Question

What is the binomial expansion of (x + 2)4? x4 + 4x3 + 6x2 + 4x + 1 8x3 + 24x2 + 32x x4 + 8x3 + 24x2 + 32x + 16 2x4 + 8x3 + 12x2 + 8x + 2

Answer 1

The answer to this question can be viewed in the images attached.

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Answer 2

`\boxed{\boxed{(x+2)^4=x^4+8x^3+24x^2+32x+16}}`

Solution-

Given expression is
`(x+2)^4`

Applying Binomial Theorem

`\left(a+b\right)^n=\sum _(i=0)^n\binom{n}{i}a^(\left(n-i\right))b^i`

Here,

a = x, b = 2 and n = 4

So,

`\left(x+2\right)^4=\sum _(i=0)^4\binom{4}{i}x^(\left(4-i\right))\cdot \:2^i`

Expanding the summation

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

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

`=1\cdot x^4\cdot \:1+4\cdot x^3\cdot \:2+6x^2\cdot \:4+4\cdot x\cdot \:8+1\cdot 1\cdot \:16`

`=x^4+8x^3+24x^2+32x+16`