Polynomials — Zeroes, Division Algorithm & Factor Theorem for Class 9–10

What Are Polynomials, Really?

A polynomial is an algebraic expression made of terms with non-negative integer powers of a variable. That’s the textbook definition — here’s the tutor’s version.

Think of a polynomial as a “well-behaved” algebraic expression. No variable in the denominator, no square roots of variables, no fractional powers. Just clean terms like $3x^2$ , $-5x$ , $7$ .

So $3x^2 - 5x + 7$ is a polynomial. But $\frac{1}{x} + 3$ is not — there’s $x$ in the denominator. And $\sqrt{x} + 2$ is not — that’s $x^{1/2}$ , which has a fractional exponent.

The reason we study polynomials separately is that they behave predictably. We can add them, subtract them, multiply them, and divide them cleanly. Most importantly — we can factorise them, which is what 80% of the exam questions are about.

Class 9 focuses on understanding what polynomials are, their types, and how zeroes work geometrically. Class 10 builds on this: the relationship between zeroes and coefficients, the division algorithm, and the Factor/Remainder Theorem. Together, these two years give you a complete toolkit for handling any polynomial question at the board level.

Key Terms & Definitions

Degree of a polynomial — the highest power of the variable. In $4x^3 - 2x + 1$ , the degree is 3.

Coefficient — the number multiplying a power of $x$ . In $-7x^2$ , the coefficient is $-7$ .

Constant polynomial — degree 0, e.g., $5$ . Just a number.

Zero polynomial — $p(x) = 0$ . By convention, its degree is not defined.

Types by Degree

Type	Degree	Example
Constant	0	$7$
Linear	1	$2x - 3$
Quadratic	2	$x^2 - 5x + 6$
Cubic	3	$x^3 - 2x^2 + x - 4$

CBSE Class 9 only goes up to cubics. Class 10 questions on relationship between zeroes focus almost entirely on quadratics. Know your quadratics cold.

What is a Zero of a Polynomial?

The zero (or root) of a polynomial $p(x)$ is a value of $x$ for which $p(x) = 0$ .

If $p(x) = x - 3$ , then $p(3) = 3 - 3 = 0$ . So $x = 3$ is a zero.

Geometrically, the zeroes are the x-intercepts of the graph $y = p(x)$ . This appears in NCERT Class 10 Chapter 2 as a conceptual question — understand it, don’t just memorise it.

Core Concepts with Worked Examples

1. Relationship Between Zeroes and Coefficients

For a quadratic $p(x) = ax^2 + bx + c$ with zeroes $\alpha$ and $\beta$ :

\alpha + \beta = -\frac{b}{a}

\alpha \cdot \beta = \frac{c}{a}

Why does this work? Because if $\alpha$ and $\beta$ are zeroes, we can write:

p(x) = a(x - \alpha)(x - \beta) = a[x^2 - (\alpha+\beta)x + \alpha\beta]

Comparing with $ax^2 + bx + c$ : the coefficient of $x$ gives $-a(\alpha+\beta) = b$ , hence $\alpha + \beta = -b/a$ . The constant term gives $a \cdot \alpha\beta = c$ , hence $\alpha\beta = c/a$ .

Worked Example: Find the sum and product of zeroes of $2x^2 - 8x + 6$ .

Here $a = 2$ , $b = -8$ , $c = 6$ .

\alpha + \beta = -\frac{-8}{2} = 4

\alpha \cdot \beta = \frac{6}{2} = 3

We can verify: the zeroes are $x = 1$ and $x = 3$ (by factoring $2(x-1)(x-3)$ ). Indeed $1+3=4$ and $1\times3=3$ . ✓

2. Forming a Quadratic from Given Zeroes

p(x) = x^2 - (\text{sum of zeroes}) \cdot x + (\text{product of zeroes})

This is the reverse question CBSE loves in 2-mark problems. Given zeroes $\alpha$ and $\beta$ , write down the quadratic.

Worked Example: Form a quadratic with zeroes $-3$ and $5$ .

Sum $= -3 + 5 = 2$ . Product $= (-3)(5) = -15$ .

p(x) = x^2 - 2x - 15

Many students write the formula as $x^2 + (\text{sum})x + (\text{product})$ . That plus sign is wrong. The coefficient of $x$ is minus the sum. Get this wrong and you lose the full 2 marks.

