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

## AP Board Class 10 Maths Chapter 3 Polynomials Ex 3.2 Textbook Solutions PDF: Download Andhra Pradesh Board STD 10th Maths Chapter 3 Polynomials Ex 3.2 Book Answers

 AP Board Class 10 Maths Chapter 3 Polynomials Ex 3.2 Textbook Solutions PDF: Download Andhra Pradesh Board STD 10th Maths Chapter 3 Polynomials Ex 3.2 Book Answers

## Andhra Pradesh State Board Class 10th Maths Chapter 3 Polynomials Ex 3.2 Books Solutions

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## AP Board Class 10th Maths Chapter 3 Polynomials Ex 3.2 Textbooks Solutions with Answer PDF Download

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### 10th Class Maths 3rd Lesson Polynomials Ex 3.2 Textbook Questions and Answers

Question 1.
The graphs of y = p(x) are given in the figure below, for some polynomials p(x). In each case, find the number of zeroes of p(x).

i) There are no zeroes as the graph does not intersect the X – axis.
ii) The number of zeroes is one as the graph intersects the X – axis at one point only.
iii) The number of zeroes is three as the graph intersects the X – axis at three points.
iv) The number of zeroes is two as the graph intersects the X – axis at two points.
v) The number of zeroes is four as the graph intersects the X – axis at four points.
vi) The number of zeroes is three as the graph intersects the X – axis at three points.

Question 2.
Find the zeroes of the given polynomials,
(i) p(x) = 3x
(ii) p(x) = x2 + 5x + 6
(iii) p(x) = (x + 2) (x + 3)
(iv) p(x) = x4 – 16
i) Given p(x) = 3x
Let p(x) = 0
So, 3x = 0
x = 03 = 0,
Zeroes of p(x) = 3x is zero.
∴ No. of zeroes is one.

ii) Given p(x) = x2 + 5x + 6 is a quadratic polynomial.
It has atmost two zeroes.
To find zeroes, let p(x) = 0
⇒ x2 + 5x + 6 = 0
⇒ x2 + 3x + 2x + 6 = 0
⇒ x(x + 3) + 2 (x + 3) = 0
⇒ (x + 3) (x + 2) = 0
⇒ x + 3 = 0 or x + 2 = 0
⇒ x = -3 or x = -2
Therefore the zeroes of the polynomial are -3 and -2.

iii) Given p(x) = (x + 2) (x + 3)
It is a quadratic polynomial.
It has atmost two zeroes.
Let p(x) = 0
⇒ (x + 2) (x + 3) = 0
⇒ (x + 2) = 0 or (x + 3) = 0
⇒ x = -2 or x = -3
Therefore the zeroes of the polynomial are -2 and – 3.

iv) Given p(x) = x4 – 16 is a biquadratic polynomial. It has atmost two zeroes.
Let p(x) = 0
⇒ x4 – 16 = 0
⇒ (x2)2 – 42 = 0
⇒ (x2 – 4) (x2 + 4) = 0
⇒ (x + 2) (x – 2) (x2 + 4) = 0
⇒ (x + 2) = 0 or (x – 2) = 0 or (x2 + 4) = 0
⇒ x = -2 (or) x = 2 (or) x2 = -4
Therefore the zeroes of the polynomial are 2, – 2, we do not consider √-4 since it is not real.

Question 3.
Draw the graphs of the given polynomial and find the zeroes. Justify the answers,
i) p(x) = x2 – x – 12
ii) p(x) = x2 – 6x + 9
iii) p(x) = x2 – 4x + 5
iv) p(x) = x2 + 3x – 4
v) p(x) = x2 – 1
i) Given polynomial p(x) = x2 – x – 12.
List of values of p(x):

Now, let’s locate the points listed above on a graph paper and draw the graph.

Result: We observe that the graph cuts the X – axis at (-3, 0) and (4, 0).
So, the zeroes of the polynomial are -3 and 4.
Justification:
Given p(x) = x2 – x – 12 = 0
⇒ x2 – 4x + 3x – 12 = 0
⇒ x(x – 4) + 3(x – 4) = 0
⇒ (x – 4) (x + 3) = 0
⇒ x – 4 = 0 and x + 3 = 0
x = 4 and x = – 3

ii) Given polynomial p(x) = x2 – 6x + 9
List of values of p(x):

Now, let’s locate the points listed above on a graph paper and draw the graph.

Result: We observe that the graph cuts the X – axis at (3, 0).
So, the zeroes of the given polynomial are same i.e., 3.
Justification:
Given p(x) = x2 – 6x + 9
⇒ x2 – 3x – 3x + 9 = 0
⇒ x(x – 3) – 3(x – 3) = 0
⇒ (x – 3) (x – 3) = 0
⇒ x – 3 = 0 and x – 3 = 0
x = 3 and x = 3

iii) Given polynomial p(x) = x2 – 4x + 5
List of values of p(x):

Now, let’s locate the points listed above on a graph paper and draw the graph.

Result: We observe that the graph does not cut the X – axis at any point.
So, the quadratic polynomial p(x) has no zeroes.
Justification: For the given p(x) = x2 – 4x + 5 not possible to split in factors.

iv) Given polynomial p(x) = x2 + 3x – 4.
List of values of p(x):

Now, let’s locate the points listed above on a graph paper and draw the graph.

Result: We observe that the graph cuts the X – axis at (-4, 0) and (1, 0).
So, the zeroes of the polynomial are -4 and 1.
Justification:
Given p(x) = x2 + 3x – 4 = 0
⇒ x2 + 4x – x – 4 = 0
⇒ x(x + 4)- 1(x + 4) = 0
⇒ (x + 4) (x – 1) = 0
⇒ x + 4 = 0 and x – 1 = 0
x = – 4 and x = 1

v) Given polynomial p(x) = x2 – 1
List of values of p(x):

Now, let’s locate the points listed above on a graph paper and draw the graph.

Result: We observe that the graph cuts the X – axis at (-1, 0) and (1,0).
So, the zeroes of the polynomial are – 1 and 1.
Justification:
Given p(x) = x2 – 1 = 0
⇒ p(x) = (x + 1) (x – 1) = 0 [∵ a2 – b2 = (a + b) (a – b)]
⇒ x + 1 = 0 and x – 1 = 0
x = -1 and x = 1

Question 4.
Why are 14 and -1 zeroes of the polynomial p(x) = 4x2 + 3x – 1 ?
Given polynomial p(x) = 4x2 + 3x – 1
Given zeroes are 14 and -1
Let x = 14

Let x = -1
⇒ p(-1) = 4(-1)2 + 3(-1)-1 = 4 – 3 – 1 = 4 – 4 = 0
∴ P(14) = 0 and p(-1) = 0
So these values are zeroes of the polynomial p(x).

## Andhra Pradesh Board Class 10th Maths Chapter 3 Polynomials Ex 3.2 Textbooks for Exam Preparations

Andhra Pradesh Board Class 10th Maths Chapter 3 Polynomials Ex 3.2 Textbook Solutions can be of great help in your Andhra Pradesh Board Class 10th Maths Chapter 3 Polynomials Ex 3.2 exam preparation. The AP Board STD 10th Maths Chapter 3 Polynomials Ex 3.2 Textbooks study material, used with the English medium textbooks, can help you complete the entire Class 10th Maths Chapter 3 Polynomials Ex 3.2 Books State Board syllabus with maximum efficiency.

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