Students can practice 10th Class Maths Solutions Telangana Chapter 3 Polynomials Ex 3.1 to get the best methods of solving problems.

## TS 10th Class Maths Solutions Chapter 3 Polynomials Exercise 3.1

Question 1.

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

(i) Coefficient of x^{5}

(ii) degree of p(x)

(iii) constant term.

a) If P(x) = 5x^{7} – 6x^{5} + 7x – 6

Solution:

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

ii) degree of p(x) = highest degree of x = 7

iii) constant term is -6

Question 2.

State which of the following statements are true and which are false ? Give reasons for your choice.

i) The degree of the polynomial

\(\sqrt{2}\)x^{2} – 3x + 1 is \(\sqrt{2}\).

Solution:

The given statement is false because \(\sqrt{2}\) is the coefficient of x^{2} but not its degree. The degree of the polynomial is 2.

ii) The coefficient of x^{2} in the polynomial p(x) = 3x^{3} – 4x^{2} + 5x + 7 is 2.

Solution:

The given statement is false because the coeffi-cient of x^{2} in the polynomial is -4 but not 2.

iii) The degree of a constant term is zero.

Solution:

The given statement is true. Because 3x^{0} = 3 the degree is 0

iv) \(\frac{1}{x^2-5 x+6}\) is a quadratic polynomial.

Solution:

The given statement is false because the variable ‘x’ appears in the denominator.

v) The degree of a polynomial is one more than the number of terms ¡n it.

Solution:

The given statement is false. There is a no relationship between the degree of the polynomial and the number of terms in it.

Question 3.

If p(t) = t^{3} – 1, find the values of p(1), p(-1), p(0), p(2). p(-2) (A.P.Mar. ’15)

Solution:

Given that p(t) = t^{3} – 1

∴ p(1) = (1)^{3} – 1 = 1 – 1 = 0

p(-1) = (-1) – 1 = -1 – 1 = -2

p(0) = (0) – 1 = 0 – 1 = -1

p(2) = (2) – 1 = 8 – 1 = 7

= (-2) – 1 = -8 – 1 = -9

Question 4.

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

Solution:

p(x) = x^{4} – 16

p(-2) = (-2)^{4} – 16 = 16 – 16 = 0

p(2) = (2^{4}) – 16 = 16 – 16 = 0

Yes, -2 and 2 are zeroes of the polynomial x^{4} – 16.

Question 5.

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

Solution:

p(x) = x^{2} – x – 6

= (3)^{2} – 3 – 6 = 9 – 3 – 6 = 9 – 9 = 0

p(-2) = (-2)^{2} – (-2) – 6 = 4 + 2 – 6 = 6 – 6 = 0

Yes, 3 and -2 are zeroes of the polynomial

p(x) = x^{2} – x – 6