This problem is straightforward if you take it step by step.

First, (x + 2)^4 = (x + 2) (x + 2) (x + 2) (x + 2)

**Step 1** is to begin with (x + 2) (x + 2). To do this, you multiply each of the elements from the first parenthesis times each of the items in the second. So, you get:

x * x = x^2

x * 2 = 2x

2 * x = 2x

2 * 2 = 4

Add these together, and you get *x^2 + 4x + 4*

**Step 2** is to multiply all that first by x:

x^3 + 4x^2 + 4x

and then by 2:

2x^2 + 8x + 8.

Add them all together and you get: *x^3 + 6 x^2 + 12x + 8*.

**Step 3** is to multiply all that by the last (x + 2).

First by x: x^4 + 6x^3 + 12x^2 + 8x

then by 2: 2x^3 + 12 x^2 + 24x +16

Add those two together and simplify for the final answer:

*x^4 + 8x^3 + 24x^2 + 32x +16*

(x+2)^4

You write (x+2) 4 times and then multiply all of them together

(x+2)(x+2)(x+2)(x+2)

You multiply the first 2 first and then when you get the answer for that, you multiply that answer with the next and then the next.

(x+2)(x+2)= x^2+4x+4

(x^2+4x+4)(x+2)= x^3+6x^2+12x+8

(x^3+6x^2+12x+8)(x+2)= x^4+8x^3+24x^2+32x+16

The answer is x^4+8x^3+24x^2+32x+16

Expansion is to basically to write (x+2) four times in a row and multiply them.

Expansion is just stretching it out so an expansion of x+2^4 would be (x+2)(x+2 )(x+2)(x+2). You then multiply. So you get (x^2 +4x+ 4)(x^2+4x+4) which equals to x^4+8x^3+24x^2+32x+16

(x + 2)^4 = (x + 2) (x + 2) (x + 2) (x + 2)

The first thing you do is foil the first 2 parentheses

(x + 2) (x + 2) l (x + 2) (x + 2)

(x + 2) (x + 2)

`x^2+2x+2x+4 `

combine like terms:P

`x^2+4x+4 `

Multiply is with the third parenthesis

`(x^2+4x+4)(x+2) `

`x^3+2x^2+4x^2+8x+4x+8 `

combine like terms

`x^3+6x^2+12x+8 `

multiply it by the last parenthesis

(x^3+6x^2+12x+8) (x+2)

`x^4+2x^3+6x^3+12x^2+12x^2+24x+8x+16 `

combine like terms

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

and that's the answer

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

(x+2)^4

Is the same as multiplying (x + 2) four times, so we get:

(x+2)(x+2)(x+2)(x+2)

Which is the same as multiplying to binomials: ((x+2)(x+2)) ((x+2)(x+2))

We can apply foil to both sides and multiply each term individually.

Use the binomal theorem:

(x+2)^4= x^4 + 4C1(x)^3(2)^1+ 4C2(x)^2(2)^2+4C3(x)^1(2)^3+(2)^4

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

This is much faster than expanding (xt2) four times.

(x+2)(x+2)(x+2)(x+2)

(x²+4x+4)(x²+4x+4)

x^4+4x³+4x²+4x³+16x²+16x+4x²+16x+16

x^4+8x³+20x²+32x+16