AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 3 Polynomials Ex 3.1 Textbook Questions and Answers.

## AP State Syllabus SSC 10th Class Maths Solutions 3rd Lesson Polynomials Exercise 3.1

### 10th Class Maths 3rd Lesson Polynomials Ex 3.1 Textbook Questions and Answers

Question 1.

a) If p(x) = 5x^{7} – 6x^{5} + 7x – 6, find

i) coefficient of x^{5}

ii) degree of p(x)

iii) constant term.

Answer:

Given p(x) = 5x^{7} – 6x^{5} + 7x – 6

i) coefficient of x^{5} is -6

ii) degree of p(x) is 7

iii) constant term is -6

b) Write three more polynomials and create three questions for each of them.

Answer:

Polynomial – 1: P(x) = x + 5

Questions:

1) What is the order of given polynomial?

2) What are maximum possible zeroes to the above polynomial?

3) What is the zero value of given polynomial?

Polynomial – 2: P(x) = x^{2} – 5x + 6

Questions:

1) What is the sum of zeroes of given polynomial?

2) What is the product of zeroes of it?

3) At how many points, do the polynomial crosses x-axis?

Polynomial – 3: P(x) = ax^{p} + bx^{2} + cx + d

Questions:

1) What will be the value of ‘p’, if the given is cubic polynomial?

2) What is the product of zeroes of it?

3) What can you say about the value of ‘a’ if the given is a cubic polynomial?

Question 3.

If p(t) = t^{3} – 1, find the values of p(1), p(-1), p(0), p(2), p(-2).

Answer:

Given polynomial p(t) = t^{3} – 1

p(1) = 1^{3} – 1 = 1 – 1 = 0

p(-1) = (-1)^{3} – 1 = – 1 – 1 = – 2

p(0) = 0^{3} – 1 = 0 – 1 = – 1

p(2) = 2^{3} – 1 = 8 – 1 = 7

p(-2) = (-2)^{3} – 1 = – 8 – 1 = – 9

Question 4.

Check whether – 2 and 2 are the zeroes of the polynomial x^{4} – 16.

Answer:

Given polynomial is x^{4} – 16

Let p(x) = x^{4} – 16

We have p(-2) = (-2)^{4} – 16

= 16 – 16 = 0 and

p(2) = (2)^{4} – 16

= 16 – 16 = 0

p(-2) = 0 and p(2) = 0.

So these are zeroes of the polynomial.

Question 5.

Check whether 3 and -2 are the zeroes of the polynomial p(x) when p(x) = x^{2} – x – 6.

Answer:

Given polynomial p(x) = x^{2} – x – 6

We have, p(3) = 3^{2} – 3 – 6

= 9 – 3 – 6

= 9 – 9

= 0 and

p(-2) = (-2)^{2} – (-2) – 6

= 4 + 2 – 6

= 6 – 6

= 0

We see that p(3) = 0 and p(-2) = 0

∴ 3 and – 2 are the zeroes of the polynomial p(x).