Polynomials

1. What is a polynomial?

Consider expressions like $x+2$, $x^{2}-5x+6$, $y^{3}-1$. Each is formed by adding terms where the variable appears with a non-negative integer exponent. These are called polynomials in one variable.

Formally, a polynomial in $x$ of degree $n$ looks like:

where $a_{n}\neq0$ and each coefficient $a_{i}$ is a real number. The individual pieces $a_{n}x^{n},\ a_{n-1}x^{n-1},\ \dots,\ a_{0}$ are called terms.

Key vocabulary:

• Coefficient — the numerical part of a term. In $-5x^{3}$, the coefficient is $-5$.
• Constant term — the term with no variable, equal to the value of the polynomial at $x=0$.
• Degree — the highest power of the variable. Degree of $3x^{4}-7x+1$ is $4$.
• Leading coefficient — the coefficient of the highest-degree term.
• Monomial — a polynomial with exactly one term, e.g. $3x^{2}$.
• Binomial — exactly two terms, e.g. $x^{2}-4$.
• Trinomial — exactly three terms, e.g. $x^{2}+3x-2$.

What is NOT a polynomial: $x^{-2}+3$ (negative power), $\sqrt{x}=x^{1/2}$ (fractional power), $\dfrac{1}{x}=x^{-1}$ (negative power). The exponent of the variable must always be a non-negative integer.

Special cases: A non-zero constant like $7$ is a polynomial of degree $0$. The zero polynomial $p(x)=0$ is given no degree (sometimes stated as $-\infty$). Every real number coefficient is allowed, including irrationals like $\sqrt{2}$, so $\sqrt{2}x+5$ is a valid polynomial.

2. Types of polynomials by degree

Degree determines the shape of the graph and the maximum number of zeroes.

Degree	Name	General form	Example
$0$	Constant	$a$, where $a\neq0$	$7$
$1$	Linear	$ax+b$, with $a\neq0$	$2x-3$
$2$	Quadratic	$ax^{2}+bx+c$, with $a\neq0$	$x^{2}-5x+6$
$3$	Cubic	$ax^{3}+bx^{2}+cx+d$, with $a\neq0$	$x^{3}-6x^{2}+11x-6$
$4$	Bi-quadratic	$ax^{4}+\cdots$, with $a\neq0$	$x^{4}-1$

Maximum zeroes: A polynomial of degree $n$ has at most $n$ real zeroes. So a linear polynomial has exactly $1$ zero, a quadratic at most $2$, and a cubic at most $3$.

3. Value of a polynomial and its zeroes

If $p(x)$ is a polynomial, its value at $x=a$ is written $p(a)$ and found by substituting $x=a$ into every term.

Example: Let $p(x)=x^{2}-3x+2$. Then:

• $p(0)=0-0+2=2$
• $p(1)=1-3+2=0$
• $p(2)=4-6+2=0$
• $p(-1)=1+3+2=6$

Notice $p(1)=0$ and $p(2)=0$. The values $x=1$ and $x=2$ make the polynomial equal to zero. These are the zeroes of the polynomial.

Definition: A real number $a$ is called a zero of a polynomial $p(x)$ if $p(a)=0$.

Zero of a linear polynomial: For $p(x)=ax+b$ with $a\neq0$, set $ax+b=0$ to get $x=-\dfrac{b}{a}$. Every linear polynomial has exactly one zero, namely $-\dfrac{b}{a}$.

Geometrically: the zeroes of $p(x)$ are the $x$-coordinates of the points where the graph of $y=p(x)$ cuts the $x$-axis. A linear graph cuts once; a quadratic graph cuts at most twice.

Number of zeroes summary: A constant non-zero polynomial has no zeroes. The zero polynomial is not assigned any zeroes (technically every real number is a zero). A polynomial of degree $n$ has at most $n$ real zeroes.

4. Remainder Theorem

When you divide a polynomial $p(x)$ by a linear divisor $(x-a)$, the division algorithm gives:

where $q(x)$ is the quotient polynomial and $r$ is the remainder (a constant, because the divisor has degree 1). The Remainder Theorem gives the exact value of $r$ without doing long division.

