\(\displaystyle \small (7x+2)(-3x+1)\)

First: \(\displaystyle \small (7x)(-3x)=-21x^2\)

Outside: \(\displaystyle \small (7x)(1)=7x\)

Inside: \(\displaystyle \small (2)(-3x)=-6x\)

Last: \(\displaystyle \small (2)(1)=2\)

Add together:

\(\displaystyle \small -21x^2+7x+(-6x)+2=-21x^2+x+2\)

\(\displaystyle (3x+2)(-x-4)\)

FOIL:

First: \(\displaystyle (3x)(-x)=-3x^2\)

Outer:\(\displaystyle (3x)(-4)=-12x\)

Inner: \(\displaystyle (2)(-x)=-2x\)

Last: \(\displaystyle (2)(-4)=-8\)

Add these together and combine like terms:

\(\displaystyle -3x^2-12x-2x-8=-3x^2-14x-8\)

This is a FOIL-ing problem. First, set up the numbers in a form we can use to create the function.

Take the opposite sign of each of the numbers and place them in this format. \(\displaystyle (x-5)(x+8)\)

Multiply the \(\displaystyle x\) in the first parentheses by the \(\displaystyle x\) and 8 in the second parentheses respectively to get \(\displaystyle x^{2}+8x\)

Multiply the \(\displaystyle -5\) in the first parentheses by the \(\displaystyle x\) and 8 in the second parentheses as well to give us \(\displaystyle -5x-40\).

Then add them together to get \(\displaystyle x^{2}+8x-5x-40\)

Combine like terms to find the answer which is \(\displaystyle x^{2}+3x-40\).

Simplify using FOIL method.

Remember that multiplying variables means adding their exponents.

F: \(\displaystyle 3x^{2}*2x^{3} = 6x^{5}\)

O: \(\displaystyle 3x^{2}*(-8) = -24x^{2}\)

I: \(\displaystyle -3 *2x^{3}=-6x^{3}\)

L: \(\displaystyle -3 *-8= 24\)

Combine the terms. Note that we cannot simplify further, as the exponents do not match and cannot be combined.

\(\displaystyle 6x^{5}-6x^{3}-24x^{2}+24\)

The FOIL method yields the products below.

First: \(\displaystyle 4x* 2x=8x^{2}\)

Outside: \(\displaystyle 4x* 6=24x\)

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

Last: \(\displaystyle -7* 6=-42\)

Add these four terms, and combine like terms, to obtain the product of the binomials.

\(\displaystyle 8x^{2}+24x+(-14x)+(-42)=8x^{2}+10x-42\)

\(\displaystyle x^3-3x^2-18x\)

First, factor out an \(\displaystyle x\), since it is present in all terms.

\(\displaystyle x(x^2-3x-18)\)

We need two factors that multiply to \(\displaystyle -18\) and add to \(\displaystyle -3\).

\(\displaystyle -6*3=-18\) and \(\displaystyle -6+3=-3\)

Our factors are \(\displaystyle -6\) and \(\displaystyle +3\).

\(\displaystyle x(x-6)(x+3)\)

We can check our answer using FOIL to get back to the original expression.

First: \(\displaystyle (x)(x)=x^2\)

Outside: \(\displaystyle (x)(3)=3x\)

Inside: \(\displaystyle (x)(-6)=-6x\)

Last: \(\displaystyle (-6)(3)=18\)

Add together and combine like terms.

\(\displaystyle x^2+3x-6x-18=x^2-3x-18\)

Distribute the \(\displaystyle x\) that was factored out first.

\(\displaystyle x(x^2-3x-18)=x^3-3x^2-18x\)

Using the FOIL method is simple. FOIL stands for First, Outside, Inside, Last. This is to help us make sure we multiply every term correctly looking at the terms inside of each parentheses. We follow FOIL to find the multiplied terms, then combine and simplify. 

First, stands for multiply each first term of the seperate polynomials. In this case, \(\displaystyle 2x * x=2x^2\). 

Inner means we multiply the two inner terms of the expression. Here it's \(\displaystyle y*x = xy\).

Outer means multiplying the two outer terms of the expression. For this expression we have \(\displaystyle 2x*3y=6xy\).

Last stands for multiplying the last terms of the polynomials. So here it's \(\displaystyle y*2y=2y^2\).

