AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 3 Polynomials Ex 3.2 Textbook Questions and Answers.

## AP State Syllabus SSC 10th Class Maths Solutions 3rd Lesson Polynomials Exercise 3.2

### 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).

Answer:

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) = x^{2} + 5x + 6

(iii) p(x) = (x + 2) (x + 3)

(iv) p(x) = x^{4} – 16

Answer:

i) Given p(x) = 3x

Let p(x) = 0

So, 3x = 0

x = \(\frac{0}{3}\) = 0,

Zeroes of p(x) = 3x is zero.

∴ No. of zeroes is one.

ii) Given p(x) = x^{2} + 5x + 6 is a quadratic polynomial.

It has atmost two zeroes.

To find zeroes, let p(x) = 0

⇒ x^{2} + 5x + 6 = 0

⇒ x^{2} + 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) = x^{4} – 16 is a biquadratic polynomial. It has atmost two zeroes.

Let p(x) = 0

⇒ x^{4} – 16 = 0

⇒ (x^{2})^{2} – 4^{2} = 0

⇒ (x^{2} – 4) (x^{2} + 4) = 0

⇒ (x + 2) (x – 2) (x^{2} + 4) = 0

⇒ (x + 2) = 0 or (x – 2) = 0 or (x^{2} + 4) = 0

⇒ x = -2 (or) x = 2 (or) x^{2} = -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) = x^{2} – x – 12

ii) p(x) = x^{2} – 6x + 9

iii) p(x) = x^{2} – 4x + 5

iv) p(x) = x^{2} + 3x – 4

v) p(x) = x^{2} – 1

Answer:

i) Given polynomial p(x) = x^{2} – 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) = x^{2} – x – 12 = 0

⇒ x^{2} – 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) = x^{2} – 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) = x^{2} – 6x + 9

⇒ x^{2} – 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) = x^{2} – 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) = x^{2} – 4x + 5 not possible to split in factors.

iv) Given polynomial p(x) = x^{2} + 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) = x^{2} + 3x – 4 = 0

⇒ x^{2} + 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) = x^{2} – 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) = x^{2} – 1 = 0

⇒ p(x) = (x + 1) (x – 1) = 0 [∵ a^{2} – b^{2} = (a + b) (a – b)]

⇒ x + 1 = 0 and x – 1 = 0

x = -1 and x = 1

Question 4.

Why are \(\frac{1}{4}\) and -1 zeroes of the polynomial p(x) = 4x^{2} + 3x – 1 ?

Answer:

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

Given zeroes are \(\frac{1}{4}\) and -1

Let x = \(\frac{1}{4}\)

Let x = -1

⇒ p(-1) = 4(-1)^{2} + 3(-1)-1 = 4 – 3 – 1 = 4 – 4 = 0

∴ P(\(\frac{1}{4}\)) = 0 and p(-1) = 0

So these values are zeroes of the polynomial p(x).