Polynomials Exercise 2.1
Math Solution Access to NCERT Class 9 Chapter 2 –
Polynomials Exercise 2.1
Polynomials in One Variable:
Let us start,
You must recall that a variable is denoted by a symbol that can
take any real value. We use the letters x, y, z, etc. to denote
variables.
Notice that 2x, 3x, – x, – x/2 are algebraic expressions.
All these expressions are of the form (a constant) × x.
Now, Suppose we want to write an
expression which is (a constant) × (a variable) and we do not know what
the constant is. In such cases, we write the constant as a, b, c, etc.
So the expression will be ax.
However, there is a difference between
a constant and a letter denoting a variable. The values of the constants remain
the same. But the values of the constants do not change in a given problem, but
the value of a variable can keep changing in problem.
1. Which of the following expressions are polynomials in one
variable, and which are not? State reasons for your answer.
(i) 4x2–3x+7
Solution:
The equation 4x2–3x+7 can be written as 4x2–3x1+7x0
Dear viewers, here is the writing process is an idea.
Since x is the only variable in the given equation and the
powers of x (2, 1 and 0) are whole numbers.
As per the Polynomial condition, say that the expression 4x2–3x+7
is a polynomial in one variable.
(ii) y2+√2
Solution:
The equation y2+√2 can be written as y2+√2y
x 0
As we can see that y is the only variable in the given
equation and the powers of y (2 and 0) are whole numbers, we can say that the
expression y2+√2 is a polynomial in one variable.
(iii) 3√t+t√2
Solution:
The equation 3√t+t√2 can be written as 3t½
+√2t
Here is the only variable in the given equation, the power of t
(i.e.,1/2) is not a whole number. Hence, we can say that the expression 3√t+t√2
is not a polynomial in one variable.
(iv) y+2/y
Solution:
The equation y+2/y can be written as y
+ y+2y = y+2y-1
Though y is the only variable in the given equation, the power of
y (i.e.,-1) is not a whole number. It is not a polynomial, because one of the
exponents of y is -1,
which is not a whole number
(v) x10+y3+t50
Solution:
Here, in the equation x10+y3+t50
Though powers 10, 3, and 50 are whole numbers, there are 3
variables used in the expression.
x10+y3+t50 Hence, it is not a polynomial in one
variable.
2. Write the coefficients of x2 in each of the
following.
(i) 2+x2+x
Solution:
The equation 2+x2+x can be written as 2+(1)x2+x
We know that the coefficient is the number which multiplies the
variable.
Here, the number that multiplies the variable x2 is 1.
The coefficient of x2 in
2+x2+x is 1.
(ii) 2–x2+x3
Solution:
The equation 2–x2+x3 can be written as 2+(–1)x2+x3
We know that the coefficient is the number (along with its sign,
i.e., – or +) which multiplies the variable.
Here, the number that multiplies the variable x2 is -1
The exact coefficient of x2 in 2–x2+x3
is -1.
(iii) (π/2)x2+x
Solution:
The equation (π/2)x2 +x can be written as (π/2)x2
+ x
We know that the coefficient is the number (along with its sign,
i.e., – or +) which multiplies the variable.
Here, the number that multiplies the variable x2 is
π/2.
The coefficients of x2 in (π/2)x2 +x is π/2.
(iii)√2x-1
Solution:
The equation √2x-1 can be written as 0x2+√2x-1 [Since 0 x 2 is 0]
We know that the coefficient is the number (along with its sign,
i.e., – or +) which multiplies the variable.
Here, the number that multiplies the variable x2 is 0.
The coefficient of x2 in √2x-1 is 0.
3. Give one example each of a binomial of
degree 35 and of a
monomial of degree 100.
Solution:
Binomial of degree 35:
A polynomial having two terms and the highest degree of 35 is called a binomial
of degree 35.
E.g., 3x35+5
Monomial of degree 100:
A polynomial having one term and the highest degree of 100 is called a monomial
of degree 100.
E.g., 4x100
4. Write the degree of each of the following polynomials.
(i) 5x3+4x2+7x
Solution:
The highest power of the variable in a polynomial is the degree of
the polynomial.
Here, 5x3+4x2+7x = 5x3+4x2+7x1
The powers of the variable x are 3, 2, 1
The degree of 5x3+4x2+7x is 3, as 3 is the highest power of x in
the equation.
(ii) 4–y2
Solution:
The highest power of the variable in a polynomial is the degree
of the polynomial.
Here, in 4–y2,
The power of the variable y is 2.
The degree of 4–y2 is 2, as 2 is the highest power of y
in the equation.
(iii) 5t–√7
Solution:
The highest power of the variable in a polynomial is the degree of
the polynomial.
Here, in 5t–√7,
The power of the variable t is 1.
(Because there is not clear given any power, so the
default power will be 1 of t)
The degree of 5t–√7 is 1, as 1 is the highest power of y in the
equation.
(iv) 3
Solution:
The highest power of the variable in a polynomial is the degree of
the polynomial.
Here, 3 = 3×1 = 3× x0
The power of the variable x here is 0.
The degree of 3 is 0.
5. Classify the following as linear, quadratic and cubic
polynomials.
Solution:
We know that,
Linear polynomial: A polynomial of
degree one is called a linear polynomial.
Quadratic polynomial: A
polynomial of degree two is called a quadratic polynomial.
Cubic polynomial: A polynomial of
degree three is called a cubic polynomial.
(i) x2+x
Solution:
The highest power of x2+x
is 2
The degree is 2.
Hence, x2+x is a quadratic polynomial.
(ii) x–x3
Solution:
The highest power of x–x3
is 3.
The degree is 3.
Hence, x–x3 is a cubic polynomial.
(iii) y+y2+4
Solution:
The highest power of y+y2+4
is 2.
the degree of Y is 2
Hence, y+y2+4is a quadratic polynomial
(iv) 1+x
Solution:
The highest power of 1+x is 1
The degree is 1.
Hence, 1+x is a linear polynomial.
(v) 3t
Solution:
The highest power of 3t or degree is 1.
The degree is 1.
Hence, 3t is a linear polynomial.
(vi) r2
Solution:
The highest power of r2 is
2.
The degree is 2.
Hence, r2 is a quadratic polynomial.
(vii) 7x3
Solution:
The highest power of 7x3 is 3.
The degree is 3.
Hence, 7x3 is a cubic polynomial.
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