TS 10th Class Maths Solutions Chapter 3 Polynomials Ex 3.1

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) = 5x⁷ – 6x⁵ + 7x – 6, find
(i) Coefficient of x⁵
(ii) degree of p(x)
(iii) constant term.
a) If P(x) = 5x⁷ – 6x⁵ + 7x – 6
Solution:
i) coefficient of x⁵ 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
\(\sqrt{2}\)x² – 3x + 1 is \(\sqrt{2}\).
Solution:
The given statement is false because \(\sqrt{2}\) is the coefficient of x² but not its degree. The degree of the polynomial is 2.

ii) The coefficient of x² in the polynomial p(x) = 3x³ – 4x² + 5x + 7 is 2.
Solution:
The given statement is false because the coeffi-cient of x² in the polynomial is -4 but not 2.

iii) The degree of a constant term is zero.
Solution:
The given statement is true. Because 3x⁰ = 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) = t³ – 1, find the values of p(1), p(-1), p(0), p(2). p(-2) (A.P.Mar. ’15)
Solution:
Given that p(t) = t³ – 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 x⁴ – 16.
Solution:
p(x) = x⁴ – 16
p(-2) = (-2)⁴ – 16 = 16 – 16 = 0
p(2) = (2⁴) – 16 = 16 – 16 = 0
Yes, -2 and 2 are zeroes of the polynomial x⁴ – 16.

Question 5.
Check whether 3 and -2 are the zeroes of the polynomial p(x) when p(x) = x² – x – 6
Solution:
p(x) = x² – 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) = x² – x – 6