NCERT MCQ Solutions for Class 9 Maths Chapter 2 Polynomials

A polynomial in one variable contains terms with non-negative whole number exponents of the variable. The expression 3x² + 4x + 5 has whole number exponents and constant coefficients which makes it a polynomial in one variable. Options (a), (b) and (d) have variables with non-whole exponents or radicals.

The degree of a polynomial is the highest exponent of the variable in the expression. In 4x³ + 3x² – x + 7, the highest exponent is 3 from the term 4x³. So, the degree of the polynomial is 3. It is not determined by the number of terms.

The coefficient of x² is the number that is multiplied by x². In the given polynomial 5x² + 3x – 1, the coefficient of x² is clearly 5. This coefficient is always the constant factor appearing with the variable raised to the corresponding power.

A polynomial must not contain variables with negative or fractional exponents. Option (c) contains x raised to –2, which is not allowed in polynomials. All other options have variables with non-negative whole number powers only, so they are valid polynomials.

The polynomial consisting of only the term 0 is known as the zero polynomial. It is unique because all its coefficients are zero. Though it is a polynomial, its degree is not defined.

The degree of a polynomial is the highest power of the variable in the expression. 
Given polynomial: 5t – 7
Identify the Degree
The term 5t has t¹ (power of t is 1)
The term -7 is a constant, meaning its power is 0.
The highest exponent in the expression is 1. 
Final Answer: The degree of 5t – 7 is 1, so the correct option is 1.

A zero ( or root) of a polynomial p(x) is the value of x that makes p(x) = 0.
Given polynomial: p(x) = 9x + 4
Solve for x when p(x) = 0
9x + 4 = 0
Isolate x 
9x = -4
x = -4/9
The zero of p(x) = 9x + 4 is -4/9.

A binomial is a polynomial with exactly two terms. In option (b), 4x² – 7 has two terms: 4x² and –7. The other options either have one term (monomial), three terms (trinomial) or no variable term at all (constant).

A constant polynomial like 6 can be written as 6x⁰. Since the exponent of x is 0, the degree is defined as 0. All non-zero constant polynomials have a degree of zero by definition, even though no variable appears explicitly.

A monomial is a polynomial consisting of only one term. Option (b), 7x, contains a single term and is thus a monomial. The other options have two or more terms, disqualifying them as monomials under the definition provided.

The degree of the zero polynomial is not defined. This is because it has no non-zero term and hence, no term with the highest power.

A polynomial of degree n can have at most n zeroes. This is based on the fundamental understanding that each zero corresponds to a root of the polynomial equation. The degree tells us the maximum number of roots possible.

The coefficient of a term in a polynomial is the numerical factor multiplied by the variable.
Given Polynomial: 2 – x² + x³
Identify the x² Term
The polynomial is written as: 2 + (-1)x² + x³
The term containing x²  is – x² .
The coefficient of x² is -1.
The correct option is -1.

The number of zeroes ( or roots) of a polynomial is determined by its degree. The degree of a polynomial is the highest exponent of the variable.
Given Polynomial: x³ + x – 3 – 3x² 
Arrange in Standard Form
Rearrange the terms in descending order of powers of x:
 x³ – 3x² + x – 3

Identify the Degree
The highest power of x is 3.
A polynomial of degree n can have at most n zeroes.
Conclusion: Since the given polynomial is of degree 3, it has three zeroes ( real or complex). Thus the correct answer is 3.

A linear polynomial has a degree of 1, which means the highest power of the variable is 1. Option (b) is of the form ax + b where a ≠ 0 and hence is a linear polynomial. The rest have degrees greater than one.

Substitute x = 2 into the polynomial:
p(2) = (2)² – 2(2)
= 4 – 4
= 0.
Since p(2) equals 0, it shows that 2 is a zero of the polynomial.
This substitution method is a common way to test for roots.

A polynomial cannot have variables with negative or fractional exponents. The expression x⁻² + 3 contains x raised to –2, which disqualifies it as a polynomial.

Every linear polynomial has exactly one zero. Option (b), 2x + 1, is a linear polynomial with only one solution when equated to zero. 

Substituting x = –1, we get:
(–1)² + 2(–1) + 1
= 1 – 2 + 1
= 0.
Since p(–1) = 0, –1 is a zero of the polynomial.
This matches the algebraic principle of evaluating expressions at specific values.