Theorem (Remainder Theorem): If a polynomial $p(x)$ is divided by the linear polynomial $(x-a)$, then the remainder is $p(a)$.

Proof: From $p(x)=(x-a)\cdot q(x)+r$ which holds for all $x$, substitute $x=a$:

So the remainder $r=p(a)$. To find the remainder, simply evaluate $p$ at $x=a$.

Division by $(ax+b)$: Write $ax+b=a\!\left(x+\dfrac{b}{a}\right)=a\!\left(x-\left(-\dfrac{b}{a}\right)\right).$ The remainder when dividing by $(ax+b)$ is $p\!\left(-\dfrac{b}{a}\right)$.

5. Factor Theorem

The Factor Theorem follows directly from the Remainder Theorem and is the most powerful tool for factorising higher-degree polynomials.

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

Two directions:

• Forward ($p(a)=0\Rightarrow$ factor): If $p(a)=0$, the remainder when dividing by $(x-a)$ is zero, so $(x-a)$ divides $p(x)$ exactly — making it a factor.
• Backward (factor $\Rightarrow p(a)=0$): If $(x-a)$ is a factor, write $p(x)=(x-a)\cdot q(x)$. Set $x=a$: $p(a)=(a-a)q(a)=0.$

Proof (combined):

$(\Rightarrow)$ Suppose $p(a)=0$. By the Remainder Theorem, $p(x)=(x-a)q(x)+p(a)=(x-a)q(x)+0=(x-a)q(x).$ So $(x-a)$ is a factor.

$(\Leftarrow)$ Suppose $(x-a)$ is a factor. Then $p(x)=(x-a)q(x)$. Setting $x=a$: $p(a)=0\cdot q(a)=0.$

How to find which values to try: For a polynomial $a_{n}x^{n}+\cdots+a_{0}$ with integer coefficients, possible rational zeroes are $\pm\dfrac{\text{(factor of }a_{0}\text{)}}{\text{(factor of }a_{n}\text{)}}.$ Try these by substitution. Once one zero $a$ is found, divide out $(x-a)$ and proceed with the quotient.

6. Factorisation of polynomials using the Factor Theorem

Strategy: (1) Find one zero $a$ by trial. (2) Write $(x-a)$ as a factor. (3) Perform long division to get the quotient $q(x)$. (4) Factorise $q(x)$ further if possible.

Splitting the middle term — for quadratics $ax^{2}+bx+c$: find two numbers $p$ and $q$ such that $p+q=b$ and $p\cdot q=ac$; split the middle term as $px+qx$ and group-factorise. This works for the quadratic factor you obtain after extracting one linear factor from a cubic.

7. Algebraic Identities — all 9 with derivations and examples

Algebraic identities are equations true for all values of the variables involved. They are derived from polynomial multiplication and are essential for fast expansion and factorisation in board exams.

Identity 1: $(a+b)^{2}=a^{2}+2ab+b^{2}$

Derivation: $(a+b)(a+b)=a\cdot a+a\cdot b+b\cdot a+b\cdot b=a^{2}+2ab+b^{2}.$

Example (expansion): $(x+4)^{2}=x^{2}+8x+16.$

Example (factorisation): $x^{2}+10x+25=(x+5)^{2}.$ Recognise as $(a+b)^{2}$ with $a=x,b=5.$

Mental maths: $103^{2}=(100+3)^{2}=10000+600+9=10609.$

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

Derivation: Replace $b$ with $-b$ in Identity 1: $(a+(-b))^{2}=a^{2}+2a(-b)+(-b)^{2}=a^{2}-2ab+b^{2}.$

Example: $(2x-3)^{2}=4x^{2}-12x+9.$

Trick: $(a-b)^{2}=(b-a)^{2}$ — squaring always gives a non-negative result.

