MathsCBSE Class 9 · Introduction to Linear Polynomials

Polynomials in One Variable: Degree and Value, Class 9

Updated 17 October 2026 · Fact-checked

A polynomial in one variable is a sum of terms of the form $ax^n$ with $n = 0, 1, 2, \dots$ and real coefficients; no negative or fractional powers and no variable in a denominator. Its degree is the highest power of $x$ with a non-zero coefficient. The value $p(a)$ is found by substituting $a$ for $x$. A linear polynomial $ax + b$ with $a \neq 0$ has exactly one zero, $-\frac{b}{a}$.

Each box is one term; the highest power of x (here 3) is the degree, and putting x = 2 into the polynomial gives its value 1.

Illustration drawn for QuestionBuddy.

Understand Polynomials in One Variable, Degree and Value

An algebraic expression is built from numbers and variables using the operations of arithmetic. A polynomial in one variable $x$ is an expression of the form $a_{n} x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{1} x + a_{0}$ , where the coefficients $a_{i}$ are real numbers and every power of $x$ is a whole number ( $0, 1, 2, \dots$ ). So $4 x^{3} - 7 x + 2$ and $5 x - 1$ are polynomials, while $2 x^{2} + \frac{3}{x}$ (power $- 1$ ) and $x^{1/2} + 3$ (power $\frac{1}{2}$ ) are not. A number such as $9$ is a constant polynomial; the number $0$ is the zero polynomial.

Terms, coefficients, degree. In $2 x^{3} - 3 x^{2} + x - 5$ the terms are $2 x^{3}, - 3 x^{2}, x, - 5$ , with coefficients $2, - 3, 1$ and constant term $- 5$ . The degree is the highest exponent whose coefficient is not zero: here $3$ . The degree of a non-zero constant is $0$ . Polynomials of degree $1, 2, 3$ are called linear, quadratic and cubic; by number of terms they are called monomial, binomial, trinomial. The degree of the zero polynomial is not defined.

Value at a point. If $p (x)$ is a polynomial, $p (a)$ is the number you get on putting $x = a$ . For $p (x) = 2 x^{3} - 3 x^{2} + x - 5$ : $p (0) = - 5$ , $p (1) = 2 - 3 + 1 - 5 = - 5$ , $p (- 1) = - 2 - 3 - 1 - 5 = - 11$ , $p (2) = 16 - 12 + 2 - 5 = 1$ . Put negative numbers in brackets so that the sign of the power is not lost.

Operations. Add or subtract by combining like terms; multiply by multiplying each term of one by each term of the other and then collecting like terms. The degree behaves predictably: $de g (p \cdot q) = de g p + de g q$ for non-zero polynomials, while $de g (p \pm q) \leq max (de g p, de g q)$ , and may be less when the leading terms cancel (for example, $p (x + 1) - p (x)$ for a quadratic is linear).

Zero of a linear polynomial. For $p (x) = a x + b$ with $a \neq = 0$ , the equation $p (x) = 0$ gives $x = - \frac{b}{a}$ , the only zero. The condition $a \neq = 0$ matters: if a parameter makes the coefficient of $x$ zero, the polynomial stops being linear. In context, the zero of a profit polynomial is the break-even point and the constant term is the value at $x = 0$ (the starting amount or fixed charge).

Typical exam checks. Decide "polynomial or not" by examining each exponent; state the degree only after simplifying (for example $(x^{2} + 1) (x^{3} - 2)$ has degree $5$ ); evaluate carefully with brackets; and in a context question interpret the answer: a negative distance or number of items is rejected.

Polynomials in One Variable, Degree and Value: formulas to remember

General polynomial: $a_{n} x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{1} x + a_{0}, a_{n} \neq = 0$; Exponents are whole numbers; degree $= n$ .
Linear polynomial: $a x + b, a \neq = 0$; Degree 1.
Zero of a linear polynomial: $a x + b = 0 \Rightarrow x = - \frac{b}{a}$; Exactly one zero.
Degree of a product: $de g (pq) = de g p + de g q$; For non-zero polynomials.
Degree of a sum: $de g (p \pm q) \leq max (de g p, de g q)$; Can be lower if leading terms cancel.
Value of a polynomial: $p (a) = a_{n} a^{n} + \dots + a_{1} a + a_{0}$; Substitute with brackets for negative $a$ .

