NCERT Solutions for Class 9 Maths Chapter 2 Polynomials — Complete Guide with All Exercises

Why Polynomials Is the Gateway to Algebra

If Chapter 1 (Number Systems) gives you the numbers, Chapter 2 (Polynomials) gives you the language to write equations with those numbers. Polynomials are the single most important algebraic structure you will encounter in school mathematics — they appear in factorisation, coordinate geometry, quadratic equations (Class 10), and calculus (Classes 11–12).

Chapter 2 typically carries 6–8 marks in the CBSE Class 9 annual exam. The questions range from 1-mark MCQs (degree of a polynomial, identifying zeroes) to 3-mark short-answer questions (remainder theorem, factorisation) and 5-mark long-answer problems (algebraic identity applications and cubic factorisation).

The chapter is built around five exercises covering a logical progression: first you learn what polynomials are (Exercise 2.1), then their zeroes (Exercise 2.2), then how division works via the Remainder Theorem (Exercise 2.3), then factorisation using the Factor Theorem (Exercise 2.4), and finally the powerful algebraic identities (Exercise 2.5).

In this guide, we solve every important problem, explain the underlying theory, flag the mistakes that cost students marks, and give you a concrete exam strategy. Let's get started!

Polynomials — Definitions and Terminology

A polynomial in one variable $x$ is an algebraic expression of the form:

where $a_0, a_1, \ldots, a_n$ are real numbers (called coefficients) and $n$ is a non-negative integer.

Each part of the form $a_k x^k$ is called a term. The highest power of $x$ that has a non-zero coefficient is called the degree of the polynomial.

Classification by Degree:

| Degree | Name | General Form | Example |
|---|---|---|---|
| 0 | Constant | $a_0$ | $7$, $-3$ |
| 1 | Linear | $ax + b$ | $3x + 5$ |
| 2 | Quadratic | $ax^2 + bx + c$ | $2x^2 - 4x + 1$ |
| 3 | Cubic | $ax^3 + bx^2 + cx + d$ | $x^3 - 6x^2 + 11x - 6$ |
| 4 | Quartic (Bi-quadratic) | $ax^4 + \ldots$ | $x^4 - 1$ |

Classification by Number of Terms:
- Monomial: One term (e.g., $5x^2$)
- Binomial: Two terms (e.g., $x^2 + 3$)
- Trinomial: Three terms (e.g., $x^2 + 3x + 2$)

What is NOT a polynomial?
- $x^{-1} + 3$ — negative exponent
- $\sqrt{x} + 1 = x^{1/2} + 1$ — fractional exponent
- $\frac{1}{x+1}$ — variable in the denominator

The Zero Polynomial

The zero polynomial is the constant polynomial $p(x) = 0$. Its degree is undefined (or sometimes stated as $-\infty$ in advanced texts). The NCERT textbook says the degree of the zero polynomial is "not defined."

Do not confuse the zero polynomial ($p(x) = 0$ for all $x$) with the zeroes of a polynomial (the values of $x$ where $p(x) = 0$). These are completely different concepts.

Exercise 2.1 — Polynomials and Their Degrees

This exercise tests your ability to identify polynomials, determine their degree, and classify them.

Problem 1: Identifying polynomials

Question: Which of the following expressions are polynomials in one variable? Which are not? Give reasons.
(i) $4x^2 - 3x + 7$
(ii) $y^2 + \sqrt{2}$
(iii) $3\sqrt{t} + t\sqrt{2}$
(iv) $y + \frac{2}{y}$
(v) $x^{10} + y^3 + t^{50}$

Solution:

(i) $4x^2 - 3x + 7$ — Yes, this is a polynomial in $x$. All exponents of $x$ are whole numbers (2, 1, 0).

(ii) $y^2 + \sqrt{2}$ — Yes, this is a polynomial in $y$. The $\sqrt{2}$ is a constant coefficient, not a variable with a fractional exponent. Degree = 2.

(iii) $3\sqrt{t} + t\sqrt{2} = 3t^{1/2} + \sqrt{2}\,t$ — No, this is NOT a polynomial because the term $3t^{1/2}$ has a fractional exponent ($\frac{1}{2}$). Polynomial exponents must be non-negative integers.

