Telangana SCERT Class 10 Maths Chapter 3 Polynomials Exercise 3.1 Maths Problems and Solution Here in this Post. Telangana SCERT Class 10 Maths Solution Chapter 3 Polynomials Exercise 3.1
Telangana SCERT Solution Class X (10) Maths Chapter 3 Polynomials Exercise 3.1
Exercise – 3.1
(Q1) In p (x) = 5x7 – 6x5 + 7x – 6, what is the
(i) Coefficient of x5
(ii) Degree of p(x)
(iii) Constant term
=> (i) Coefficient of x5 = – 6
(ii) Degree of p (x) = 7
[The highest power of given polynomial is degree of p (x)]
(ii) Constant term = – 6
(Q2) State which of the following statements are true and which are false? Give reasons for choice.
(i) The degree of the polynomial √2 x2 – 3x + 1 is √2
=> False
The degree of polynomial is 2. But here he has given √2. √2 is the coefficient of x2
∴ It is a false statement.
(ii) The coefficient of x2 in the polynomial p (x) = 3x3 – 4x2 + 5x + 7 is 2
=> False
The coefficient of x2 in the given polynomial is – 4. But here he has given 2.
Hence, it is a false statement
(iii) The degree of a constant term is zero.
=> e.g. 2, – 3, 1/2
2xo – 3xo + 1/2 xo
The degree of a constant term is always zero
Hence, it is a true statement.
(iv) 1/x2 – 5x + 6 is a quadratic polynomial
= (x2 – 5x + 6)-1 = x-2 – 5x-1 + 6
Definition of polynomial: An Algebraic expression in which the powers of variables or non-negative integers.
But here the powers of variables are negative integers. So, here x-2, x-1
Hence, it is false statement
(v) The degree of a polynomial is one more than the number of terms in it.
=> False
There is no relation between the degree of polynomial and the number of terms of polynomial.
Degree is the highest power of variable in the given polynomial.
But he was given degree is one more than number of terms.
So, it is false statement.
(Q3) If p (t) = t3 – 1, find the values of p (1), p (-1), p (o), p (2), p (-2)
=> Solution:
Give polynomial is –
P (t) = t3 – 1
If t = 1 => substitute t = 1 in given polynomial.
P (1) = (1)3 – 1
= 1 – 1
P (1) = 0
If t = -1 => Substitute t = -1 in given polynomial.
P (-1) = (-1)3 – 1
= -1 -1
P (-1) = -2
If t = 0 => Substitute t = 0 in given polynomial
P (0) = (0)3 – 1
P (0) = – 1
If t = 2 => substitute t = 2 in given polynomial.
P (2) = (2)3 – 1
= 8 – 1
P (2) = 7
If t = -2 => Substitute t = – 2 in given polynomial
P (-2) = (-2)3 – 1
= -8 -1
P (-2) = -9
(Q4) Check whether – 2 and 2 are the zeroes of the of the polynomial
=> Solution:
Let p (x) = x4 – 16
Substituting x = – 2
P (-2) = (-2)4 – 16
= 16 – 16
P (-2) = 0
‘-2’ is zero of x4 – 16
Substituting x = 2
P (2) = (2)4 – 16
= 16 – 16
P (2) = 0
∴ ‘2’ is zero of x4 – 16
∴ – 2 and 2 are the zero’s of x4 – 16
(Q5) Check whether 3 and – 2 are the zeroes of the polynomial p (x) when p (x) = x2 – x – 6
=> Solution:
Given: p (x) = x2 – x – 6 —— (1)
Substituting x = 3 in ——- (1)
P (3) = (3)3 – (3) – 6
= 9 – 3 – 6
= 6 – 6
P (3) = 0
∴ ‘3’ is a zero of p (x)
Substituting x = – 2 in (1)
P (-2) = (-2)2 – (-2) – 6
= 4 + 2 – 6
= 6 – 6
P (x) = 0
∴ ‘-2’ is zero of p (x)
∴ 3 and – 2 are the zeroes of P (x)
Here is your solution of Telangana SCERT Class 10 Math Chapter 3 Polynomials Exercise 3.1
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