Step 1: Evaluate \(\displaystyle (x+2)^2\).

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

Step 2. Evaluate \(\displaystyle (x+2)^3\)

From the previous step, we already know what \(\displaystyle (x+2)^2\) is. 

\(\displaystyle (x+2)^3\) is just multiplying by another \(\displaystyle (x+2)\)

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

Step 3: Evaluate \(\displaystyle (x+2)^4\). 

\(\displaystyle (x+2)^4=(x+2)^3(x+2)\)

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

The expansion of \(\displaystyle (x+2)^4\) is \(\displaystyle x^4+8x^3+24x^2+32x+16\)

Solution:

We can look at Pascal's Triangle, which is a quick way to do Binomial Expansion. We read each row (across, left to right)

For the first row, we only have a constant.
For the second row, we get \(\displaystyle x+constant\).
...
For the 7th row, we will start with an \(\displaystyle x^6\) term and end with a constant.

Step 1: We need to locate the 7th row of the triangle and write the numbers in that row out.

The 7th row is \(\displaystyle 1,6,15,20,15,6,1\).

Step 2: If we translate the 7th row into an equation, we get:

 \(\displaystyle x^6+6x^5+15x^4+20x^3+15x^2+6x+1\). This is the solution.

Method One:

We will start expanding slowly, and we will end up at exponent \(\displaystyle 7\)

Step 1: Expand: \(\displaystyle (x-1)^2\)

\(\displaystyle (x-1)^2=(x-1)(x-1)\)
\(\displaystyle (x-1)(x-1)=x^2-x-x+1\)
\(\displaystyle (x-1)^2={\color{Blue} x^2-2x+1}\)

Step 2: Multiply \(\displaystyle (x-1)\) by the product of \(\displaystyle (x-1)^2\). By doing this, we are now expanding \(\displaystyle (x-1)^3\).

\(\displaystyle (x-1)^3=(x^2-2x+1)(x-1)\)
\(\displaystyle (x-1)^3=x^3-x^2-2x^2+2x+x-1\)
\(\displaystyle (x-1)^3={\color{Red} x^3-3x^2+3x-1}\)

Step 3: Multiply by \(\displaystyle (x-1)\) again

\(\displaystyle (x-1)^3(x-1)=(x-1)^4\)
\(\displaystyle (x-1)^4=(x^3-3x^2+3x-1)(x-1)\)
\(\displaystyle (x-1)^4=x^4-x^3-3x^3+3x^2+3x^2-3x-x+1\)
\(\displaystyle (x-1)^4={\color{Magenta} x^4-4x^3+6x^2-4x+1}\).

After Step 4: \(\displaystyle (x-1)^5={\color{Orange}x^5-5x^4+10x^3-10x^2+5x-1}\)
After Step 5: \(\displaystyle (x-1)^6={\color{Pink} x^6-6x^5+15x^4-20x^3+15x^2-6x+1}\)

After Step 6, the final answer is:

\(\displaystyle (x-1)^7={\color{DarkGreen} x^7-7x^6+21x^5-35x^4+35x^3-21x^2+7x-1}\).

Method Two:

You can find the expansion of this binomial by using the Pascal's Triangle (shown below)

If you look at Row \(\displaystyle 8\) of the triangle above, the row that starts with \(\displaystyle 1,7,...\).

We need to negate every \(\displaystyle 2\)nd term, as the answer in Method One has every even term negative. 

We will still get the answer: \(\displaystyle x^6-7x^6+21x^5-35x^34+35x^3-21x^2+7x-1\).

Step 1: Expand \(\displaystyle (x+1)^3\)

\(\displaystyle (x+1)^3=(x+1)(x+1)(x+1)\)

Step 2: FOIL the first two parentheses:

\(\displaystyle (x+1)(x+1)=x^2+x+x+1=x^2+2x+1\)

Step 3: Multiply the expansion in step 2 by \(\displaystyle (x+1)\):

\(\displaystyle (x^2+2x+1)(x+1)=x^3+x^2+2x^2+2x+x+1=x^3+3x^2+3x+1\)

The expanded form of \(\displaystyle (x+1)^3\) is \(\displaystyle x^3+3x^2+3x+1\).

Step 1: Multiply \(\displaystyle (x-3)^2\)

\(\displaystyle (x-3)^2=(x-3)(x-3)=x^2-3x-3x+9={\color{Red} x^2-6x+9}\)

Step 2: Multiply the result in step 1 by \(\displaystyle (x-3)\)

\(\displaystyle (x^2-6x+9)(x-3)=x^3-3x^2-6x^2+18x+9x-27={\color{Blue} x^3-9x^2+27x-27}\)

Step 3: Multiply the result of step 2 by \(\displaystyle (x-3)\)

\(\displaystyle (x^3-9x^2+27x-27)(x-3)=x^4-3x^3-9x^3+27x^2+27x^2-81x-27x+81\)
Simplify:

\(\displaystyle x^4-12x^2+54x^2-108x+81\)