(iv) $y + \frac{2}{y} = y + 2y^{-1}$ — No, this is NOT a polynomial because $2y^{-1}$ has a negative exponent.

(v) $x^{10} + y^3 + t^{50}$ — This is a polynomial, but it is a polynomial in three variables ($x$, $y$, $t$), not in one variable.

Key Takeaway: For a polynomial in one variable, every exponent of that variable must be a whole number (0, 1, 2, 3, ...).

Problem 2: Degree of polynomials

Question: Write the degree of each of the following polynomials:
(i) $5x^3 + 4x^2 + 7x$
(ii) $4 - y^2$
(iii) $5t - \sqrt{7}$
(iv) $3$

Solution:

(i) Highest power of $x$ is 3. Degree = 3 (cubic polynomial).

(ii) Highest power of $y$ is 2. Degree = 2 (quadratic polynomial).

(iii) Highest power of $t$ is 1. Degree = 1 (linear polynomial).

(iv) $3 = 3x^0$. Highest power of $x$ is 0. Degree = 0 (constant polynomial).

Common Error: Students sometimes say the degree of $3$ is "no degree" or "undefined." Only the zero polynomial has an undefined degree. The constant $3$ has degree 0.

Problem 3: Classifying polynomials

Question: Classify the following as linear, quadratic, and cubic polynomials:
(i) $x^2 + x$
(ii) $x - x^3$
(iii) $y + y^2 + 4$
(iv) $1 + x$
(v) $3t$
(vi) $r^2$
(vii) $7x^3$

Solution:

| Polynomial | Degree | Type |
|---|---|---|
| $x^2 + x$ | 2 | Quadratic |
| $x - x^3$ | 3 | Cubic |
| $y + y^2 + 4$ | 2 | Quadratic |
| $1 + x$ | 1 | Linear |
| $3t$ | 1 | Linear |
| $r^2$ | 2 | Quadratic |
| $7x^3$ | 3 | Cubic |

Exercise 2.2 — Zeroes of a Polynomial

A zero (or root) of a polynomial $p(x)$ is a value $x = a$ such that $p(a) = 0$. Geometrically, the zeroes are the $x$-coordinates where the graph of $y = p(x)$ crosses or touches the $x$-axis.

Maximum number of zeroes:
- A linear polynomial has exactly 1 zero.
- A quadratic polynomial has at most 2 zeroes.
- A cubic polynomial has at most 3 zeroes.
- In general, a polynomial of degree $n$ has at most $n$ zeroes.

Problem 1: Evaluating a polynomial

Question: Find the value of the polynomial $p(x) = 5x - 4x^2 + 3$ at (i) $x = 0$ and (ii) $x = -1$.

Solution:

(i) $p(0) = 5(0) - 4(0)^2 + 3 = 0 - 0 + 3 = 3$

(ii) $p(-1) = 5(-1) - 4(-1)^2 + 3 = -5 - 4(1) + 3 = -5 - 4 + 3 = -6$

Key Technique: Substitute the value carefully, paying special attention to negative signs. A common error is computing $(-1)^2$ as $-1$ instead of $+1$.

Problem 2: Verifying zeroes

Question: Verify whether the following are zeroes of the polynomial indicated against them:
(i) $p(x) = 3x + 1$, $x = -\frac{1}{3}$
(ii) $p(x) = 5x - \pi$, $x = \frac{4}{5}$
(iii) $p(x) = x^2 - 1$, $x = 1$, $x = -1$
(iv) $p(x) = (x+1)(x-2)$, $x = -1$, $x = 2$
(v) $p(x) = x^2$, $x = 0$

Solution:

(i) $p\left(-\frac{1}{3}\right) = 3\left(-\frac{1}{3}\right) + 1 = -1 + 1 = 0$ ✓ Yes, it is a zero.

(ii) $p\left(\frac{4}{5}\right) = 5 \times \frac{4}{5} - \pi = 4 - \pi \neq 0$ (since $\pi \approx 3.14$). Not a zero.