Finally we combine the like terms together to get

\(\displaystyle 2x^2+xy+6xy+3y^2 = 2x^2+7xy+3y^2\).

This problem involves multiplying two binomials. To solve, we will need to use the FOIL method.

\(\displaystyle (a+b)(c+d)=ac+ad+bc+bd\)

Comparing this with our original equation, \(\displaystyle a=2x\), \(\displaystyle b=-5y\), \(\displaystyle c=x\), and \(\displaystyle d=6y\).

Using these values, we can substitute for the FOIL equation.

\(\displaystyle (2x-5y)(x+6y)=(2x)(x)+(2x)(6y)+(-5y)(x)+(-5y)(6y)\)

\(\displaystyle (2x-5y)(x+6y)=2xx+2x6y-5yx-5y6y\)

\(\displaystyle (2x-5y)(x+6y)=2x^2+12xy-5xy-30y^2\)

Notice that the two center terms use the same variables; this allows us to combine like terms.

\(\displaystyle (2x-5y)(x+6y)=2x^2+7xy-30y^2\)

To solve, it may be easier to convert the radicals to exponents.

\(\displaystyle (r^{\frac{1}{2}}-3^{\frac{1}{2}})(r^{\frac{1}{2}}+3^\frac{1}{2})\)

Remember, the method used in multiplying two binomials is given by the equation:

\(\displaystyle (a+b)(c+d)=ac+ad+bc+bd\)

Comparing this with our expression, we can identify the following variables:

\(\displaystyle a=r^{\frac{1}{2}}\)

\(\displaystyle b=-3^{\frac{1}{2}}\)

\(\displaystyle c=r^{\frac{1}{2}}\)

\(\displaystyle d=3^{\frac{1}{2}}\)

We can substitute these values into the FOIL expression. Multiply to simplify.

\(\displaystyle (r^{\frac{1}{2}}-3^{\frac{1}{2}})(r^{\frac{1}{2}}+3^\frac{1}{2})=(r^{\frac{1}{2}})(r^{\frac{1}{2}})+(r^{\frac{1}{2}})(3^{\frac{1}{2}})+(-3^{\frac{1}{2}})(r^{\frac{1}{2}})+(-3^{\frac{1}{2}})(3^{\frac{1}{2}})\)

\(\displaystyle (r^{\frac{1}{2}}-3^{\frac{1}{2}})(r^{\frac{1}{2}}+3^\frac{1}{2})=r^{\frac{1}{2}}r^{\frac{1}{2}}+3^{\frac{1}{2}}r^{\frac{1}{2}}-3^{\frac{1}{2}}r^{\frac{1}{2}}-3^{\frac{1}{2}}3^{\frac{1}{2}}\)

Simplify by combining like terms. The center terms are equal and opposite, allowing them to cancel to zero.

\(\displaystyle (r^{\frac{1}{2}}-3^{\frac{1}{2}})(r^{\frac{1}{2}}+3^\frac{1}{2})=r^{\frac{1}{2}}r^{\frac{1}{2}} -3^{\frac{1}{2}}3^{\frac{1}{2}}\)

A term to a given power can be combined with another term with the same base using the identity \(\displaystyle a^ba^c=a^{b+c}\). This allows us to adjust our final answer.

\(\displaystyle (r^{\frac{1}{2}}-3^{\frac{1}{2}})(r^{\frac{1}{2}}+3^\frac{1}{2})=r^{\frac{1}{2}+\frac{1}{2}} -3^{\frac{1}{2}+\frac{1}{2}}\)

\(\displaystyle (r^{\frac{1}{2}}-3^{\frac{1}{2}})(r^{\frac{1}{2}}+3^\frac{1}{2})=r -3\)

Using the FOIL distribution method:

\(\displaystyle (x+2)(x-2)\)

First: \(\displaystyle x*x = x^2\)

Outer: \(\displaystyle x*-2= -2x\)

Inner: \(\displaystyle 2*x = 2x\)

Last: \(\displaystyle 2*-2 = -4\)

Resulting in: \(\displaystyle x^2 - 2x + 2x -4\)

Combining like terms, the \(\displaystyle x\)'s cancel for a final answer of:

\(\displaystyle x^{2} - 4\)