MathsCBSE Class 10 · Polynomials

Polynomials and Their Degree for Class 10

Updated 13 October 2026 · Fact-checked

A polynomial in x is a sum of terms where each power of x is a whole number and each coefficient is real. Its degree is the highest power with a non-zero coefficient. Board questions ask you to identify polynomials, state degree and type, find values like p(2), and find an unknown k that makes a polynomial linear or quadratic.

Look at the power on each term: the highest one is the degree, and it tells you the type, here 2 so quadratic.

Illustration drawn for QuestionBuddy.

Understand Polynomials and Their Degree

A **polynomial in $x$ ** is an expression made by adding terms of the form $a x^{n}$ , where $a$ is a real number and $n$ is a whole number ( $0, 1, 2, \dots$ ). So $3 x^{2} - 5 x + 7$ is a polynomial, and so is the constant $9$ (think of it as $9 x^{0}$ ).

The test is the powers of the variable. $x + 2$ is not a polynomial, because $x = x^{\frac{1}{2}}$ . $x + \frac{1}{x}$ is not one either, because $\frac{1}{x} = x^{- 1}$ . But $2 x^{2} + 1$ is a polynomial: the surd sits in the coefficient, and coefficients may be any real number.

The degree is the highest power of $x$ whose coefficient is not zero. Always simplify first. For example, $(x + 1)^{2} - x^{2} = 2 x + 1$ has degree 1, not 2.

By degree, you name the type:

Degree 1, linear: $a x + b$ , $a \neq = 0$ .
Degree 2, quadratic: $a x^{2} + b x + c$ , $a \neq = 0$ .
Degree 3, cubic: $a x^{3} + b x^{2} + c x + d$ , $a \neq = 0$ .

A non-zero constant has degree 0. The zero polynomial has no defined degree.

Why does the condition $a \neq = 0$ matter? If the leading coefficient is zero, that term vanishes and the polynomial drops to a lower type. This is how board questions with an unknown $k$ work: for $(k - 3) x^{2} + 5 x + 1$ to be quadratic you need $k \neq = 3$ ; for it to be linear you need $k = 3$ .

The value of $p (x)$ at $x = k$ , written $p (k)$ , is found by substitution. This links to the rest of the chapter: a zero is a value where $p (k) = 0$ , and a polynomial of degree $n$ has at most $n$ zeros. So the degree tells you in advance how many zeros to look for and what shape the graph will have.

Polynomials and Their Degree: formulas to remember

General polynomial: $p (x) = a_{n} x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{1} x + a_{0}, a_{n} \neq = 0$; Every power n is a whole number; the degree is n.
Linear polynomial: $a x + b, a \neq = 0$; Degree 1, at most one zero.
Quadratic polynomial: $a x^{2} + b x + c, a \neq = 0$; Degree 2, at most two zeros.
Cubic polynomial: $a x^{3} + b x^{2} + c x + d, a \neq = 0$; Degree 3, at most three zeros.
Value of a polynomial: $p (k) = value of p (x) when x = k$; Substitute carefully with brackets, especially negative k.
Degree and zeros: $de g p = n \Rightarrow at most n zeros$; Links degree to the rest of the chapter.

How to solve Polynomials and Their Degree questions

Use this routine for any question on identifying, classifying or finding an unknown in a polynomial.

1Simplify the expression fully: expand brackets and collect like terms.
2Check every power of the variable: each must be a whole number (no roots of x, no x in a denominator).
3Write the terms in descending powers of x.
4Find the highest power whose coefficient is non-zero: that is the degree.
5If a coefficient contains k, set it equal to 0 (term vanishes) or not equal to 0 (term stays), as the required degree demands.
6Substitute the found k back and check the degree really is what was asked.
7For a value p(k), substitute using brackets and simplify step by step.

Polynomials and Their Degree 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.

Polynomials and their degree

Suppose you cut equal squares of side x cm from each corner of a 20 cm by 12 cm cardboard sheet and fold up the sides to make an open box for a school craft project.

Write the volume V(x) of the box as a polynomial in standard form, state its degree and type, and find the volume when x = 2 cm.