How to solve Polynomials in One Variable, Degree and Value questions

Most questions are: is it a polynomial, what is its degree, what is its value, what is its zero.

1Check each exponent of the variable: a polynomial has only whole-number exponents and no variable in a denominator or under a root.
2Simplify (expand brackets, combine like terms) before reading off the degree.
3For $p (a)$ , substitute $a$ in brackets, evaluate powers first, then multiply, then add.
4For operations, line up like terms; for a product, multiply term by term and then collect.
5For a zero of $a x + b$ solve $a x + b = 0$ ; for an unknown coefficient, substitute the given zero and solve.
6In word problems, interpret the answer and reject values that make no sense.

Polynomials in One Variable, Degree and Value in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Linear polynomial as a cost

Suppose a ticketing app charges 150 rupees per ticket plus a fixed booking fee of 50 rupees, so the cost of $n$ tickets is $C (n) = 150 n + 50$ .

Find the cost of 4 tickets and the number of tickets that can be bought for 1250 rupees.

Try it first, then show the solutionSolution

$C (4) = 150 (4) + 50 = 600 + 50 = 650$ rupees.
Set $C (n) = 1250$ : $150 n = 1200$ .
$n = 8$ .

Answer: 4 tickets cost 650 rupees; 1250 rupees buys 8 tickets.

Memory hook: The constant term is the cost of zero tickets: the fixed fee.

Tricks to remember Polynomials in One Variable, Degree and Value

Short memory aids. Say each one aloud once, then use it on a question the same day.

Whole-number powers only
Scan every exponent: if it is negative or a fraction, or $x$ sits in a denominator or under a root, it is not a polynomial.
Use it when: Polynomial-or-not questions.
Highest power wins
Degree is the largest exponent after simplifying, with a non-zero coefficient.
Use it when: Degree questions, especially after multiplying out.
Brackets for negatives
Write $p (- 2)$ as substitution in brackets before doing any arithmetic.
Use it when: Sign errors in evaluation.

Polynomials in One Variable, Degree and Value: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Polynomial or not · 3 marks · Short answerBoard level

State which of the following are polynomials in $x$ , and give the degree of each polynomial: (a) $4 x^{3} - 7 x + 2$ (b) $2 x^{2} + \frac{3}{x}$ (c) $5 x - 1$ (d) $x^{1/2} + 3$ (e) $9$ .

Try it first, then show the solutionSolution

(a) All exponents are whole numbers: a polynomial of degree 3. (½)
(b) $\frac{3}{x} = 3 x^{- 1}$ has a negative exponent: not a polynomial. (½)
(c) $5$ is a real coefficient and the exponent of $x$ is 1: a polynomial of degree 1. (1)
(d) The exponent $\frac{1}{2}$ is not a whole number: not a polynomial. (½)
(e) A non-zero constant: a polynomial of degree 0. (½)

Answer: (a) degree 3, (c) degree 1 and (e) degree 0 are polynomials; (b) and (d) are not.

Example 2: Evaluating · 3 marks · Short answerBoard level

For $p (x) = 2 x^{3} - 3 x^{2} + x - 5$ find $p (0)$ , $p (- 1)$ , $p (2)$ and $p (2) - p (- 1)$ .

Try it first, then show the solutionSolution

$p (0) = - 5$ . (½)
$p (- 1) = 2 (- 1)^{3} - 3 (- 1)^{2} + (- 1) - 5 = - 2 - 3 - 1 - 5 = - 11$ . (1)
$p (2) = 2 (8) - 3 (4) + 2 - 5 = 16 - 12 + 2 - 5 = 1$ . (1)
$p (2) - p (- 1) = 1 - (- 11) = 12$ . (½)

Answer: $p (0) = - 5$ , $p (- 1) = - 11$ , $p (2) = 1$ and $p (2) - p (- 1) = 12$ .

Example 3: Sum, difference and product · 5 marks · Long answerHarder board level

Let $p (x) = 3 x^{2} - 2 x + 5$ and $q (x) = x^{2} + 4 x - 1$ . Find $p + q$ , $p - q$ and $p \cdot q$ , state the degrees, and verify the product at $x = 2$ .

