# Binomial expansion

algebraic expansion of powers of a binomial

Binomial expansion uses an expression to make a series. It uses a bracket expression like ${\displaystyle (x+y)^{n}}$. There are three binomial expansions.

## The formulas

There are basically three binomial expansion formulas:

 ${\displaystyle (a+b)^{2}=a^{2}+2ab+b^{2}}$ 1st (Plus) ${\displaystyle (a-b)^{2}=a^{2}-2ab+b^{2}}$ 2nd (Minus) ${\displaystyle (a+b)\cdot (a-b)=a^{2}-b^{2}}$ 3rd (Plus-Minus)

We can explain why there are such 3 formulas with a simple expansion of the product:

${\displaystyle (a+b)^{2}=(a+b)\cdot (a+b)=a\cdot a+a\cdot b+b\cdot a+b\cdot b=a^{2}+2\cdot a\cdot b+b^{2}}$
${\displaystyle (a-b)^{2}=(a-b)\cdot (a-b)=a\cdot a-a\cdot b-b\cdot a+b\cdot b=a^{2}-2\cdot a\cdot b+b^{2}}$
${\displaystyle (a+b)\cdot (a-b)=a\cdot a-a\cdot b+b\cdot a-b\cdot b=a^{2}-b^{2}}$

## Using Pascal's triangle

If ${\displaystyle n}$  is an integer (${\displaystyle n\in \mathbb {Z} }$ ), we use Pascal's triangle.

To expand ${\displaystyle (x+y)^{2}}$ :

• find row 2 of Pascal's triangle (1, 2, 1)
• expand ${\displaystyle x}$  and ${\displaystyle y}$  so the ${\displaystyle x}$  power goes down by 1 each time from ${\displaystyle n}$  to 0 and the ${\displaystyle y}$  power goes up by 1 each time from 0 up to ${\displaystyle n}$
• times the numbers from Pascal's triangle with the right terms.

So ${\displaystyle (x+y)^{2}=1x^{2}y^{0}+2x^{1}y^{1}+1x^{0}y^{2}}$

For example:

${\displaystyle (3+2x)^{2}=1\cdot 3^{2}\cdot (2x)^{0}+2\cdot 3^{1}\cdot (2x)^{1}+1\cdot 3^{0}\cdot (2x)^{2}=9+12x+4x^{2}}$

So as a rule:

${\displaystyle (x+y)^{n}=a_{0}x^{n}y^{0}+a_{1}x^{n-1}y^{1}+a_{2}x^{n-2}y^{2}+\cdots +a_{n-1}x^{1}y^{n-1}+a_{n}x^{0}y^{n}}$

where ${\displaystyle a_{i}}$  is the number at row ${\displaystyle n}$  and position ${\displaystyle i}$  in Pascal's triangle.

### Examples

${\displaystyle (5+3x)^{3}=1\cdot 5^{3}\cdot (3x)^{0}+3\cdot 5^{2}\cdot (3x)^{1}+3\cdot 5^{1}\cdot (3x)^{2}+1\cdot 5^{0}\cdot (3x)^{3}}$
${\displaystyle =125+75\cdot 3x+15\cdot 9x^{2}+1\cdot 27x^{3}=125+225x+135x^{2}+27x^{3}}$

${\displaystyle (5-3x)^{3}=1\cdot 5^{3}\cdot (-3x)^{0}+3\cdot 5^{2}\cdot (-3x)^{1}+3\cdot 5^{1}\cdot (-3x)^{2}+1\cdot 5^{0}\cdot (-3x)^{3}}$
${\displaystyle =125+75\cdot (-3x)+15\cdot 9x^{2}+1\cdot (-27x^{3})=125-223x+135x^{2}-27x^{3}}$

${\displaystyle (7+4x^{2})^{5}=1\cdot 7^{5}\cdot (4x^{2})^{0}+5\cdot 7^{4}\cdot (4x^{2})^{1}+10\cdot 7^{3}\cdot (4x^{2})^{2}+10\cdot 7^{2}\cdot (4x^{2})^{3}+5\cdot 7^{1}\cdot (4x^{2})^{4}+1\cdot 7^{0}\cdot (4x^{2})^{5}}$
${\displaystyle =16807+12005\cdot 4x^{2}+3430\cdot 16x^{4}+490\cdot 64x^{6}+35\cdot 256x^{8}+1\cdot 1024x^{10}}$
${\displaystyle \,=16807+48020x^{2}+54880x^{4}+31360x^{6}+8960x^{8}+1024x^{10}}$