Try it first, then show the solutionSolution

Length of base $= 20 - 2 x$ , breadth $= 12 - 2 x$ , height $= x$ .
$V (x) = x (20 - 2 x) (12 - 2 x)$ .
$(20 - 2 x) (12 - 2 x) = 240 - 64 x + 4 x^{2}$ .
$V (x) = 4 x^{3} - 64 x^{2} + 240 x$ .
Highest power 3 with coefficient $4 \neq = 0$ , so degree 3: cubic.
$V (2) = 4 (8) - 64 (4) + 240 (2) = 32 - 256 + 480 = 256$ .

Answer: $V (x) = 4 x^{3} - 64 x^{2} + 240 x$ , a cubic of degree 3; volume at x = 2 is 256 cubic cm.

Memory hook: Three dimensions of a box multiply three x's: a cubic.

Polynomials and their degree

Imagine two mobile data plans. Plan A charges a fixed Rs 150 a month plus Rs 12 for each extra GB; Plan B charges a fixed Rs 250 plus Rs 8 for each extra GB. You use x extra GB in a month.

Write the bills A(x) and B(x) and the difference D(x) = A(x) - B(x) as polynomials, and state the degree of D(x). Find the zero of D(x) and explain what it means. Which plan is cheaper if you use 30 extra GB, and by how much?

Try it first, then show the solutionSolution

$A (x) = 12 x + 150$ and $B (x) = 8 x + 250$ .
$D (x) = (12 x + 150) - (8 x + 250) = 4 x - 100$ .
Coefficient of $x$ is $4 \neq = 0$ , so $D (x)$ has degree 1: linear.
$4 x - 100 = 0 \Rightarrow x = 25$ .
At 25 extra GB both bills are equal: $A (25) = 300 + 150 = 450$ and $B (25) = 200 + 250 = 450$ .
$D (30) = 120 - 100 = 20 > 0$ , so $A (30)$ is Rs 20 more than $B (30)$ .
Check: $A (30) = 360 + 150 = 510$ , $B (30) = 240 + 250 = 490$ .

Answer: $D (x) = 4 x - 100$ , degree 1; its zero x = 25 GB is where both bills are equal (Rs 450); at 30 extra GB Plan B is cheaper by Rs 20 (Rs 490 against Rs 510).

Memory hook: Two straight-line bills meet at the zero of their difference.

Tricks to remember Polynomials and Their Degree

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

Whole powers only
Polynomial powers are 'whole and on top': no roots of x, no x downstairs.
Use it when: Instantly rejects expressions like root x or 1/x.
Simplify before you judge
Expand first, degree second.
Use it when: Stops you calling (x+1)^2 - x^2 a quadratic.
L-Q-C = 1-2-3
Linear, Quadratic, Cubic go 1, 2, 3, and the degree caps the zeros.
Use it when: Classify type and maximum number of zeros together.

Polynomials and Their Degree: solved examples

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

Example 1: Identify polynomials · 1 mark · MCQBoard level

Which of the following is a polynomial?
(a) $x^{2} + \frac{2}{x}$
(b) $x + 3 x$
(c) $3 x^{2} - \frac{x}{2} + 5$
(d) $x^{\frac{3}{2}} - 1$

Try it first, then show the solutionSolution

(a) has $\frac{2}{x} = 2 x^{- 1}$ : power $- 1$ , not allowed.
(b) has $x = x^{\frac{1}{2}}$ : not allowed.
(d) has power $\frac{3}{2}$ : not allowed.
(c) has powers 2, 1, 0; the surd is only in a coefficient: allowed.

Answer: (c) $3 x^{2} - \frac{x}{2} + 5$

Example 2: Degree after simplifying · 2 marks · Very short answerBoard level

Find the degree of $p (x) = (2 x - 1) (x + 3) - 2 x (x - 1)$ and name its type. Also find $p (- 2)$ .

