Recall the mnemonic FOIL:

When multiplying together two binomials, according to the FOIL method you must find the product of the first terms of both binomials, the product of the terms on the outside of the entire expression, the product of the terms on the inside of the expression, and the product of the last terms of both binomials. The sum of these products is the expanded form of the original expression.

Let us apply the FOIL method to the expression . The first two terms of  are  and . Their product is .

The terms on the outside of the expression  are  and . Their product is .

The terms on the inside of the expression  are  and . Their product is .

The last two terms of  are  and . Their product is .

Adding together each of these products yields

.

Hence, the expanded form of  is .

Use FOIL:

Given

First FOIL the first equation. FOIL means to multiply each term in the first binomial with each term in the second binomial.

Now solve the second equation for .

Substitute the equation for  into the first equation to get a new equation only in terms of .

Distribute to eliminate the parentheses and simplify.

Now use the quadratic formula.

Given a quadratic in the form, 

For this particular question,

From here recall that if the value under the radical sign equals zero than  results in having just one solution.

Therefore, set the value that is under the radical equal to zero and solve for .

Since the binomials containing  are the same, set one equal to zero and solve.

If a problem asks you to expand an expression, you must use the Distributive property. If you are using the FOIL method, you first multiply the first term in each parentheses by each other, followed by the outside terms, then the inside terms, and then the last terms. This is illustrated below. 

First, you multiply  which equals 

Second, you multiply 

Third, you multiply 

Last, you multiply 

Then you simply rearrange them in order of exponents to get 

The problem is basically asking for two binomial expressions to be multiplied and the product to be simplified.

Multiply the two binomials in the definition of the operation using the FOIL method - multiplying each term in the first binomial by each term in the second, as follows:

F(irst): 

O(uter): 

I(nner): 

L(last): 

Add the terms and collect like terms:

The domain means what real number can you plug in that would still make the function work. For this case, we have to worry about the denominator so that it does not equal 0, so we solve the following. 7x – 1 = 0, 7x = 1, x = 1/7, so when x ≠ 1/7 the function will work.

The domain of a function or relation is the set of all possible x-values. Here that is simply a list of the x-coordinates of all of the coordinate pairs. So the domain is {–2, –3, 2, 4}.

Because x and y must be negative integers greater than negative five, then x and y can only be equal to the following values:

x can equal -4, -3, -2, or -1

y can equal -4, -3, -2, or -1

Now we can try all of the combinations of x and y, and see what x # y would equal. It is helpful to note that x # y is the same as y # x because 2yx + y + x = 2xy + x  + y. This means that the order of x and y doesn't matter. 

-4 # -4 = 2(-4)(-4) + -4 + -4 = 24

-4 # -3 = 2(-4)(-3) + -4 + -3 = 17

-4 # -2 = 10

-4 # -1 = 3

-3 # -3 = 12

-3 # -2 = 7

-3 # -1 = 2

-2 # -2 = 4

-2 # -1 = 1

-1 # -1 = 0

We don't need to find -3 # -4, -2 # -4, etc, because x # y = y # x . 

The smallest value of x # y must be 0.

The domain of a function includes all of the values of x for which that function is defined. In other words, the domain is all of the real values of x that will produce a real number. Let's look at the domains of each function one at a time.

First, let's examine

In general, when we are examining the domain of a function, we want to find places where we end up with zeros in the denominators or square-roots of negative numbers. For example, in the function , if we let x = 4, then we would be forced to evaluate 1/0, which isn't possible. We can never divide by zero. Thus, this function is not defined over all real values of x. We can eliminate it from the answer choices.

Next, let's look at . Let's set the denominator equal to zero to see if there are any values of x which might give us a zero in the denominator.

Subtract one from both sides.

Take the cube root of both sides.

Thus, if x = –1, then f(x) will be equal to 1/0, which isn't defined, because we can't divide by zero. Therefore, we can eliminate this answer choice.

Now, let's analyze .

We can never take the square root of a negative number. Thus, if , then f(x) won't be defined. For example, if x = 4, then , which would produce an imaginary number. Therefore, this function can't be the correct answer.

Next, let's look at . It will help us to rewrite f(x) in a form using square roots. In general, . As a result, we can rewrite f(x) as follows:

. In this form, we can see that if is negative, then f(x) won't be defined. Thus, if x = –2, for example, we would be forced to find the square root of –8, which produces an imaginary result. So, this function isn't the correct answer either.

By process of elimination the answer must be . The reasons that this function is defined for all values of x is because the denominator can never be zero. Thus, we can pick any value of x from negative to positive infinity, and we will get a real value for f(x).

The answer is