The degree of a polynomial is the highest power of the variable in the given expression. 
Given Expression: 3 
Since 3 is a constant, it can be written as:
3 = 3x⁰
where x⁰ = 1.
Identify the Degree 
A constant term always has a degree of 0, because it does not contain a variable. 
The degree of 3 is 0, so the correct option is zero.

We need to find the value of: 95 × 96
Use the Difference of Squares Formula
We express the number in a way that simplifies multiplication:
95 × 96 = (95 × (95 +1))
We use this identify: 
(a – 1) (a) = a² – a
Here, a = 96: 95 × 96² – 96
Compute 96² 
96² = 9216
Subtract 96 
9216 – 96 = 9120
The correct option is 9120.

To factor a quadratic polynomial, find two numbers whose product is the constant term and whose sum is the middle coefficient. Here, –2 and –3 multiply to 6 and add to –5. Hence, (x – 2)(x – 3) is the correct factorisation.

The identity (x – y)² = x² – 2xy + y² is explicitly listed in the textbook. It shows how to square a binomial difference. This formula is often used in expansion and simplification of algebraic expressions.

The identity (a + b)³ expands to a³ + 3a²b + 3ab² + b³. This is one of the key algebraic identities discussed under cubes of binomials. It is used in both expansion and factorisation techniques.

The polynomial x³ – 3x + 2 has three separate terms: x³, –3x and 2. Each term is separated by a plus or minus sign. The total count of such individual parts gives the number of terms.

Degree of a polynomial is determined by the highest exponent of the variable. In option (b), the term 2x⁴ gives it a degree of 4, fulfilling the condition for a polynomial of degree 4.

A zero (or root) of a polynomial p(x) is the value of x that makes p(x) = 0.
Given Polynomial: p(x) = 2x – 7
Solve for x when p(x) = 0 
2x – 7 = 0
Isolate x 
2x = 7
x = 7/2

We need to find the value of: 104 x 96
Use the Difference of Squares Formula
We rewrite the numbers as: 104 x 96 = (100 + 4) (100 – 4)
Using the identify: (a + b) (a – b) = a² – b²
where a = 100 and b = 4:
104 × 96 = 100² – 4²
Compute the Values
100² = 10000
4² = 16
10000 – 16 = 9984
The correct option is 9984.

A quadratic polynomial is a polynomial of degree 2. In option (b), the term with the highest power is x², hence the degree is 2. Other options are either linear or cubic and do not qualify.

The identity used to simplify (x + y)(x – y) is x² – y². This identity is derived from the difference of squares and is listed directly in the chapter for quick algebraic simplification.

A trinomial is a polynomial with exactly three terms. Option (b) contains three terms: 2x³, 3x² and 4x. Other options have fewer terms and don’t qualify.

Using the identity (a + b)² = a² + 2ab + b², substitute a = 2x and b = 3y.
We get: (2x)² + 2(2x)(3y) + (3y)² = 4x² + 12xy + 9y².
This matches option (c), which is the correct expansion.

The product (x – 1)(x + 1) is a standard identity representing the difference of squares. It equals x² – 1. This result is frequently used in simplification and factorisation of polynomials.

A zero (or root) of a polynomial p(x) is the value of x that makes p(x) = 0.
Given Polynomial: p(x) = 1x + m
Solve for x when p(x) = 0
x + m = 0
Isolate x
x = -m
Since there is no l in the given polynomial, it seems like the problem might have a typo. However, if the correct polynomial was: 
p(x) = lx + m
Then solving lx + m = 0 would give: lx = -m
x = -m/l

A polynomial consists of terms where the variable has only non-negative integer exponents.
Analyzing Each Option:
(a). 2x – The exponent of x is 1, which is a non-negative integer. This is a valid term of a polynomial.
(b). 3/x – This can be rewritten as 3x ⁻¹. Since the exponent is negative, this is not a polynomial term.
(c). x√x – We rewrite √x as x¹/², so:
x√x = x.x¹/² = x ³/² Since the exponent 3/2 is not an integer, this is not a polynomial terms.
(d). √x – since √x = x ¹/², and the exponent 1/2 is not a integer, this is not a polynomial term. 
The correct answer is 2x.