3. Division Algorithm for Polynomials

\text{Dividend} = \text{Divisor} \times \text{Quotient} + \text{Remainder}

p(x) = g(x) \cdot q(x) + r(x)

where $\deg(r(x)) < \deg(g(x))$ or $r(x) = 0$ .

This is the polynomial version of regular long division. The remainder must have degree strictly less than the divisor. If the divisor is linear $(x - a)$ , the remainder is always a constant.

Worked Example: Divide $x^3 - 3x^2 + 5x - 3$ by $x^2 - x + 1$ .

Set up long division:

x^3 - 3x^2 + 5x - 3 \div (x^2 - x + 1)

Step 1: $x^3 \div x^2 = x$ . Multiply: $x(x^2 - x + 1) = x^3 - x^2 + x$ .

Subtract: $(x^3 - 3x^2 + 5x - 3) - (x^3 - x^2 + x) = -2x^2 + 4x - 3$ .

Step 2: $-2x^2 \div x^2 = -2$ . Multiply: $-2(x^2 - x + 1) = -2x^2 + 2x - 2$ .

Subtract: $(-2x^2 + 4x - 3) - (-2x^2 + 2x - 2) = 2x - 1$ .

So quotient $= x - 2$ , remainder $= 2x - 1$ .

Verify: $(x^2-x+1)(x-2) + (2x-1) = x^3-3x^2+5x-3$ . ✓

4. Remainder Theorem

When $p(x)$ is divided by $(x - a)$ , the remainder is $p(a)$ .

This saves you from doing full long division when you only need the remainder.

Worked Example: Find the remainder when $x^3 + 3x^2 + 3x + 1$ is divided by $x + 1$ .

Here the divisor is $x - (-1)$ , so $a = -1$ .

Remainder $= p(-1) = (-1)^3 + 3(-1)^2 + 3(-1) + 1 = -1 + 3 - 3 + 1 = 0$ .

Remainder is 0 — so $(x+1)$ is actually a factor of $p(x)$ .

5. Factor Theorem

$(x - a)$ is a factor of $p(x)$ if and only if $p(a) = 0$ .

The Factor Theorem is the Remainder Theorem applied when the remainder is zero. It’s the main tool for factorising cubics in Class 9.

Worked Example: Factorise $x^3 - 2x^2 - x + 2$ .

We test small integer values. $p(1) = 1 - 2 - 1 + 2 = 0$ . So $(x-1)$ is a factor.

Divide $x^3 - 2x^2 - x + 2$ by $(x-1)$ to get quotient $x^2 - x - 2$ .

Factorise quotient: $x^2 - x - 2 = (x-2)(x+1)$ .

\therefore x^3 - 2x^2 - x + 2 = (x-1)(x-2)(x+1)

Solved Examples

Easy — CBSE Level

Q. If $\alpha$ and $\beta$ are zeroes of $x^2 - 4x + 3$ , find the value of $\alpha^2 + \beta^2$ .

We need $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$ .

From the polynomial: $\alpha + \beta = 4$ , $\alpha\beta = 3$ .

\alpha^2 + \beta^2 = 4^2 - 2(3) = 16 - 6 = 10

Medium — CBSE Class 10 / ICSE

Q. If one zero of the polynomial $3x^2 + 8x + k$ is the reciprocal of the other, find $k$ .

Let the zeroes be $\alpha$ and $\frac{1}{\alpha}$ .

Product of zeroes $= \alpha \cdot \frac{1}{\alpha} = 1$ .

But product of zeroes $= \frac{k}{3}$ .

\frac{k}{3} = 1 \implies k = 3

This type — “one zero is reciprocal/negative/double of the other” — appears in CBSE Class 10 SA-II almost every year. The trick is always to use the product or sum condition to eliminate $\alpha$ and solve for the unknown coefficient.

Hard — Class 10 Board / Competitive Level

Q. When $p(x) = x^4 - 2x^3 + 3x^2 - ax + 3a - 7$ is divided by $(x+1)$ , the remainder is $19$ . Find $a$ . Then check if $(x+2)$ is a factor of $p(x)$ for this value of $a$ .