(iii) $p(1) = 1^2 - 1 = 0$ ✓; $p(-1) = (-1)^2 - 1 = 1 - 1 = 0$ ✓. Both are zeroes.

(iv) $p(-1) = (-1+1)(-1-2) = 0 \times (-3) = 0$ ✓; $p(2) = (2+1)(2-2) = 3 \times 0 = 0$ ✓. Both are zeroes.

(v) $p(0) = 0^2 = 0$ ✓. Yes, $x = 0$ is a zero (with multiplicity 2).

Problem 3: Finding the zero of a linear polynomial

Question: Find the zero of the polynomial in each of the following cases:
(i) $p(x) = x + 5$
(ii) $p(x) = x - 5$
(iii) $p(x) = 2x + 5$
(iv) $p(x) = 3x - 2$
(v) $p(x) = 3x$
(vi) $p(x) = ax$, $a \neq 0$
(vii) $p(x) = cx + d$, $c \neq 0$

Solution:
For a linear polynomial $p(x) = ax + b$, the zero is $x = -\frac{b}{a}$.

| Polynomial | Zero |
|---|---|
| $x + 5$ | $x = -5$ |
| $x - 5$ | $x = 5$ |
| $2x + 5$ | $x = -\frac{5}{2}$ |
| $3x - 2$ | $x = \frac{2}{3}$ |
| $3x$ | $x = 0$ |
| $ax$ ($a \neq 0$) | $x = 0$ |
| $cx + d$ ($c \neq 0$) | $x = -\frac{d}{c}$ |

The Remainder Theorem — Theory and Proof

The Remainder Theorem is one of the most elegant results in polynomial algebra, and it saves enormous time compared to long division.

Statement: If $p(x)$ is any polynomial of degree $\geq 1$ and $a$ is any real number, then when $p(x)$ is divided by the linear polynomial $(x - a)$, the remainder is $p(a)$.

Proof:
By the division algorithm for polynomials:

where $q(x)$ is the quotient and $r$ is a constant (since the divisor is linear, the remainder has degree 0).

Substitute $x = a$:

Therefore, the remainder $r = p(a)$. $\square$

What if the divisor is $(ax - b)$ instead of $(x - a)$?
Rewrite $(ax - b) = a\left(x - \frac{b}{a}\right)$. The remainder when $p(x)$ is divided by $(ax - b)$ is $p\left(\frac{b}{a}\right)$.

For example, the remainder when $p(x)$ is divided by $(2x - 3)$ is $p\left(\frac{3}{2}\right)$.

The Factor Theorem

The Factor Theorem is a special case of the Remainder Theorem.

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

Proof: By the Remainder Theorem, the remainder when $p(x)$ is divided by $(x - a)$ is $p(a)$. If $p(a) = 0$, then:

So $(x - a)$ is a factor. Conversely, if $(x - a)$ is a factor, then $p(x) = (x - a) \cdot q(x)$, so $p(a) = 0$. $\square$

The Factor Theorem is your primary tool for factorising cubic (and higher-degree) polynomials. The strategy is:
1. Guess a root $a$ by trying small integers ($0, \pm 1, \pm 2, \ldots$).
2. Confirm that $p(a) = 0$.
3. Divide $p(x)$ by $(x - a)$ to get the remaining quadratic.
4. Factorise the quadratic.

Exercise 2.3 — Remainder Theorem Problems

This exercise applies the Remainder Theorem to find remainders without performing long division.

Problem 1: Remainder when dividing by (x + 1)

Question: Find the remainder when $x^3 + 3x^2 + 3x + 1$ is divided by:
(i) $x + 1$
(ii) $x - \frac{1}{2}$
(iii) $x$
(iv) $x + \pi$
(v) $5 + 2x$

Solution:

(i) Divisor: $(x + 1) = (x - (-1))$, so $a = -1$.

Remainder = 0. (This means $(x+1)$ is a factor!)

(ii) Divisor: $\left(x - \frac{1}{2}\right)$, so $a = \frac{1}{2}$.

$$= \frac{1}{8} + \frac{3}{4} + \frac{3}{2} + 1 = \frac{1}{8} + \frac{6}{8} + \frac{12}{8} + \frac{8}{8} = \frac{27}{8}$$

Remainder = $\frac{27}{8}$.