Mental maths: $98^{2}=(100-2)^{2}=10000-400+4=9604.$

Identity 3: $a^{2}-b^{2}=(a+b)(a-b)$

Derivation: $(a+b)(a-b)=a\cdot a-a\cdot b+b\cdot a-b\cdot b=a^{2}-b^{2}.$ The middle terms cancel.

Example: $x^{2}-16=(x+4)(x-4).$ Also: $4x^{2}-9=(2x+3)(2x-3).$

Mental maths: $97\times103=(100-3)(100+3)=10000-9=9991.$

Also used for: simplifying $\dfrac{a^{2}-b^{2}}{a-b}=a+b$ (as long as $a\neq b$).

Identity 4: $(a+b)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3}$

Derivation: $(a+b)^{3}=(a+b)\cdot(a+b)^{2}=(a+b)(a^{2}+2ab+b^{2})$

$=a^{3}+2a^{2}b+ab^{2}+a^{2}b+2ab^{2}+b^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3}.$

Compact form: $a^{3}+b^{3}+3ab(a+b).$ This form shows how $a^{3}+b^{3}$ relates to the cube.

Example: $(x+2)^{3}=x^{3}+6x^{2}+12x+8.$

Mental maths: $101^{3}=(100+1)^{3}=1000000+30000+300+1=1030301.$

Identity 5: $(a-b)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3}$

Derivation: Replace $b$ with $-b$ in Identity 4:

$(a+(-b))^{3}=a^{3}+3a^{2}(-b)+3a(-b)^{2}+(-b)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3}.$

Compact form: $a^{3}-b^{3}-3ab(a-b).$

Example: $(2x-1)^{3}=8x^{3}-12x^{2}+6x-1.$

Mental maths: $99^{3}=(100-1)^{3}=1000000-30000+300-1=970299.$

Identity 6: $a^{3}+b^{3}=(a+b)(a^{2}-ab+b^{2})$

Derivation: Expand the right side:

$(a+b)(a^{2}-ab+b^{2})=a^{3}-a^{2}b+ab^{2}+a^{2}b-ab^{2}+b^{3}=a^{3}+b^{3}.$ Confirmed.

Alternative: From Identity 4, $(a+b)^{3}=a^{3}+b^{3}+3ab(a+b),$ so $a^{3}+b^{3}=(a+b)^{3}-3ab(a+b)=(a+b)[(a+b)^{2}-3ab]=(a+b)(a^{2}-ab+b^{2}).$

Example: $x^{3}+8=x^{3}+2^{3}=(x+2)(x^{2}-2x+4).$

Example: $27+125y^{3}=3^{3}+(5y)^{3}=(3+5y)(9-15y+25y^{2}).$

Identity 7: $a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2})$

Derivation: Replace $b$ with $-b$ in Identity 6: $a^{3}+(-b)^{3}=(a-b)(a^{2}+ab+b^{2})$, so $a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2}).$

Alternatively verify directly: $(a-b)(a^{2}+ab+b^{2})=a^{3}+a^{2}b+ab^{2}-a^{2}b-ab^{2}-b^{3}=a^{3}-b^{3}.$

Example: $8x^{3}-27=(2x)^{3}-3^{3}=(2x-3)(4x^{2}+6x+9).$

Example: $a^{3}-64b^{3}=(a-4b)(a^{2}+4ab+16b^{2}).$

Identity 8: $(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2ab+2bc+2ca$

Derivation: Write $(a+b+c)^{2}=((a+b)+c)^{2}=(a+b)^{2}+2(a+b)c+c^{2}$

$=a^{2}+2ab+b^{2}+2ac+2bc+c^{2}=a^{2}+b^{2}+c^{2}+2ab+2bc+2ca.$

Memory tip: "Square each, then add twice every pair-product." There are $\binom{3}{2}=3$ pairs: $(ab),(bc),(ca).$

Example: $(x+2y-3z)^{2}=x^{2}+4y^{2}+9z^{2}+4xy-12yz-6xz.$

Example: $(3a-2b-5c)^{2}=9a^{2}+4b^{2}+25c^{2}-12ab+20bc-30ca.$

Useful corollary: If $a+b+c=0$, squaring both sides: $a^{2}+b^{2}+c^{2}+2(ab+bc+ca)=0$, so $ab+bc+ca=-\dfrac{a^{2}+b^{2}+c^{2}}{2}.$

Reverse use: Given $a+b+c$ and $ab+bc+ca$, find $a^{2}+b^{2}+c^{2}=(a+b+c)^{2}-2(ab+bc+ca).$ Very common board question!

