2.4 Textbooks Solutions PDF

Andhra Pradesh State Board STD 9th Maths Chapter 2 Polynomials and Factorisation Ex 2.4 Books Solutions with Answers are prepared and published by the Andhra Pradesh Board Publishers. It is an autonomous organization to advise and assist qualitative improvements in school education. If you are in search of AP Board Class 9th Maths Chapter 2 Polynomials and Factorisation Ex 2.4 Books Answers Solutions, then you are in the right place. Here is a complete hub of Andhra Pradesh State Board Class 9th Maths Chapter 2 Polynomials and Factorisation Ex 2.4 solutions that are available here for free PDF downloads to help students for their adequate preparation. You can find all the subjects of Andhra Pradesh Board STD 9th Maths Chapter 2 Polynomials and Factorisation Ex 2.4 Textbooks. These Andhra Pradesh State Board Class 9th Maths Chapter 2 Polynomials and Factorisation Ex 2.4 Textbooks Solutions English PDF will be helpful for effective education, and a maximum number of questions in exams are chosen from Andhra Pradesh Board.

Andhra Pradesh State Board Class 9th Maths Chapter 2 Polynomials and Factorisation Ex 2.4 Books Solutions

Board	AP Board
Materials	Textbook Solutions/Guide
Format	DOC/PDF
Class	9th
Subject	Maths
Chapters	Maths Chapter 2 Polynomials and Factorisation Ex 2.4
Provider	Hsslive

How to download Andhra Pradesh Board Class 9th Maths Chapter 2 Polynomials and Factorisation Ex 2.4 Textbook Solutions Answers PDF Online?

Visit our website - Hsslive
Click on the Andhra Pradesh Board Class 9th Maths Chapter 2 Polynomials and Factorisation Ex 2.4 Answers.
Look for your Andhra Pradesh Board STD 9th Maths Chapter 2 Polynomials and Factorisation Ex 2.4 Textbooks PDF.
Now download or read the Andhra Pradesh Board Class 9th Maths Chapter 2 Polynomials and Factorisation Ex 2.4 Textbook Solutions for PDF Free.

AP Board Class 9th Maths Chapter 2 Polynomials and Factorisation Ex 2.4 Textbooks Solutions with Answer PDF Download

Find below the list of all AP Board Class 9th Maths Chapter 2 Polynomials and Factorisation Ex 2.4 Textbook Solutions for PDF’s for you to download and prepare for the upcoming exams:

AP State Syllabus 9th Class Maths Solutions 2nd Lesson Polynomials and Factorisation Exercise 2.4

Question 1.
Determine which of the following polynomials has (x + 1) as a factor.
i) x³ – x² – x + 1
Solution:
f(- 1) = (- 1)³ – (- 1)² – (- 1) + 1
= -1 – 1 + 1 + 1 = 0
∴ (x + 1) is a factor.

ii) x⁴ -x³ +x² – x + 1
Solution:
f(- 1) = (- 1)⁴ – (- 1)³ + (- 1)² – (- 1) + 1
= 1 + 1 + 1 + 1 + 1= 5
∴ (x + 1) is not a factor.

iii) x⁴ + 2x³ + 2x² + x + 1
Solution:
f(- 1) = (-1)⁴ + 2 (- 1)³ + 2 (- 1)² + (-1) + 1
= 1 – 2 + 2 – 1 + 1 = 1
∴ (x + 1) is not a factor.

iv) x³ – x² – (3 – √3)x + √3
Solution:
f(- 1) = (- 1)³ – (- 1)²– (3 – √3)(-1) + √3
= – 1 – 1 + 3 – √3 + √3 = 1
∴ (x + 1) is not a factor.

Question 2.
Use the factor theorem to determine whether g(x) is a factor of f(x) in each of the following cases:
i) f(x) = 5x³ + x² – 5x – 1; g(x) = x + 1
[Factor theorem : If f(x) is a polynomial; f(a) = 0 then (x – a) is a factor of f(x); a ∈ R]
Solution:
g(x) = x+ 1 = x- a say
∴ a = – 1
f(a) = f(- 1) = 5 (- 1)³ + (- 1)² – 5 (- 1) – 1
= -5 + 1 + 5 – 1 = 0
∴ x + 1 is a factor of f(x).

ii) f(x) = x³ + 3x² + 3x + 1; g(x) = x + 1
Solution:
g(x) = x + 1 = x – a
∴ a = – 1
f(a) = f(- 1) = (- 1)³ + 3 (- 1)² + 3(-1) + 1
= -1 + 3 – 3 + 1 =0
∴ f(x) is a factor of g(x).

iii) f(x) = x³ – 4x² + x + 6;
g(x) = x – 2
Solution:
g(x) = x- 2 = x- a
∴ a = 2
f(a) = f(2) = 2³ – 4(2)² + 2 + 6
= 8 – 16 + 2 + 6 = 0
∴ g(x) is a factor of f(x).

iv) f(x) = 3x³+ x² – 20x +12; g(x) = 3x – 2
Solution:
g(x) = 3x – 2 = 𝑥−23 = x – a
∴ a = 2/3

v) f(x) = 4x³+ 20x²+ 33x + 18; g(x) = 2x + 3
Solution:
g(x) = 2x + 3 = x + 32= = x – a
∴ a = -3/2

∴ g(x) is a factor of f(x).