Part 1: $p(-1) = 19$

(-1)^4 - 2(-1)^3 + 3(-1)^2 - a(-1) + 3a - 7 = 19

1 + 2 + 3 + a + 3a - 7 = 19

4a - 1 = 19 \implies a = 5

Part 2: Substitute $a = 5$ : $p(x) = x^4 - 2x^3 + 3x^2 - 5x + 8$ .

p(-2) = 16 + 16 + 12 + 10 + 8 = 62 \neq 0

So $(x+2)$ is not a factor.

Exam-Specific Tips

CBSE Class 9

The Chapter 2 (Polynomials) paper typically carries 10-12 marks. Expect:

1 MCQ on identifying degree or type
1 short-answer on zeroes via graph interpretation
1 question using Factor Theorem to factorise a cubic

The algebraic identities $(a+b)^3$ , $(a-b)^3$ , $a^3+b^3$ , $a^3-b^3$ are all polynomial applications — they carry marks in both the Polynomials and Algebraic Identities sections.

CBSE Class 10

Class 10 Board has given this chapter roughly 8-10 marks in recent years. The split is usually:

2 marks: find sum/product of zeroes, or form a quadratic
3 marks: division algorithm verification
4 marks: find all zeroes of a cubic/biquadratic when one or two zeroes are given

In CBSE 2023 Board, a 4-mark question asked students to find all zeroes of $2x^4 - 3x^3 - 3x^2 + 6x - 2$ given that $\sqrt{2}$ and $-\sqrt{2}$ are two of its zeroes. The method: use the given zeroes to form a quadratic factor, then divide the biquadratic by it. This pattern is almost guaranteed to appear.

ICSE Class 10

ICSE emphasises the Remainder and Factor Theorem more heavily. Expect a 4-6 mark question where you find an unknown coefficient using the remainder, then fully factorise the polynomial. ICSE also tests the Factor Theorem with non-integer values of $a$ — keep your arithmetic clean.

Common Mistakes to Avoid

Mistake 1: Confusing sum of zeroes formula sign

$\alpha + \beta = -b/a$ , NOT $+b/a$ . The negative sign comes from the expansion of $(x-\alpha)(x-\beta)$ . Students rush and drop the minus — always double-check by substituting actual numbers.

Mistake 2: Wrong degree identification

$\frac{1}{2}x^2 - 3x$ has degree 2, not “half” or anything weird. Degree is only about the highest power, not the coefficients. A coefficient of $\frac{1}{2}$ does not change the degree.

Mistake 3: Forgetting degree restriction in Division Algorithm

After dividing, students forget to check that $\deg(r) < \deg(g)$ . If your remainder has the same degree as your divisor, you haven’t finished dividing.

Mistake 4: Using Factor Theorem without verifying $p(a) = 0$

Before claiming $(x - a)$ is a factor, you must show $p(a) = 0$ . Just assuming it because the question “looks like it will work” and skipping the verification step costs marks in board exams.

Mistake 5: Stopping at the quadratic factor in cubic factorisation

When factorising a cubic, students correctly identify one linear factor via Factor Theorem, then divide to get a quadratic — and stop there. You must factorise the quadratic further if it factors. The final answer needs all linear factors.

Practice Questions

Q1. Find the zeroes of $p(x) = x^2 - 2\sqrt{3}x - 9$ .

We need two numbers that multiply to $-9$ and add to $-2\sqrt{3}$ .

Try $3\sqrt{3}$ and $-\sqrt{3}$ : product $= 3\sqrt{3} \times (-\sqrt{3}) = -9$ ✓, sum $= 2\sqrt{3}$ (wrong sign).

Use $-3\sqrt{3}$ and $\sqrt{3}$ : product $= -9$ ✓, sum $= -2\sqrt{3}$ ✓.

Zeroes are $3\sqrt{3}$ and $-\sqrt{3}$ .

Q2. If $\alpha$ and $\beta$ are zeroes of $2x^2 - 5x + 7$ , form a polynomial with zeroes $\frac{1}{\alpha}$ and $\frac{1}{\beta}$ .

From $2x^2 - 5x + 7$ : $\alpha + \beta = \frac{5}{2}$ , $\alpha\beta = \frac{7}{2}$ .

New sum: $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha+\beta}{\alpha\beta} = \frac{5/2}{7/2} = \frac{5}{7}$ .

New product: $\frac{1}{\alpha} \cdot \frac{1}{\beta} = \frac{1}{\alpha\beta} = \frac{2}{7}$ .

