Polynomial Identities and the Binomial Theorem

Equations that are always true

Most equations are true for some values and false for others: $2x + 1 = 7$ holds only when $x = 3$. An identity holds for every value.

$$(a + b)^2 = a^2 + 2ab + b^2$$

Pick any $a$ and $b$ and both sides agree. That is a claim about infinitely many cases, which is why checking a few numbers cannot establish it.

Proving an identity means rewriting one side into the other, one justified step at a time:

$$(a + b)^2 = (a + b)(a + b) = a^2 + ab + ba + b^2 = a^2 + 2ab + b^2 \quad\checkmark$$

Each step used the distributive property or the commutative property, both of which hold for every number. So the result holds for every number too.

Identities worth knowing

Name	Identity
square of a sum	$(a + b)^2 = a^2 + 2ab + b^2$
square of a difference	$(a - b)^2 = a^2 - 2ab + b^2$
difference of squares	$a^2 - b^2 = (a - b)(a + b)$
sum of cubes	$a^3 + b^3 = (a + b)(a^2 - ab + b^2)$
difference of cubes	$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$

The cube identities are proved the same way, by multiplying out the right side:

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

The middle terms cancel in pairs — that cancellation is what the particular middle factor $a^2 - ab + b^2$ was chosen to produce.

Why identities explain numerical patterns

Identities do real work on numbers.

Mental multiplication. $49 \times 51 = (50 - 1)(50 + 1)$, a difference of squares:

$$50^2 - 1^2 = 2500 - 1 = 2499$$

Pythagorean triples. Expand both sides of

$$(x^2 - y^2)^2 + (2xy)^2 = (x^2 + y^2)^2$$

The left is $x^4 - 2x^2y^2 + y^4 + 4x^2y^2 = x^4 + 2x^2y^2 + y^4$, which is exactly the right side. So for any whole numbers $x > y$, the three numbers $x^2 - y^2$, $2xy$ and $x^2 + y^2$ satisfy the Pythagorean theorem.

$x$	$y$	$x^2 - y^2$	$2xy$	$x^2 + y^2$
$2$	$1$	$3$	$4$	$5$
$3$	$2$	$5$	$12$	$13$
$4$	$1$	$15$	$8$	$17$
$4$	$3$	$7$	$24$	$25$

One line of algebra produces an unlimited supply of right triangles with whole-number sides. The identity does more than confirm the pattern; it is the reason the pattern exists.

Pascal’s triangle

Expand $(x + y)^n$ for small $n$ and look only at the coefficients:

$n$	Coefficients of $(x + y)^n$
$0$	$1$
$1$	$1 \;\; 1$
$2$	$1 \;\; 2 \;\; 1$
$3$	$1 \;\; 3 \;\; 3 \;\; 1$
$4$	$1 \;\; 4 \;\; 6 \;\; 4 \;\; 1$
$5$	$1 \;\; 5 \;\; 10 \;\; 10 \;\; 5 \;\; 1$

This is Pascal’s triangle. Each row starts and ends with $1$, and each inner number is the sum of the two above it: $4 + 6 = 10$.

The addition rule comes from multiplying by one more $(x + y)$. Each term of the next power is made in two ways — from a term times $x$ and from a neighboring term times $y$ — so its coefficient is the sum of the two coefficients above it.

The binomial theorem

$$(x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k$$

The coefficients are the combinations $\binom{n}{k}$. Why combinations? Multiplying out $(x + y)^n$ means choosing $x$ or $y$ from each of the $n$ factors. The term $x^{n-k}y^k$ appears once for every way of choosing which $k$ factors supply the $y$ — and that number is $\binom{n}{k}$.

Expand $(x + 2)^4$.

Row $4$ gives $1, 4, 6, 4, 1$. The powers of $x$ fall from $4$ to $0$ while the powers of $2$ rise from $0$ to $4$:

$$x^4 + 4x^3(2) + 6x^2(2^2) + 4x(2^3) + 2^4 = x^4 + 8x^3 + 24x^2 + 32x + 16$$

Worked examples

Common mistakes

Practice problems