(iii) Divisor: $x = (x - 0)$, so $a = 0$.

Remainder = 1.

(iv) Divisor: $(x + \pi) = (x - (-\pi))$, so $a = -\pi$.

Remainder = $1 - 3\pi + 3\pi^2 - \pi^3$.

Notice this equals $(1 - \pi)^3$ by the identity $(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$.

(v) Divisor: $(5 + 2x) = 2\left(x + \frac{5}{2}\right) = 2\left(x - \left(-\frac{5}{2}\right)\right)$, so $a = -\frac{5}{2}$.

$$= -\frac{125}{8} + \frac{75}{4} - \frac{15}{2} + 1 = -\frac{125}{8} + \frac{150}{8} - \frac{60}{8} + \frac{8}{8} = -\frac{27}{8}$$

Remainder = $-\frac{27}{8}$.

Problem 2: Remainder with algebraic coefficients

Question: Find the remainder when $x^3 - ax^2 + 6x - a$ is divided by $(x - a)$.

Solution:
By the Remainder Theorem, remainder $= p(a)$.

Remainder = $5a$.

Key Insight: Even when the polynomial has parameters (like $a$), the Remainder Theorem works the same way. Just substitute and simplify.

Problem 3: Checking factors using the Factor Theorem

Question: Check whether $7 + 3x$ is a factor of $3x^3 + 7x$.

Solution:
$(7 + 3x) = 3\left(x + \frac{7}{3}\right)$, so the zero is $x = -\frac{7}{3}$.

$$= 3 \times \left(-\frac{343}{27}\right) - \frac{49}{3}$$

$$= -\frac{343}{9} - \frac{49}{3} = -\frac{343}{9} - \frac{147}{9} = -\frac{490}{9}$$

Since $p\left(-\frac{7}{3}\right) \neq 0$, $(7 + 3x)$ is not a factor of $3x^3 + 7x$.

Exercise 2.4 — Factorisation Using the Factor Theorem

This is the most exam-important exercise in the chapter. You will use the Factor Theorem to find one factor of a cubic polynomial, then divide to get the remaining quadratic, and factorise that too.

Strategy for finding the first root:
If $p(x) = a_n x^n + \ldots + a_0$, then by the Rational Root Theorem, any rational root $\frac{p}{q}$ must have $p$ dividing $a_0$ (the constant term) and $q$ dividing $a_n$ (the leading coefficient). For most NCERT problems, the leading coefficient is 1, so you only need to try factors of the constant term: $\pm 1, \pm 2, \ldots$

Problem 1: Determine factors

Question: Determine which of the following polynomials has $(x + 1)$ as a factor:
(i) $x^3 + x^2 + x + 1$
(ii) $x^4 + x^3 + x^2 + x + 1$
(iii) $x^4 + 3x^3 + 3x^2 + x + 1$
(iv) $x^3 - x^2 - (2 + \sqrt{2})x + \sqrt{2}$

Solution:
For $(x+1)$ to be a factor, we need $p(-1) = 0$.

(i) $p(-1) = -1 + 1 - 1 + 1 = 0$ ✓ Yes.

(ii) $p(-1) = 1 - 1 + 1 - 1 + 1 = 1 \neq 0$. No.

(iii) $p(-1) = 1 - 3 + 3 - 1 + 1 = 1 \neq 0$. No.

(iv) $p(-1) = -1 - 1 + 2 + \sqrt{2} + \sqrt{2} = 2\sqrt{2} \neq 0$. No.

Problem 2: Factorise x³ - 2x² - x + 2

Question: Use the Factor Theorem to factorise $x^3 - 2x^2 - x + 2$.

Solution:
Let $p(x) = x^3 - 2x^2 - x + 2$.

Step 1: Find a root. Constant term = 2, so try $x = \pm 1, \pm 2$.

$p(1) = 1 - 2 - 1 + 2 = 0$ ✓

So $(x - 1)$ is a factor.

Step 2: Divide by $(x - 1)$.
Using synthetic division or long division:

Step 3: Factorise the quadratic.
$x^2 - x - 2$: we need two numbers that multiply to $-2$ and add to $-1$. Those are $-2$ and $+1$.