Try it first, then show the solutionSolution

$p + q = 4 x^{2} + 2 x + 4$ and $p - q = 2 x^{2} - 6 x + 6$ ; both have degree 2. (1½)
$p \cdot q = 3 x^{2} (x^{2} + 4 x - 1) - 2 x (x^{2} + 4 x - 1) + 5 (x^{2} + 4 x - 1)$ . (1)
$= 3 x^{4} + 12 x^{3} - 3 x^{2} - 2 x^{3} - 8 x^{2} + 2 x + 5 x^{2} + 20 x - 5$ . (1)
$= 3 x^{4} + 10 x^{3} - 6 x^{2} + 22 x - 5$ , of degree $4 = 2 + 2$ . (½)
Check: $p (2) = 13$ , $q (2) = 11$ , so $p (2) q (2) = 143$ ; the product polynomial at 2: $48 + 80 - 24 + 44 - 5 = 143$ . (1)

Answer: $p + q = 4 x^{2} + 2 x + 4$ , $p - q = 2 x^{2} - 6 x + 6$ , $pq = 3 x^{4} + 10 x^{3} - 6 x^{2} + 22 x - 5$ .

Example 4: Unknown in a linear polynomial · 3 marks · Short answerHarder board level

The polynomial $p (x) = (3 k + 2) x - (k + 5)$ has $x = 1$ as a zero. Find $k$ , write $p (x)$ , and say for which value of $k$ it would fail to be linear.

Try it first, then show the solutionSolution

$p (1) = 0$ : $(3 k + 2) - (k + 5) = 0$ , so $2 k - 3 = 0$ and $k = \frac{3}{2}$ . (1)
Then $3 k + 2 = 6.5$ and $k + 5 = 6.5$ , so $p (x) = 6.5 x - 6.5$ , zero $= 1$ as required. (1)
$p$ fails to be linear when the coefficient $3 k + 2 = 0$ , that is $k = - \frac{2}{3}$ . (1)

Answer: $k = \frac{3}{2}$ , $p (x) = 6.5 x - 6.5$ ; not linear if $k = - \frac{2}{3}$ .

Example 5: Taxi fare as a polynomial · 4 marks · Case-basedHOTS / competency

Case: a taxi charges a fixed 40 rupees plus 12 rupees per kilometre, so the fare for $x$ km is $C (x) = 12 x + 40$ . (i) State the degree. (ii) Find the fare for 7.5 km. (iii) A fare of 262 rupees is charged: find the distance. (iv) What does the zero of $C (x)$ mean, and why does it have no use here?

Try it first, then show the solutionSolution

(i) $C (x) = 12 x + 40$ is a linear polynomial, degree 1. (½)
(ii) $C (7.5) = 12 (7.5) + 40 = 90 + 40 = 130$ rupees. (1)
(iii) $12 x + 40 = 262$ , so $12 x = 222$ and $x = 18.5$ km. (1)
(iv) The zero is $x = - \frac{40}{12} = - \frac{10}{3}$ km, a negative distance, so it has no meaning in the situation: fares are defined only for $x \geq 0$ . (1½)

Answer: (i) degree 1; (ii) 130 rupees; (iii) 18.5 km; (iv) the zero is the negative distance $- \frac{10}{3}$ km, which is not meaningful.

Where students lose marks in Polynomials in One Variable, Degree and Value

The slips examiners see most often, and what to write instead.

Mistake: Calling a negative power a polynomial term
Ignoring the exponent of $x$ in a fraction.
Wrong working
$x^{2} + \frac{1}{x}$ is a polynomial of degree 2.
Right working
$\frac{1}{x} = x^{- 1}$ , so it is not a polynomial.
Do this: Rewrite fractions with the variable as powers.
Usually costs: 1 mark.
Mistake: Degree read from the first term
Not simplifying or ordering the terms.
Wrong working
$5 + 3 x - 2 x^{4}$ has degree 0 because it starts with 5.
Right working
The highest power is $x^{4}$ , so the degree is 4.
Do this: Arrange in descending powers.
Usually costs: 1 mark.
Mistake: Sign error when substituting a negative number
Dropping brackets.
Wrong working
$p (- 2) = - 2^{2} + 3 (- 2)$ for $x^{2} + 3 x$ .
Right working
$p (- 2) = (- 2)^{2} + 3 (- 2) = 4 - 6 = - 2$ .
Do this: Always put the substituted number in brackets.
Usually costs: 1 to 2 marks.

