Question

What is the expansion of (3+x)^4

Answer 1

`\left(3+x\right)^4:\quad x^4+12x^3+54x^2+108x+81`

Explanation:

Considering the expression

`\left(3+x\right)^4`

Lets determine the expansion of the expression

`\left(3+x\right)^4`

`\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _(i=0)^n\binom{n}{i}a^(\left(n-i\right))b^i`

`a=3,\:\:b=x`

`=\sum _(i=0)^4\binom{4}{i}\cdot \:3^(\left(4-i\right))x^i`

Expanding summation

`\binom{n}{i}=(n!)/(i!\left(n-i\right)!)`

`i=0\quad :\quad (4!)/(0!\left(4-0\right)!)3^4x^0`

`i=1\quad :\quad (4!)/(1!\left(4-1\right)!)3^3x^1`

`i=2\quad :\quad (4!)/(2!\left(4-2\right)!)3^2x^2`

`i=3\quad :\quad (4!)/(3!\left(4-3\right)!)3^1x^3`

`i=4\quad :\quad (4!)/(4!\left(4-4\right)!)3^0x^4`

`=(4!)/(0!\left(4-0\right)!)\cdot \:3^4x^0+(4!)/(1!\left(4-1\right)!)\cdot \:3^3x^1+(4!)/(2!\left(4-2\right)!)\cdot \:3^2x^2+(4!)/(3!\left(4-3\right)!)\cdot \:3^1x^3+(4!)/(4!\left(4-4\right)!)\cdot \:3^0x^4`

`=(4!)/(0!\left(4-0\right)!)\cdot \:3^4x^0+(4!)/(1!\left(4-1\right)!)\cdot \:3^3x^1+(4!)/(2!\left(4-2\right)!)\cdot \:3^2x^2+(4!)/(3!\left(4-3\right)!)\cdot \:3^1x^3+(4!)/(4!\left(4-4\right)!)\cdot \:3^0x^4`

as

`(4!)/(0!\left(4-0\right)!)\cdot \:\:3^4x^0:\:\:\:\:\:\:81`

`(4!)/(1!\left(4-1\right)!)\cdot \:3^3x^1:\quad 108x`

`(4!)/(2!\left(4-2\right)!)\cdot \:3^2x^2:\quad 54x^2`

`(4!)/(3!\left(4-3\right)!)\cdot \:3^1x^3:\quad 12x^3`

`(4!)/(4!\left(4-4\right)!)\cdot \:3^0x^4:\quad x^4`

so equation becomes

`=81+108x+54x^2+12x^3+x^4`

`=x^4+12x^3+54x^2+108x+81`

Therefore,

• 
  `\left(3+x\right)^4:\quad x^4+12x^3+54x^2+108x+81`