Final Answer:

Verification: The zeroes are $x = 1, 2, -1$. Check: $p(2) = 8 - 8 - 2 + 2 = 0$ ✓, $p(-1) = -1 - 2 + 1 + 2 = 0$ ✓.

Problem 3: Factorise x³ - 3x² - 9x - 5

Question: Factorise $x^3 - 3x^2 - 9x - 5$.

Solution:
Let $p(x) = x^3 - 3x^2 - 9x - 5$. Constant term = $-5$, so try $x = \pm 1, \pm 5$.

$p(-1) = -1 - 3 + 9 - 5 = 0$ ✓

So $(x + 1)$ is a factor. Dividing:

Factorise $x^2 - 4x - 5$: numbers that multiply to $-5$ and add to $-4$ are $-5$ and $+1$.

Final Answer:

Note: $x = -1$ is a repeated root (multiplicity 2).

Problem 4: Factorise x³ + 13x² + 32x + 20

Question: Factorise $x^3 + 13x^2 + 32x + 20$.

Solution:
Let $p(x) = x^3 + 13x^2 + 32x + 20$. Constant term = 20, so try factors of 20.

$p(-1) = -1 + 13 - 32 + 20 = 0$ ✓

So $(x + 1)$ is a factor. Dividing:

Factorise $x^2 + 12x + 20$: numbers that multiply to 20 and add to 12 are 10 and 2.

Final Answer:

Problem 5: Factorise 2y³ + y² - 2y - 1

Question: Factorise $2y^3 + y^2 - 2y - 1$.

Solution:
Let $p(y) = 2y^3 + y^2 - 2y - 1$. Leading coefficient = 2, constant = $-1$.

Possible rational roots: $\pm 1, \pm \frac{1}{2}$.

$p(1) = 2 + 1 - 2 - 1 = 0$ ✓

So $(y - 1)$ is a factor. Dividing:

Factorise $2y^2 + 3y + 1$: we need numbers that multiply to $2 \times 1 = 2$ and add to 3. Those are 2 and 1.

Final Answer:

Exercise 2.5 — Algebraic Identities

This exercise is about the eight algebraic identities prescribed in the NCERT syllabus. These identities are powerful tools for expanding, factorising, and evaluating expressions. You MUST memorise all eight.

The Eight Identities:

Identity I: $(a + b)^2 = a^2 + 2ab + b^2$

Identity II: $(a - b)^2 = a^2 - 2ab + b^2$

Identity III: $a^2 - b^2 = (a + b)(a - b)$

Identity IV: $(x + a)(x + b) = x^2 + (a + b)x + ab$

Identity V: $(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$

Identity VI: $(a + b)^3 = a^3 + b^3 + 3ab(a + b) = a^3 + 3a^2b + 3ab^2 + b^3$

Identity VII: $(a - b)^3 = a^3 - b^3 - 3ab(a - b) = a^3 - 3a^2b + 3ab^2 - b^3$

Identity VIII: $a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$

Problem 1: Expansion using identities

Question: Use suitable identities to find the following products:
(i) $(x + 4)(x + 10)$
(ii) $(x + 8)(x - 10)$
(iii) $(3x + 4)(3x - 5)$
(iv) $(y^2 + \frac{3}{2})(y^2 - \frac{3}{2})$
(v) $(3 - 2x)(3 + 2x)$

Solution:

(i) Using Identity IV: $(x + a)(x + b) = x^2 + (a+b)x + ab$

(ii) $(x + 8)(x - 10) = x^2 + (8 + (-10))x + 8 \times (-10) = x^2 - 2x - 80$

(iii) Let $y = 3x$: $(y + 4)(y - 5) = y^2 + (4-5)y + 4(-5) = y^2 - y - 20$

(iv) Using Identity III: $a^2 - b^2 = (a+b)(a-b)$

(v) $(3 - 2x)(3 + 2x) = 3^2 - (2x)^2 = 9 - 4x^2$

Problem 2: Evaluating numbers using identities

Question: Evaluate the following using suitable identities:
(i) $103^2$
(ii) $99^2$
(iii) $105 \times 95$
(iv) $9.8 \times 10.2$