Question 3.
Show that (x – 2), (x + 3) and (x – 4) are factors of x³ – 3x² – 10x + 24.
Solution:
Given f(x) = x³ – 3x² – 10x + 24
To check whether (x – 2), (x + 3) and (x – 4) are factors of f(x), let f(2), f(- 3) and f(4)
f(2) = 2³ – 3(2)² – 10(2) + 24
= 8- 12-20 + 24 = 0
∴ (x – 2) is a factor of f(x).

f(- 3) = (- 3)³ – 3(- 3)²– 10(- 3) + 24
= – 27 – 27 + 30 + 24 = 0
∴ (x + 3) is a factor of f(x).

f(4) = (4)³ – 3 (4)² – 10 (4) + 24
= 64 – 48 – 40 + 24
= 88 – 88
= 0
∴ (x – 4) is a factor of f(x).

Question 4.
Show that (x + 4), (x – 3) and (x – 7) are factors of x³ – 6x² – 19x + 84.
Solution:
Let f(x) = x³ – 6x² – 19x + 84
To verify whether (x + 4), (x – 3) and (x – 7) are factors of f(x) we use factor theorem.

Let f(- 4), f(3) and f(7)
f(- 4) = (- 4)³ – 6 (- 4)² – 19 (- 4) + 84
= -64 – 96 + 76 + 84
= 0 .
∴ (x + 4) is a factor of f(x).

f(3) = 3³ – 6(3)² – 19(3) + 84
= 27 – 54 – 57 + 84
= 0
∴ (x – 3) is a factor of f(x).

f(7) = 7³ – 6(7)² – 19(7) + 84
= 343 – 294 – 133 + 84
= 427 – 427
= 0
∴ (x – 7) is a factor of f(x).

Question 5.
If both (x – 2) and (𝑥−12) of px² + 5x + r, show that p = r.
Solution:
Let f(x) = px²+ 5x + r
As (x – 2) and (𝑥−12) are factor of f(x), we have f(2) = 0 and f(1/2) = 0
∴ f(2) = p(2)² + 5(2) + r
= 4p + 10 + r = 0
= 4p + r
= – 10 ………………(1)

⇒ p + 10 + 4r = 0
⇒ p + 4r = – 10 ………………. (2)
From (1) and (2);
4p + r = p + 4r
4p – p = 4r – r
3p = 3r
∴ P = r

Question 6.
If (x² – 1) is a factor of ax⁴ + bx³ + cx² + dx + e, show that a + c + e = b + d = 0.
Solution:
Let f(x) = ax⁴ + bx³ + cx² + dx + e
As (x – 1) is a factor of f(x) we have
x² – 1 = (x + 1) (x – 1) hence f(1) = 0 and f(-1) = 0
f(1) = a + b + c + d + e = 0 ……………. (1)
and f(-1) = a- b + c- d + e = 0
⇒ a + c + e = b + d
Substitute this value in equation (1)
a + c + e + b + d=0
b + d + b + d=0
2 (b + d) = 0
⇒ b + d = 0
∴ a + c + e = b + d = 0

Question 7.
Factorise
i) x³ – 2x² – x + 2
Solution:
Let f(x) = x³ – 2x² – x + 2
By trial, we find f(l) = 1³ – 2(1)² – 1 + 2
= 1 – 2 – 1 + 2
= 0 .
∴ (x – 1) is a factor of f(x).
[by factor theorem]
Now dividing f(x) by (x – 1).

f(x) = (x – 1) (x² – x – 2)
= (x – 1) [x² – 2x + x- 2]
= (x – 1) [x (x – 2) + 1 (x – 2)]
= (x – 1) (x – 2) (x + 1)

ii) x³ – 3x² – 9x – 5
Solution:
Let f(x) = x³ – 3x² – 9x – 5By trial,
f(- 1) = (- 1)³ – 3(- 1)² – 9(- 1) – 5
=-1 – 3 + 9 – 5
=0
∴ (x + 1) is a factor of f(x).
[ ∵ by factor theorem]
Now dividing f(x) by (x + 1).

f(x)=(x + 1)(x² – 4x – 5)
But x²– 4x – 5 = x² – 5x + x – 5
= x (x – 5) + 1 (x – 5)
=(x – 5)(x + 1)
∴ f(x)=(x + 1)(x + 1)(x – 5)

iii) x³ + 13x² + 32x + 20
Solution:
Let f(x) = x³ + 13x² + 32x + 20
Let f(- 1)
= (- 1)³ + 13 (- 1)² + 32 (- 1) + 20
= – 1 + 13 – 32 + 20 = 33 – 33 = 0
∴ (x + 1) is a factor of f(x).
[ ∵ by factor theorem] Now dividing f(x) by (x + 1).

iv) y³ + y² – y – 1
Let f(y) = y³ + y² – y – 1
f(1) = 1³+ 1²– 1 – 1 = 0
(y – 1) is a factor of f(y).
Now dividing f(y) by (y – 1).