Try it first, then show the solutionSolution

$(2 x - 1) (x + 3) = 2 x^{2} + 5 x - 3$ . (½)
$2 x (x - 1) = 2 x^{2} - 2 x$ .
$p (x) = 2 x^{2} + 5 x - 3 - 2 x^{2} + 2 x = 7 x - 3$ . (½)
Degree 1, so $p (x)$ is linear. (½)
$p (- 2) = 7 (- 2) - 3 = - 17$ . (½)

Answer: Degree 1 (linear); $p (- 2) = - 17$ .

Example 3: Unknown k to fix the type · 3 marks · Short answerHarder board level

For $p (x) = (k^{2} - 4) x^{3} + (k + 2) x^{2} + 3 x - k$ , find the value(s) of $k$ for which $p (x)$ is (i) a quadratic, (ii) a linear polynomial. In case (i), find $p (1)$ .

Try it first, then show the solutionSolution

(i) Quadratic needs coefficient of $x^{3} = 0$ and of $x^{2} \neq = 0$ . (½)
$k^{2} - 4 = 0 \Rightarrow k = 2$ or $k = - 2$ . (½)
$k + 2 \neq = 0 \Rightarrow k \neq = - 2$ , so $k = 2$ . (½)
Then $p (x) = 4 x^{2} + 3 x - 2$ , so $p (1) = 4 + 3 - 2 = 5$ . (½)
(ii) Linear needs $k^{2} - 4 = 0$ and $k + 2 = 0$ , so $k = - 2$ . (½)
Then $p (x) = 3 x + 2$ , coefficient of $x$ is $3 \neq = 0$ , so linear. (½)

Answer: (i) $k = 2$ , $p (1) = 5$ ; (ii) $k = - 2$ , giving $3 x + 2$ .

Example 4: Two conditions on values · 5 marks · Long answerHarder board level

$p (x) = a x^{2} + b x + 6$ is a polynomial with $p (1) = 3$ and $p (- 1) = 13$ . Find $a$ and $b$ , state the degree of $p (x)$ , and find $p (2)$ . Does $p (x)$ have any real zero? Justify.

Try it first, then show the solutionSolution

$p (1) = a + b + 6 = 3 \Rightarrow a + b = - 3$ . (1)
$p (- 1) = a - b + 6 = 13 \Rightarrow a - b = 7$ . (½)
Adding: $2 a = 4$ , $a = 2$ ; then $b = - 5$ . (1)
So $p (x) = 2 x^{2} - 5 x + 6$ . Check: $p (1) = 2 - 5 + 6 = 3$ and $p (- 1) = 2 + 5 + 6 = 13$ . (½)
Coefficient of $x^{2}$ is $2 \neq = 0$ : degree 2, quadratic. (½)
$p (2) = 8 - 10 + 6 = 4$ . (½)
Degree 2 means at most 2 zeros. Discriminant $D = b^{2} - 4 a c = (- 5)^{2} - 4 (2) (6) = 25 - 48 = - 23$ . (½)
$D < 0$ , so $2 x^{2} - 5 x + 6 = 0$ has no real root: $p (x)$ has no real zero. (½)

Answer: $a = 2$ , $b = - 5$ ; $p (x) = 2 x^{2} - 5 x + 6$ , degree 2; $p (2) = 4$ ; no real zero, since $D = - 23 < 0$ .

Example 5: Find the error: degree of a sum · 4 marks · Short answerHOTS / competency

Arjun says: 'If $p (x)$ has degree 3 and $q (x)$ has degree 3, then $p (x) + q (x)$ always has degree 3.' Test his claim with $p (x) = 2 x^{3} - x + 4$ and $q (x) = - 2 x^{3} + 5 x^{2} + 1$ . Then find $k$ so that $p (x) + k q (x)$ is quadratic, and state that quadratic.

Try it first, then show the solutionSolution

$p (x) + q (x) = 5 x^{2} - x + 5$ : degree 2. (1)
So the claim is false: the $x^{3}$ terms can cancel. (1)
$p (x) + k q (x)$ has $x^{3}$ coefficient $2 - 2 k$ ; set $2 - 2 k = 0$ , so $k = 1$ . (1)
With $k = 1$ , $x^{2}$ coefficient is $5 \neq = 0$ , giving $5 x^{2} - x + 5$ . (1)

Answer: Claim false (sum is $5 x^{2} - x + 5$ , degree 2); $k = 1$ , quadratic $5 x^{2} - x + 5$ .