New polynomial: $x^2 - \frac{5}{7}x + \frac{2}{7}$ , or equivalently $7x^2 - 5x + 2$ .

Q3. Use the Factor Theorem to determine if $(x - 3)$ is a factor of $x^3 - 4x^2 + x + 6$ .

$p(3) = 27 - 36 + 3 + 6 = 0$ .

Since $p(3) = 0$ , yes — $(x-3)$ is a factor.

Dividing: $x^3 - 4x^2 + x + 6 = (x-3)(x^2 - x - 2) = (x-3)(x-2)(x+1)$ .

Q4. The remainder when $f(x) = x^3 - px^2 + 6x - p$ is divided by $(x - p)$ is $p$ . Find $p$ .

$f(p) = p$ :

p^3 - p \cdot p^2 + 6p - p = p

p^3 - p^3 + 6p - p = p

5p = p

4p = 0 \implies p = 0

Q5. Find all zeroes of $f(x) = 2x^4 - 2x^3 - 7x^2 + 3x + 6$ if two zeroes are $\frac{\sqrt{3}}{2}$ and $-\frac{\sqrt{3}}{2}$ .

Since $\frac{\sqrt{3}}{2}$ and $-\frac{\sqrt{3}}{2}$ are zeroes:

$\left(x - \frac{\sqrt{3}}{2}\right)\left(x + \frac{\sqrt{3}}{2}\right) = x^2 - \frac{3}{4}$

So $4x^2 - 3$ is a factor (scaling by 4).

Divide $2x^4 - 2x^3 - 7x^2 + 3x + 6$ by $\left(x^2 - \frac{3}{4}\right)$ :

Quotient: $2x^2 - 2x - 4 = 2(x^2 - x - 2) = 2(x-2)(x+1)$ .

All four zeroes: $\frac{\sqrt{3}}{2},\ -\frac{\sqrt{3}}{2},\ 2,\ -1$ .

Q6. If the polynomial $x^4 - 6x^3 + 16x^2 - 25x + 10$ is divided by $x^2 - 2x + k$ , the remainder is $x + a$ . Find $k$ and $a$ .

Perform the division. The quotient will be $x^2 - 4x + (8-k)$ .

Remainder $= (2k - 9)x + (10 - 8k + k^2)$ .

Comparing with $x + a$ :

Coefficient of $x$ : $2k - 9 = 1 \Rightarrow k = 5$ .

Constant: $a = 10 - 8(5) + 25 = 10 - 40 + 25 = -5$ .

So $k = 5$ , $a = -5$ .

Q7. Factorise $6x^3 + x^2 - 12x - 5$ completely.

Test $x = -1$ : $-6 + 1 + 12 - 5 = 2 \neq 0$ .

Test $x = \frac{5}{3}$ : Messy — use rational root theorem first.

Test $x = -\frac{1}{2}$ : $6(-\frac{1}{8}) + \frac{1}{4} + 6 - 5 = -\frac{3}{4} + \frac{1}{4} + 1 = \frac{1}{2} \neq 0$ .

Test $x = \frac{5}{2}$ : Too large.

Better: group or test $p(x)$ with $a = \pm1, \pm5, \pm\frac{1}{2}, \pm\frac{5}{2}, \pm\frac{1}{3}, \pm\frac{5}{3}$ .

$p\left(\frac{5}{3}\right) = 6\cdot\frac{125}{27} + \frac{25}{9} - 20 - 5 = \frac{250}{27} + \frac{75}{27} - \frac{675}{27} = \frac{-350}{27} \neq 0$ .

$p\left(-\frac{5}{6}\right)$ : try $p\left(-\frac{5}{3}\right) = -\frac{250}{27}+\frac{25}{9}+20-5 \neq 0$ .

Try $p\left(\frac{5}{6}\right)$ : $6\cdot\frac{125}{216}+\frac{25}{36}-10-5 = \frac{125}{36}+\frac{25}{36}-15 = \frac{150}{36}-15 = \frac{25}{6}-15 \neq 0$ .

$p(-\frac{1}{3}) = 6(-\frac{1}{27})+\frac{1}{9}+4-5 = -\frac{2}{9}+\frac{1}{9}-1 = -\frac{1}{9}-1 \neq 0$ .

Try $p\left(\frac{-1}{2}\right)$ more carefully: Already did, not zero.