\(\displaystyle (x-3)^4=x^4-12x^2+54x^2-108x+81\)

\(\displaystyle In\ order\ to\ square\ a\ binomial\ one\ must\ rewrite\ it\ twice\ and\ then\ distribute\ all\ terms.\)

\(\displaystyle (3x+4)^2=(3x+4)(3x+4)\)

\(\displaystyle (3x+4)(3x+4)=3x(3x+4)+4(3x+4)\)

\(\displaystyle 3x(3x+4)+4(3x+4)=9x^2+12x+12x+16\)

\(\displaystyle 9x^2+12x+12x+16=9x^2+24x+16\)

\(\displaystyle The\ expansion\ is: 9x^2+24x+16\)

The easiest way to expand binomials raised to higher powers is to use Pascal's Triangle. 

Pascal's Triangle is used to find the multipliers for each level of exponent. 

It follows the pattern listed below. 

To complete the expansion we will take the row that corresponds to the 4th exponent for this problem. 

\(\displaystyle (x+3)^4\)

We will now organize this into columns and rows.

The second and third rows are organized by taking the left term from the highest to lowest power, and the right term from lowest to highest power. 

Since the x is on the left, it is raised to the 4th power, 3rd power and so on. 

Since the 3 is on the right, it is raised to the 0 power, 1 power and so on. 

\(\displaystyle 1\ \ \ \ 4\ \ \ 6\ \ \ 4\ \ \ \ 1\)

\(\displaystyle x^4\ \ x^3\ \ x^2\ \ x^1\ \ x^0\)

\(\displaystyle 3^0\ \ 3^1\ \ 3^2\ \ 3^3\ \ 3^4\)

We now simplify each of the terms, the bottom row is the only one to be simplified in this case. Anything to the zero power except zero is 1. 

\(\displaystyle 1\ \ \ \ 4\ \ \ 6\ \ \ 4\ \ \ \ 1\)

\(\displaystyle x^4\ \ x^3\ \ x^2\ \ x^1\ \ 1\)

\(\displaystyle 1\ \ \ 3\ \ \ 9\ \ \ 27\ \ 81\)

Now we multiply each column together to obtain the full expansion. 

For example to obtain the third term we multiplied everything in the 3rd column: 

\(\displaystyle 6*x^2*9=54x^2\)

We did this for all of the columns to get the below final answer. 

\(\displaystyle x^4+12x^3+54x^2+108x+81\)

Let's start with a smaller expansion:

\(\displaystyle \large \large (1-x)^2=(1-2x+x^2)\)
We multiply the expansion of \(\displaystyle \large (1-x)^2\) by \(\displaystyle \large (1-x)\):

\(\displaystyle \large (1-2x+x^2)(1-x)=1-x-2x+2x^2+x^2-x^3\)

\(\displaystyle \large \large =-x^3+3x^2-3x+1\Rightarrow (1-x)^3\)

Multiply again by \(\displaystyle \large (1-x)\):

\(\displaystyle \large (1-x)^3\cdot(1-x)= (-x^3+3x^2-3x+1)(1-x)\)

\(\displaystyle \large =-x^3+x^4-3x^2-3x^3-3x+3x^2+1-x\)

\(\displaystyle \large =x^4-4x^3+6x^2-4x+1\)

Multiply by \(\displaystyle \large (1-x)^2\):

\(\displaystyle \large (x^4-4x^3+6x^2-4x+1)\cdot(1-2x+x^2)\)

\(\displaystyle \large =x^4-x^5+x^6-4x^3+8x^4-4x^5+6x^2-12x^3+6x^4-4x+8x^2-8x^3+1-2x+x^2\)

\(\displaystyle \large =x^6-6x^5+15x^4-20x^3+15x^2-6x+1=(1-x)^6\)

\(\displaystyle (3x+7)(5x+4)\)

Expand by distributing each of the factors

\(\displaystyle 3x(5x)+3x(4)+7(5x)+7(4)\)

Simplify

\(\displaystyle 15x^{2}+12x+35x+28\)

Simplify

\(\displaystyle 15x^{2}+47x+28\)

Step 1: Let's start small, expand \(\displaystyle \large (x+1)^2\).

\(\displaystyle \large (x+1)^2=(x+1)(x+1)=x^2+x+x+1=x^2+2x+1\)

Step 2: Expand \(\displaystyle \large (x+1)^3\)

Take the final answer of Step 1 and multiply it by \(\displaystyle \large (x+1)\)...

\(\displaystyle \large (x^2+2x+1)(x+1)=x^3+x^2+2x^2+2x+x+1=x^3+3x^2+3x+1\)

Step 3: Multiply again by \(\displaystyle \large (x+1)\) to the final answer of Step 2...

\(\displaystyle \large (x^3+3x^2+3x+1)(x+1)=x^4+x^3+3x^3+3x^2+3x^2+3x+x+1\)

\(\displaystyle \large =x^4+4x^3+6x^2+4x+1\)