Example 6: Degree after a cube cancels · 1 mark · MCQBoard level

The degree of the polynomial $(x - 1)^{3} - x^{3} + 3 x^{2}$ is
(a) 3
(b) 2
(c) 1
(d) 0

Try it first, then show the solutionSolution

$(x - 1)^{3} = x^{3} - 3 x^{2} + 3 x - 1$ .
$(x - 1)^{3} - x^{3} + 3 x^{2} = 3 x - 1$ .
Highest power with non-zero coefficient is 1.

Answer: (c) 1

Example 7: Which one is quadratic? · 1 mark · MCQBoard level

Which of the following is a quadratic polynomial when $k = 3$ ?
(a) $(k - 3) x^{2} + k x + 1$
(b) $(k^{2} - 9) x^{3} + (k - 1) x^{2} + 4$
(c) $(k - 3) x^{3} + (k - 3) x^{2} + x$
(d) $k x^{3} - 9 x^{2} + 1$

Try it first, then show the solutionSolution

(a) becomes $3 x + 1$ : linear.
(b) becomes $0 \cdot x^{3} + 2 x^{2} + 4 = 2 x^{2} + 4$ : quadratic.
(c) becomes $x$ : linear.
(d) becomes $3 x^{3} - 9 x^{2} + 1$ : cubic.

Answer: (b) $(k^{2} - 9) x^{3} + (k - 1) x^{2} + 4$ , which becomes $2 x^{2} + 4$

Example 8: A hidden constant polynomial · 1 mark · MCQBoard level

The degree of $p (x) = (x + 2)^{2} - (x^{2} + 4 x)$ is
(a) 2
(b) 1
(c) 0
(d) not defined

Try it first, then show the solutionSolution

$(x + 2)^{2} = x^{2} + 4 x + 4$ .
$p (x) = x^{2} + 4 x + 4 - x^{2} - 4 x = 4$ .
$4$ is a non-zero constant, so its degree is 0 (only the zero polynomial has no degree).

Answer: (c) 0

Example 9: Degree 3 and the number of zeros · 1 mark · Assertion-reasonHOTS / competency

Assertion (A): The polynomial $x^{3} + x$ has degree 3 but only one real zero.
Reason (R): A polynomial of degree 3 always has exactly three real zeros.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Try it first, then show the solutionSolution

$x^{3} + x = x (x^{2} + 1)$ .
$x^{2} + 1 \geq 1 > 0$ for every real $x$ , so the only real zero is $x = 0$ . A is true.
Degree 3 means at most three zeros, not exactly three; this very example shows it. R is false.

Answer: (c) A is true but R is false.

Example 10: Case study: a water tank · 4 marks · Case-basedHOTS / competency

Suppose your housing society builds a cuboid water tank with a square base of side $x$ m and height $(x + 1)$ m. The four side walls are to be painted.
(i) Write the volume $V (x)$ of the tank as a polynomial and state its degree. (1)
(ii) Write the total area $A (x)$ of the four side walls as a polynomial and name its type. (1)
(iii) If the four walls have a total area of 48 square metres, find $x$ and the volume of the tank. (2)

Try it first, then show the solutionSolution

(i) $V (x) = x \cdot x \cdot (x + 1) = x^{3} + x^{2}$ ; highest power 3, so degree 3 (cubic). (1)
(ii) Each wall is $x$ by $(x + 1)$ , so $A (x) = 4 x (x + 1) = 4 x^{2} + 4 x$ ; degree 2, quadratic. (1)
(iii) $4 x^{2} + 4 x = 48 \Rightarrow x^{2} + x - 12 = 0$ . (½)
$(x + 4) (x - 3) = 0$ , so $x = 3$ or $x = - 4$ . (½)
A length cannot be negative, so $x = 3$ m. (½)
$V (3) = 27 + 9 = 36$ , so the volume is 36 cubic metres. Check: $A (3) = 36 + 12 = 48$ . (½)

Answer: (i) $V (x) = x^{3} + x^{2}$ , degree 3; (ii) $A (x) = 4 x^{2} + 4 x$ , quadratic; (iii) $x = 3$ m, volume 36 cubic metres.

