Step One: Expand

\(\displaystyle (5m+2)^2=(5m+2)(5m+2)\)

Step 2: Use the FOIL method

First: \(\displaystyle 5m\cdot 5m=25m^2\)

Outside: \(\displaystyle 5m\cdot 2=10m\)

Inside: \(\displaystyle 5m\cdot 2=10m\)

Last: \(\displaystyle 2\cdot 2=4\)

Sum the products:

\(\displaystyle 25m^2+10m+10m+4\)

\(\displaystyle 25m^2+20m+4\)

Use the FOIL method to find the product.  The answer must be in standard form.

\(\displaystyle (x^3+2x)(x^6+9x^2)\)

Step 1: Use the FOIL method

First: \(\displaystyle x^3\cdot x^6=x^9\)

Outside: \(\displaystyle x^3\cdot9x^2=9x^5\)

Inside: \(\displaystyle 2x\cdot x^6=2x^7\)

Last: \(\displaystyle 2x \cdot 9x^2=18x^3\)

Add these to find the product:

\(\displaystyle x^9+9x^5+2x^7+18x^3\)

Step 2: Write the product in standard form

Standard form means the terms are written from highest degree to lowest degree. You find the degree of a term by adding the exponents in the term.

Therefore:

\(\displaystyle x^9+2x^7+9x^5+18x^3\)

Simplify:

\(\displaystyle (4r^2+2s^3)^2\)

Expand the equation and use the FOIL method:

\(\displaystyle (4r^2+2s^3)(4r^2+2s^3)\)

First: \(\displaystyle 4r^2*4r^2=16r^4\)

Outside: \(\displaystyle 4r^2*2s^3=8s^3r^2\)

Inside: \(\displaystyle 2s^3*4r^2=8s^3r^2\)

Last: \(\displaystyle 2s^3*2s^3=4s^6\)

Sum the terms:

\(\displaystyle 16r^4+8s^3r^2+8s^3r^2+4s^6\)

\(\displaystyle 16r^4+16s^3r^2+4s^6\)

FOIL stands for First, Inside, Outside, Last. So multiply the first terms together:
\(\displaystyle x*2x = 2x^2\)

The outside terms:

\(\displaystyle x*(-3)=-3x\)

The inside terms:

\(\displaystyle -4*2x = -8x\)

and the last terms:

\(\displaystyle (-4)*(-3)=12\)

Now combine like terms:

\(\displaystyle 2x^2 + -3x + -8x +12 = 2x^2 -11x+12\)

In order to get the \(\displaystyle ax^2 + bx + c\) form, we must FOIL out \(\displaystyle (2x-5)^2\).

FOIL is technique for distributing two binomials. The letters stand for First, Outer, Inner, and Last.

"First" stands for multiply the terms which occur first in each binomial.

"Outer" stands for multiply the outermost terms in the product.

"Inner" stands for multiply the innermost two terms.

"Last" stands for multiply the terms which occur last in each binomial.

Then, we must simplify the like terms, as shown below:

\(\displaystyle (2x-5)(2x-5) = 4x^2 - 10x - 10x +25 = 4x^2 - 20x +25\).

Here, \(\displaystyle a=4\), \(\displaystyle b=-20\), while \(\displaystyle c=25\), so \(\displaystyle (4) +(-20) + (25) = 9\).

Use FOIL to multiply the products. 
First terms: \(\displaystyle x*2x=2x^2\)

Outside terms: \(\displaystyle x*(-4)=-4x\)

Inside terms: \(\displaystyle 7*2x = 14x\)

Last terms: \(\displaystyle 7*(-4)=-28\)
Now combine like terms: \(\displaystyle 2x^2-4x+14x -28 = 2x^2+10x-28\)

Use FOIL (First Outside Inside Last) to multiply the binomials:

\(\displaystyle 2x*x = 2x^2\)

\(\displaystyle 2x*8 = 16x\)

\(\displaystyle -4*x = -4x\)

\(\displaystyle -4*8 = -32\)

now combine like terms and you get the result: \(\displaystyle 2x^2+12x-32\)

To reach the solution, use the FOIL method of the distributive property. This stands for First, Outer, Inner, Last. In this particular problem, we begin by multiplying the first two \(\displaystyle \textup{x's}\) together, yielding \(\displaystyle {\textup{x}}^{2}\). Then, we multiply \(\displaystyle \textup{x}\) by the outer number, which is \(\displaystyle 3\), yielding \(\displaystyle \textup{3x}\). Next, we multiply the two inner numbers, \(\displaystyle 5\) and \(\displaystyle \textup{x}\), yielding \(\displaystyle \textup{5x}\). Finally, we multiply the last two numbers, \(\displaystyle 5\) and \(\displaystyle 3\), yielding \(\displaystyle 15\). Thus, when we add all those together, the answer is  \(\displaystyle \textup{x}^{2}+\textup{8x}+15\).

To FOIL using the distributive property, multiply the first terms together, then the outer terms, then the inner terms, and finally, the last terms. 

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

To FOIL using the distributive property, multiply the first terms together, then the outer terms, then the inner terms, and finally, the last terms.

\(\displaystyle (6x-2)^2=36x^2-12x-12x+4=36x^2-24x+4\)