Solution:

(i) $103^2 = (100 + 3)^2 = 100^2 + 2(100)(3) + 3^2 = 10000 + 600 + 9 = 10609$

(ii) $99^2 = (100 - 1)^2 = 100^2 - 2(100)(1) + 1^2 = 10000 - 200 + 1 = 9801$

(iii) $105 \times 95 = (100 + 5)(100 - 5) = 100^2 - 5^2 = 10000 - 25 = 9975$

(iv) $9.8 \times 10.2 = (10 - 0.2)(10 + 0.2) = 10^2 - 0.2^2 = 100 - 0.04 = 99.96$

Problem 3: Factorising using identities

Question: Factorise the following:
(i) $4x^2 + 20x + 25$
(ii) $9y^2 - 66yz + 121z^2$
(iii) $(2a - 3b)^2 - (3a - 2b)^2$

Solution:

(i) $4x^2 + 20x + 25 = (2x)^2 + 2(2x)(5) + 5^2 = (2x + 5)^2$ [Identity I]

(ii) $9y^2 - 66yz + 121z^2 = (3y)^2 - 2(3y)(11z) + (11z)^2 = (3y - 11z)^2$ [Identity II]

(iii) Using $A^2 - B^2 = (A+B)(A-B)$ with $A = 2a - 3b$ and $B = 3a - 2b$:

$A + B = (2a - 3b) + (3a - 2b) = 5a - 5b = 5(a - b)$
$A - B = (2a - 3b) - (3a - 2b) = -a - b = -(a + b)$

Or equivalently: $= -5(a^2 - b^2)$.

Problem 4: Expanding cubes

Question: Expand:
(i) $(2x + 1)^3$
(ii) $(2a - 3b)^3$

Solution:

(i) Using $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$:

$$= 8x^3 + 12x^2 + 6x + 1$$

(ii) Using $(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$:

$$= 8a^3 - 36a^2b + 54ab^2 - 27b^3$$

Problem 5: Factorising cubes and the special identity

Question: Factorise $8x^3 + 27y^3 + 36x^2y + 54xy^2$.

Solution:
Rearrange: $8x^3 + 36x^2y + 54xy^2 + 27y^3$

Recognise this as $(a + b)^3$ where $a = 2x$ and $b = 3y$:

Question: If $x + y + z = 0$, show that $x^3 + y^3 + z^3 = 3xyz$.

Solution:
Using Identity VIII:

Since $x + y + z = 0$:

$$\therefore x^3 + y^3 + z^3 = 3xyz \quad \square$$

Application: If $a + b + c = 0$, find $a^3 + b^3 + c^3$.

Using the result above, $a^3 + b^3 + c^3 = 3abc$. This is a guaranteed 1-mark question in exams.

Problem 6: Using Identity V with three terms

Question: Factorise $4x^2 + y^2 + z^2 - 4xy - 2yz + 4xz$.

Solution:
Rewrite as:

This matches $(a + b + c)^2$ with $a = 2x$, $b = -y$, $c = z$.

Question: Evaluate $(-12)^3 + (7)^3 + (5)^3$ without direct computation.

Solution:
Note that $-12 + 7 + 5 = 0$.

By the identity, when $a + b + c = 0$: $a^3 + b^3 + c^3 = 3abc$.

Polynomial Long Division — Step-by-Step Method

After finding one factor using the Factor Theorem, you need to divide the polynomial by that factor to get the quotient. Here is the complete method:

Example: Divide $x^3 - 2x^2 - x + 2$ by $(x - 1)$.

Step 1: Write the dividend in descending powers: $x^3 - 2x^2 - x + 2$.

Step 2: Divide the first term of the dividend by the first term of the divisor: $\frac{x^3}{x} = x^2$. Write $x^2$ in the quotient.

Step 3: Multiply the divisor by $x^2$: $x^2(x - 1) = x^3 - x^2$.

Step 4: Subtract: $(x^3 - 2x^2) - (x^3 - x^2) = -x^2$. Bring down $-x$: $-x^2 - x$.