Exam tips

Check every exponent before calling something a polynomial.
Expand and collect like terms before stating the degree.
Show the substitution line with brackets; method marks are given for it.
In a context question, interpret the zero and reject meaningless values.

Practice questions on Polynomials in One Variable, Degree and Value

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The degree of the polynomial $(x^{2} + 1) (x^{3} - 2)$ is
(a) 3
(b) 5
(c) 6
(d) 2
Hint
Multiply the leading terms.
Answer
(b) 5. The leading term is $x^{2} \cdot x^{3} = x^{5}$ .
Q2 · 3 marks · Short answerBoard level
If $p (t) = t^{3} - 4 t + 1$ , find $p (0)$ , $p (2)$ and $p (- 3)$ .
Hint
Use brackets when substituting $- 3$ .
Answer
$p (0) = 1$ . $p (2) = 8 - 8 + 1 = 1$ . $p (- 3) = (- 3)^{3} - 4 (- 3) + 1 = - 27 + 12 + 1 = - 14$ .
Q3 · 3 marks · Short answerHarder board level
If $p (x) = x^{2} - 3 x + 2$ , find and simplify $p (x + 1) - p (x)$ . What is its degree?
Hint
Expand $p (x + 1)$ first.
Answer
$p (x + 1) = (x + 1)^{2} - 3 (x + 1) + 2 = x^{2} - x$ . So $p (x + 1) - p (x) = x^{2} - x - (x^{2} - 3 x + 2) = 2 x - 2$ , a polynomial of degree 1 (the $x^{2}$ terms cancel).
Q4 · 5 marks · Long answerHarder board level
A quadratic polynomial $p (x) = a x^{2} + b x + c$ satisfies $p (0) = - 3$ , $p (1) = 0$ and $p (- 1) = - 4$ . Find $a$ , $b$ , $c$ , write $p (x)$ and find its zeros by factorising.
Hint
Substitute $0$ , $1$ , $- 1$ to get three equations; $c$ comes first.
Answer
$c = - 3$ . $p (1) = a + b - 3 = 0$ gives $a + b = 3$ . $p (- 1) = a - b - 3 = - 4$ gives $a - b = - 1$ . Adding: $2 a = 2$ , so $a = 1$ , $b = 2$ . $p (x) = x^{2} + 2 x - 3 = (x + 3) (x - 1)$ , with zeros $- 3$ and $1$ . Check $p (- 1) = 1 - 2 - 3 = - 4$ .
Q5 · 4 marks · Case-basedHOTS / competency
Case: a shopkeeper's profit (in rupees) from selling $x$ notebooks in a week is $P (x) = 8 x - 600$ . (i) Find the break-even number of notebooks. (ii) Find the profit when 120 are sold. (iii) What is $P (60)$ and what does the sign tell you? (iv) For which values of $x$ is there a profit?
Hint
Break-even is the zero of $P$ .
Answer
(i) $8 x - 600 = 0$ gives $x = 75$ . (ii) $P (120) = 960 - 600 = 360$ rupees. (iii) $P (60) = 480 - 600 = - 120$ , a loss of 120 rupees. (iv) $8 x - 600 > 0$ gives $x > 75$ , so a profit needs more than 75 notebooks.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Degree of a non-zero constant?Show answer
0.
2.Degree of $(x^{2} + 1) (x^{3} - 2)$ ?Show answer
5.
3.Zero of $a x + b$ ?Show answer
$- \frac{b}{a}$ .
4.Is $x + 1$ a polynomial?Show answer
No, the exponent $\frac{1}{2}$ is not a whole number.
5.What is $p (a)$ ?Show answer
The value of the polynomial at $x = a$ .

Polynomials in One Variable, Degree and Value: frequently asked questions

Is $\sqrt{2}x^2$ a polynomial term?

Yes. The coefficient \sqrt{2} is a real number and the exponent of x is 2. The root is not on the variable.

What is the degree of the zero polynomial?

It is not defined, since no power of x has a non-zero coefficient.

Can a polynomial have two variables?

Yes, but this page deals only with polynomials in one variable.