Identity 9: $a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$

Derivation: Expand the right-hand side by distributing $a$, $b$, $c$:

$a(a^{2}+b^{2}+c^{2}-ab-bc-ca)=a^{3}+ab^{2}+ac^{2}-a^{2}b-abc-a^{2}c$

$b(a^{2}+b^{2}+c^{2}-ab-bc-ca)=a^{2}b+b^{3}+bc^{2}-ab^{2}-b^{2}c-abc$

$c(a^{2}+b^{2}+c^{2}-ab-bc-ca)=a^{2}c+b^{2}c+c^{3}-abc-bc^{2}-ac^{2}$

Adding: $a^{3}+b^{3}+c^{3}-3abc.$ All other terms cancel in pairs. Confirmed.

Elegant rewrite: $a^{2}+b^{2}+c^{2}-ab-bc-ca=\dfrac{1}{2}\!\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right].$

This is always non-negative, so the sign of $a^{3}+b^{3}+c^{3}-3abc$ matches the sign of $(a+b+c)$.

Most important special case: If $a+b+c=0$, then $a^{3}+b^{3}+c^{3}-3abc=0$, giving:

Example 1: If $x+y+z=0$, find $x^{3}+y^{3}+z^{3}$. Answer: $3xyz.$

Example 2: Let $a=1,b=\omega,c=\omega^{2}$ where $\omega$ is a cube root of unity — then $a+b+c=0$, so $1+\omega^{3}+\omega^{6}=3\omega^{3}=3.$

8. Using identities for expansion — NCERT examples

9. Common mistakes to avoid

• $(a+b)^{2}\neq a^{2}+b^{2}$ — the cross-term $2ab$ is almost always forgotten. Write the full identity.
• $(a+b)^{3}\neq a^{3}+b^{3}$ — differ by $3ab(a+b)$; always use the full expansion.
• Confusing Remainder Theorem with Factor Theorem — the remainder is $p(a)$; it is zero exactly when $(x-a)$ is a factor.
• $\sqrt{x}$ and $x^{-1}$ are NOT polynomials — always verify that every exponent is a non-negative integer.
• In trial-and-error factorisation, only testing $x=\pm1$ and missing other rational candidates like $\pm2,\pm3,\pm\dfrac{1}{2}$.
• Applying $a^{3}+b^{3}+c^{3}=3abc$ without first checking that $a+b+c=0$ — this is a conditional result, not always true.
• In Identity 8, forgetting the sign: $(a+b-c)^{2}$ gives $-2bc-2ca$ but $+2ab$.

10. Quick revision checklist

• Polynomial: all exponents are non-negative integers; degree = highest power.
• Zero of $p(x)$: value $a$ where $p(a)=0$; linear poly has exactly one zero $-b/a$.
• Remainder Theorem: remainder when dividing $p(x)$ by $(x-a)$ is simply $p(a)$.
• Factor Theorem: $(x-a)$ is a factor of $p(x)$ if and only if $p(a)=0$.
• Factorise cubics: find one zero by trial from factors of constant term, divide, factorise the resulting quadratic.
• Memorise all 9 identities; immediately recognise $(a+b+c)^{2}$ and the $a^{3}+b^{3}+c^{3}-3abc$ pattern.
• If $a+b+c=0$ then $a^{3}+b^{3}+c^{3}=3abc$ — a favourite 2-mark board question.
• $a^{2}+b^{2}+c^{2}=(a+b+c)^{2}-2(ab+bc+ca)$ — derive $a^{2}+b^{2}+c^{2}$ given the sum and sum-of-products.