Where students lose marks in Polynomials and Their Degree

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

Mistake: Judging degree before simplifying
The highest-looking power may cancel.
Wrong working
$(x + 2)^{2} - x^{2}$ has degree 2.
Right working
$(x + 2)^{2} - x^{2} = 4 x + 4$ , degree 1.
Do this: Always expand and collect like terms first.
Usually costs: The full 1 mark in an MCQ.
Mistake: Forgetting the non-zero condition on the next coefficient
Setting the top coefficient to 0 is only half the job.
Wrong working
$(k^{2} - 9) x^{2} + (k - 3) x + 1$ linear when $k^{2} = 9$ , so $k = \pm 3$ .
Right working
Also need $k - 3 \neq = 0$ , so only $k = - 3$ .
Do this: Write both conditions: top coefficient $= 0$ , next $\neq = 0$ .
Usually costs: ½ to 1 mark in a 2- or 3-mark answer.
Mistake: Rejecting surd coefficients
Confusing $x$ with $2 x$ .
Wrong working
$5 x^{2} + x$ is not a polynomial.
Right working
Powers are 2 and 1; $5$ is just a real coefficient, so it is a polynomial.
Do this: Only the powers of x must be whole numbers.
Usually costs: 1 mark.
Mistake: Sign slips when finding p(negative)
Substituting without brackets.
Wrong working
$p (x) = x^{2} - 3 x$ , $p (- 2) = - 2^{2} - 3 (- 2) = - 4 + 6 = 2$ .
Right working
$p (- 2) = (- 2)^{2} - 3 (- 2) = 4 + 6 = 10$ .
Do this: Always put the substituted value in brackets.
Usually costs: The value mark and any step built on it.

Exam tips

Show the simplified standard form before stating degree; the examiner gives a step mark for it.
In 'find k' questions, write each coefficient condition on its own line, then check your k by substituting back.
Remember degree caps the number of zeros; MCQs on graphs often hinge on this.
Cubic polynomials appear only for degree, values and reading zeros from a graph; do not use cubic zero-coefficient relations.