∴ f(x) = (x + 1)(x² + 12x + 20)
But (x² + 12x + 20) = x²+ 10x + 2x + 20
=x(x + 10)+2(x + 10)
=(x + 10)(x + 2)
∴f(x) = (x + 1)(x + 2)(x + 10)

Question 8.
If ax² + bx + c and bx² + ax + c have a common factor x + 1 then show that c = 0 and a = b.
Solution:
Let f(x) = ax² + bx + c and g(x) = bx² + ax + c given that (x + 1) is a common factor for both f(x) and g(x).
∴ f(-1) = g(- 1)
⇒a(- 1)² + b(- 1) + c
= b(- 1)² + a (- 1) + c
⇒ a – b + c = b – a + c
⇒ a + a = b + b
⇒ 2a = 2b
⇒ a = b
Also f(- 1) = a – b + c = 0
⇒ b – b + c = 0
⇒ c = 0

Question 9.
If x² – x – 6 and x² + 3x – 18 have a common factor x – a then find the value of a.
Solution:
Let f(x) = x² – x – 6 and
g(x) = x² + 3x – 18
Given that (x – a) is a factor of both f(x) and g(x).
f(a) = g(a) = 0
⇒ a² – a – 6 = a² + 3a – 18
⇒ – 4a = – 18 + 6
⇒ – 4a = – 12
∴ a = 3

Question 10.
If (y – 3) is a factor of y³– 2y²– 9y + 18, then find the other two factors.
Solution:
Let f(y) = y³– 2y²– 9y + 18
Given that (y – 3) is a factor of f(y).
Dividing f(y) by (y – 3)

∴ f(y) = (y – 3) (y + y – 6)
But y² + y – 6
= y² + 3y – 2y – 6
= y (y + 3) – 2 (y + 3)
= (y + 3) (y – 2)
∴ f(y) = (y – 2)(y – 3)(y + 3)
The other two factors are (y – 2) and (y + 3).

AP Board Textbook Solutions PDF for Class 9th Maths

Andhra Pradesh Board Class 9th Maths Chapter 2 Polynomials and Factorisation Ex 2.4 Textbooks for Exam Preparations

Andhra Pradesh Board Class 9th Maths Chapter 2 Polynomials and Factorisation Ex 2.4 Textbook Solutions can be of great help in your Andhra Pradesh Board Class 9th Maths Chapter 2 Polynomials and Factorisation Ex 2.4 exam preparation. The AP Board STD 9th Maths Chapter 2 Polynomials and Factorisation Ex 2.4 Textbooks study material, used with the English medium textbooks, can help you complete the entire Class 9th Maths Chapter 2 Polynomials and Factorisation Ex 2.4 Books State Board syllabus with maximum efficiency.

FAQs Regarding Andhra Pradesh Board Class 9th Maths Chapter 2 Polynomials and Factorisation Ex 2.4 Textbook Solutions

How to get AP Board Class 9th Maths Chapter 2 Polynomials and Factorisation Ex 2.4 Textbook Answers??

Students can download the Andhra Pradesh Board Class 9 Maths Chapter 2 Polynomials and Factorisation Ex 2.4 Answers PDF from the links provided above.

Can we get a Andhra Pradesh State Board Book PDF for all Classes?

Yes you can get Andhra Pradesh Board Text Book PDF for all classes using the links provided in the above article.

Important Terms

Andhra Pradesh Board Class 9th Maths Chapter 2 Polynomials and Factorisation Ex 2.4, AP Board Class 9th Maths Chapter 2 Polynomials and Factorisation Ex 2.4 Textbooks, Andhra Pradesh State Board Class 9th Maths Chapter 2 Polynomials and Factorisation Ex 2.4, Andhra Pradesh State Board Class 9th Maths Chapter 2 Polynomials and Factorisation Ex 2.4 Textbook solutions, AP Board Class 9th Maths Chapter 2 Polynomials and Factorisation Ex 2.4 Textbooks Solutions, Andhra Pradesh Board STD 9th Maths Chapter 2 Polynomials and Factorisation Ex 2.4, AP Board STD 9th Maths Chapter 2 Polynomials and Factorisation Ex 2.4 Textbooks, Andhra Pradesh State Board STD 9th Maths Chapter 2 Polynomials and Factorisation Ex 2.4, Andhra Pradesh State Board STD 9th Maths Chapter 2 Polynomials and Factorisation Ex 2.4 Textbook solutions, AP Board STD 9th Maths Chapter 2 Polynomials and Factorisation Ex 2.4 Textbooks Solutions,

HSSlive: Plus One & Plus Two Notes & Solutions for Kerala State Board

Hsslive.co.in: Kerala Higher Secondary News, Plus Two Notes, Plus One Notes, Plus two study material, Higher Secondary Question Paper.

Thursday, July 14, 2022