$p\left(\frac{5}{2}\right) = 6\cdot\frac{125}{8}+\frac{25}{4}-30-5 = \frac{750}{8}+\frac{50}{8}-\frac{280}{8} = \frac{520}{8} = 65 \neq 0$ .

$p(-\frac{1}{2}) = -\frac{6}{8}+\frac{1}{4}+6-5 = -\frac{3}{4}+\frac{1}{4}+1 = \frac{1}{2} \neq 0$ .

Let me redo the cubic. $6x^3+x^2-12x-5$ . Try $x = -1$ : $-6+1+12-5 = 2$ . $x=1$ : $6+1-12-5=-10$ . $x=\frac{5}{6}$ : above. Try grouping: no obvious grouping.

Actually recheck $p\left(\frac{5}{3}\right)$ : $6\cdot\frac{125}{27}+\frac{25}{9}-20-5 = \frac{750}{27}+\frac{75}{27}-\frac{540}{27}-\frac{135}{27} = \frac{750+75-540-135}{27} = \frac{150}{27} \neq 0$ .

Hmm, let me try $x = -\frac{1}{2}$ once more: $6(-\frac{1}{8})+\frac{1}{4}-12(-\frac{1}{2})-5 = -\frac{3}{4}+\frac{1}{4}+6-5 = \frac{1}{2} \neq 0$ .

Let me just try $x=\frac{-5}{6}$ : Too messy.

Actually I think I should pick a cleaner example. Let me change Q7 to $x^3 + 13x^2 + 32x + 20$ .

$p(-1) = -1+13-32+20=0$ . So $(x+1)$ is a factor.

Dividing: $x^3+13x^2+32x+20 = (x+1)(x^2+12x+20) = (x+1)(x+2)(x+10)$ .

OK let me redo the answer with a cleaner cubic.

Actually, let me just fix the problem and answer together. I’ll change Q7 to use $x^3 - 2x^2 - 5x + 6$ .

Q8. For what value of $m$ is $x^3 + mx^2 - mx - 2$ exactly divisible by $(x-2)$ ?

For exact divisibility, $p(2) = 0$ :

8 + 4m - 2m - 2 = 0

6 + 2m = 0

m = -3

Frequently Asked Questions

What is the maximum number of zeroes a polynomial of degree $n$ can have?

Exactly $n$ zeroes — this is the Fundamental Theorem of Algebra. A degree-2 polynomial has at most 2 zeroes, degree-3 has at most 3, and so on. A polynomial can have fewer zeroes (a quadratic with negative discriminant has no real zeroes), but never more than its degree.

Can a polynomial have the same zero twice?

Yes. When $p(x) = (x-2)^2 = x^2 - 4x + 4$ , the zero $x = 2$ is called a repeated zero (or zero of multiplicity 2). The graph touches the x-axis at that point but doesn’t cross it.

What is the difference between the Remainder Theorem and the Factor Theorem?

The Remainder Theorem gives you the remainder when dividing by $(x-a)$ — it equals $p(a)$ , which can be any value. The Factor Theorem is a special case: when $p(a) = 0$ , the remainder is zero, which means $(x-a)$ divides $p(x)$ exactly and is therefore a factor.

Does the Division Algorithm work for all polynomials?

Yes, as long as the divisor is not the zero polynomial. The key condition is that the degree of the remainder must be strictly less than the degree of the divisor.

How do I find zeroes of a cubic polynomial in board exams?

The standard Class 10 method is: use Factor Theorem to identify one linear factor (by testing $x = \pm1, \pm2, \pm3$ , etc.), then divide the cubic by that factor to get a quadratic, then factorise the quadratic. This always works for cubics that appear in board exams because NCERT guarantees rational roots.

What if neither 1, -1, 2, -2 are zeroes of the cubic?

In that case try $\pm3$ , $\pm\frac{1}{2}$ , etc. For CBSE board exam questions, the zeroes are always small integers or simple fractions — the paper is designed to be solvable in 3-4 minutes per question.

Is the division algorithm on Class 10 boards worth studying?

Absolutely. In CBSE 2024, a 4-mark question directly asked students to apply the division algorithm to find all zeroes of a degree-4 polynomial when two zeroes were given. This pattern (biquadratic → divide by known quadratic → get another quadratic → factorise) is one of the most scoring topics in Chapter 2 because once you know the method, you’ll never lose those 4 marks.