Practice questions on Polynomials and Their Degree

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

Q1 · 2 marks · Very short answerBoard level
Find the degree and type of $p (x) = (x^{2} + 1) (x - 2) - x^{3}$ , and find $p (3)$ .
Hint
Expand and let the $x^{3}$ terms cancel.
Answer
$p (x) = - 2 x^{2} + x - 2$ , degree 2, quadratic; $p (3) = - 18 + 3 - 2 = - 17$ .
Q2 · 3 marks · Short answerHarder board level
For $p (x) = (m - 5) x^{2} + (m^{2} - 25) x + 7$ , find the value of $m$ for which $p (x)$ is a constant polynomial, and state that constant. Show that there is no value of $m$ for which $p (x)$ is linear.
Hint
Constant: both coefficients zero. Linear: $x^{2}$ coefficient zero, $x$ coefficient non-zero.
Answer
Constant: $m - 5 = 0$ and $m^{2} - 25 = 0$ give $m = 5$ , and then $p (x) = 7$ . Linear needs $m - 5 = 0$ , so $m = 5$ , but then $m^{2} - 25 = 0$ too, so the $x$ term also vanishes. Hence no value of $m$ makes $p (x)$ linear.
Q3 · 3 marks · Short answerHarder board level
$p (x) = x^{3} + a x^{2} + b x - 6$ satisfies $p (1) = 0$ and $p (2) = 0$ . Find $a$ and $b$ , then find $p (- 1)$ .
Hint
Form two linear equations in $a$ and $b$ .
Answer
$1 + a + b - 6 = 0 \Rightarrow a + b = 5$ ; $8 + 4 a + 2 b - 6 = 0 \Rightarrow 2 a + b = - 1$ . So $a = - 6$ , $b = 11$ . $p (- 1) = - 1 - 6 - 11 - 6 = - 24$ .
Q4 · 3 marks · Short answerBoard level
If $p (x) = (k + 1) x^{2} - 4 x + k$ and $p (2) = 6$ , find $k$ . State the degree of $p (x)$ and find $p (- 1)$ .
Hint
Substitute $x = 2$ , solve for $k$ , then check the $x^{2}$ coefficient.
Answer
$4 (k + 1) - 8 + k = 6 \Rightarrow 5 k = 10$ , $k = 2$ . $p (x) = 3 x^{2} - 4 x + 2$ , degree 2; $p (- 1) = 3 + 4 + 2 = 9$ .
Q5 · 4 marks · Case-basedHOTS / competency
A rectangular garden is $x$ m long and $(x - 4)$ m wide. A path 1 m wide runs along the outside on all sides. Write the area of the path alone as a polynomial, state its degree, and find it when $x = 10$ .
Hint
Outer rectangle is $(x + 2)$ by $(x - 2)$ ; subtract the garden.
Answer
$(x + 2) (x - 2) - x (x - 4) = x^{2} - 4 - x^{2} + 4 x = 4 x - 4$ . Degree 1 (linear). At $x = 10$ : 36 square metres.
Q6 · 3 marks · Short answerHarder board level
$p (x) = (a - 2) x^{3} + (a + b) x^{2} + b x + 4$ is a linear polynomial. Find $a$ and $b$ . Then find the zero of $p (x)$ and the value $p (- 3)$ .
Hint
Set the $x^{3}$ and $x^{2}$ coefficients to zero one after the other, then check the $x$ coefficient is non-zero.
Answer
$a - 2 = 0 \Rightarrow a = 2$ ; $a + b = 0 \Rightarrow b = - 2$ . Then $p (x) = - 2 x + 4$ , and the $x$ coefficient $- 2 \neq = 0$ , so it is linear. Zero: $- 2 x + 4 = 0 \Rightarrow x = 2$ . $p (- 3) = 6 + 4 = 10$ .
Q7 · 3 marks · Short answerHOTS / competency
For $q (x) = (x + k)^{3} - x^{3} - 9 x^{2}$ , find $k$ so that $q (x)$ is a linear polynomial. Write $q (x)$ for this $k$ and find its zero.
Hint
Expand $(x + k)^{3}$ ; the $x^{3}$ terms cancel, so look at the $x^{2}$ coefficient.
Answer
$(x + k)^{3} - x^{3} = 3 k x^{2} + 3 k^{2} x + k^{3}$ , so $q (x) = (3 k - 9) x^{2} + 3 k^{2} x + k^{3}$ . Linear needs $3 k - 9 = 0$ , so $k = 3$ , and then the $x$ coefficient is $27 \neq = 0$ . $q (x) = 27 x + 27$ , with zero $x = - 1$ .

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 $7$ ?Show answer
0 (non-zero constant).
2.Is $x^{2} + \frac{1}{x ^{2}}$ a polynomial?Show answer
No, it has power $- 2$ .
3.Standard form of a quadratic?Show answer
$a x^{2} + b x + c$ , $a \neq = 0$ .
4.Maximum zeros of a cubic?Show answer
3.
5.Degree of the zero polynomial?Show answer
Not defined.

Polynomials and Their Degree: frequently asked questions

Can a polynomial have a surd like root 3 in it?

Yes, as long as the surd is in a coefficient. Root 3 times x squared is fine, but root x is not, because then the power of x is a half.

Why is the degree of the zero polynomial not defined?

Every coefficient is zero, so there is no highest power with a non-zero coefficient. Hence no degree can be given.

Do I need to know cubic polynomials for the board exam?

You need their standard form, degree and how many zeros they can have. The relationship between zeros and coefficients of a cubic, and the division algorithm, are not in the 2026-27 syllabus.

How is degree linked to the graph?

Degree 1 gives a straight line and degree 2 gives a parabola. The degree also tells you the most times the graph can meet the x-axis.