Step 5: Divide: $\frac{-x^2}{x} = -x$. Multiply: $-x(x-1) = -x^2 + x$.

Step 6: Subtract: $(-x^2 - x) - (-x^2 + x) = -2x$. Bring down $+2$: $-2x + 2$.

Step 7: Divide: $\frac{-2x}{x} = -2$. Multiply: $-2(x-1) = -2x + 2$.

Step 8: Subtract: $(-2x + 2) - (-2x + 2) = 0$.

Quotient: $x^2 - x - 2$. Remainder: $0$.

Synthetic Division (Shortcut)

For dividing by a linear factor $(x - a)$, synthetic division is faster:

Example: Divide $x^3 - 2x^2 - x + 2$ by $(x - 1)$ (i.e., $a = 1$).

Write the coefficients: $1, -2, -1, 2$.

```
1 | 1 -2 -1 2
| 1 -1 -2
| 1 -1 -2 0
```

Process: Bring down 1. Multiply by $a = 1$: $1 \times 1 = 1$. Add: $-2 + 1 = -1$. Multiply: $-1 \times 1 = -1$. Add: $-1 + (-1) = -2$. Multiply: $-2 \times 1 = -2$. Add: $2 + (-2) = 0$.

Quotient coefficients: $1, -1, -2$ → $x^2 - x - 2$. Remainder: $0$.

Synthetic division is much faster than long division once you are comfortable with it.

Additional Solved Problems — Exam-Level Difficulty

These problems go beyond the basic NCERT exercises and match CBSE exam and HOTS difficulty.

Problem 1: Finding unknown coefficients

Question: If $(x - 2)$ is a factor of $x^3 + ax^2 + bx + 16$ and $b = 4a$, find $a$ and $b$.

Solution:
Since $(x - 2)$ is a factor: $p(2) = 0$.

$$4a + 2b = -24$$

$$2a + b = -12 \quad \cdots (1)$$

Given: $b = 4a \quad \cdots (2)$

Substituting (2) into (1):

$$6a = -12$$

$$a = -2, \quad b = 4(-2) = -8$$

Verification: $p(x) = x^3 - 2x^2 - 8x + 16$. $p(2) = 8 - 8 - 16 + 16 = 0$ ✓.

Problem 2: If a + b + c = 9 and ab + bc + ca = 26

Question: If $a + b + c = 9$ and $ab + bc + ca = 26$, find $a^2 + b^2 + c^2$.

Solution:
Using Identity V: $(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$

$$81 = a^2 + b^2 + c^2 + 52$$

$$a^2 + b^2 + c^2 = 29$$

Problem 3: Finding a³ + b³ + c³

Question: If $a + b + c = 5$, $a^2 + b^2 + c^2 = 29$, find $ab + bc + ca$ and also $a^3 + b^3 + c^3$ given that $abc = -12$.

Solution:
Part 1: $(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$

$$ab + bc + ca = \frac{25 - 29}{2} = -2$$

Part 2: Using Identity VIII:

$$a^3 + b^3 + c^3 - 3(-12) = 5(29 - (-2))$$

$$a^3 + b^3 + c^3 + 36 = 5(31) = 155$$

$$a^3 + b^3 + c^3 = 155 - 36 = 119$$

Common Mistakes Students Make in Polynomials

Avoid these frequent errors to secure full marks:

1. Confusing "zero polynomial" with "zeroes of a polynomial":
2. * Mistake: Saying "the zero polynomial has no zeroes."
3. * Fix: The zero polynomial $p(x) = 0$ is zero for EVERY value of $x$, so every real number is a zero of the zero polynomial. The term "zero of a polynomial" means a root, not the polynomial itself.
4. Fractional exponents = polynomial:
5. * Mistake: Accepting $\sqrt{x} + 3$ as a polynomial.
6. * Fix: $\sqrt{x} = x^{1/2}$, which has a fractional exponent. This is NOT a polynomial.
7. Sign errors in the Factor Theorem:
8. * Mistake: Testing $(x + 2)$ by substituting $x = 2$ instead of $x = -2$.
9. * Fix: $(x + 2) = (x - (-2))$, so $a = -2$. Always negate the constant in the factor.
10. Wrong identity applied:
11. * Mistake: Using $(a+b)^2 = a^2 + b^2$ (forgetting the $2ab$ term).
12. * Fix: $(a+b)^2 = a^2 + 2ab + b^2$. The middle term $2ab$ is NEVER zero (unless $a$ or $b$ is zero).
13. Arithmetic errors in long division:
14. * Mistake: Getting the wrong quadratic after dividing by the linear factor.
15. * Fix: Verify by multiplying the quotient by the divisor and adding the remainder. The result should equal the original polynomial.
16. Forgetting to check all possible roots:
17. * Mistake: Trying only positive integers as roots.
18. * Fix: Try both positive and negative factors of the constant term. For $x^3 + x^2 - x - 1$, trying $x = -1$ gives $p(-1) = -1 + 1 + 1 - 1 = 0$, but trying only positive integers misses this.
19. Not stating the identity used:
20. * Mistake: Just writing the answer without naming the identity.
21. * Fix: CBSE marking schemes award marks for stating the identity. Write "Using $(a+b)^2 = a^2 + 2ab + b^2$" before applying it.

Board Exam Strategy for Polynomials

Quick Revision: All Identities at a Glance

Bookmark this for last-minute revision:

Identity I: $(a + b)^2 = a^2 + 2ab + b^2$

Identity II: $(a - b)^2 = a^2 - 2ab + b^2$

Identity III: $a^2 - b^2 = (a + b)(a - b)$

Identity IV: $(x + a)(x + b) = x^2 + (a+b)x + ab$

Identity V: $(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$

Identity VI: $(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$

Identity VII: $(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$

Identity VIII: $a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)$

Derived Results:
- $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$
- $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$
- If $a + b + c = 0$, then $a^3 + b^3 + c^3 = 3abc$

Key Theorems:
- Remainder Theorem: remainder of $p(x) \div (x-a)$ is $p(a)$
- Factor Theorem: $(x-a)$ is a factor of $p(x)$ iff $p(a) = 0$
- A polynomial of degree $n$ has at most $n$ zeroes.

Practice Problems for Self-Assessment

Try these before checking the answers:

Level 1 (Basic):
1. What is the degree of $5 - 3x + 2x^3$?
2. Find the zero of $p(x) = 4x - 8$.
3. Find the remainder when $x^4 + 1$ is divided by $x + 1$.

Level 2 (Intermediate):
4. Factorise $x^3 - 6x^2 + 11x - 6$.
5. Evaluate $102^3$ using a suitable identity.
6. If $a + b + c = 0$, find $\frac{a^3 + b^3 + c^3}{abc}$ (given $abc \neq 0$).

Level 3 (Advanced/HOTS):
7. If $x + \frac{1}{x} = 5$, find $x^3 + \frac{1}{x^3}$.
8. Find the value of $k$ if $(x - 1)$ is a factor of $4x^3 + 3x^2 - 4x + k$.
9. Factorise $a^3 - b^3 - a + b$.

Answers:
1. Degree 3
2. $x = 2$
3. $p(-1) = 1 + 1 = 2$
4. Try $x = 1$: $1 - 6 + 11 - 6 = 0$. Divide: $(x-1)(x^2 - 5x + 6) = (x-1)(x-2)(x-3)$.
5. $102^3 = (100+2)^3 = 100^3 + 3(100)^2(2) + 3(100)(4) + 8 = 1000000 + 60000 + 1200 + 8 = 1061208$.
6. Since $a+b+c=0$, $a^3+b^3+c^3 = 3abc$. So $\frac{3abc}{abc} = 3$.
7. Cube both sides using $(a+b)^3 = a^3 + b^3 + 3ab(a+b)$: $(x + \frac{1}{x})^3 = x^3 + \frac{1}{x^3} + 3(x + \frac{1}{x})$. So $125 = x^3 + \frac{1}{x^3} + 15$, giving $x^3 + \frac{1}{x^3} = 110$.
8. $p(1) = 4 + 3 - 4 + k = 0 \implies k = -3$.
9. $= (a-b)(a^2+ab+b^2) - (a-b) = (a-b)(a^2+ab+b^2-1)$.

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