Find all factors of 24

1, 2, 3,4, 6, 8, 12, 24

Now find two factors that add up to  and multiply to ;  and  are the two factors.

By factoring, you can set the equation to be 

If you FOIL it out, it gives you .

Set each part of the equation equal to 0, and solve for .

 and

 and

The numbers by which x⁶ – y⁶ is divisible will be all of its factors. In other words, we need to find all of the factors of x⁶ – y⁶ , which essentially means we must factor x⁶ – y⁶ as much as we can.

First, we will want to apply the difference of squares rule, which states that, in general, a² – b² = (a – b)(a + b). Notice that a and b are the square roots of the values of a² and b², because √a² = a, and √b² = b (assuming a and b are positive). In other words, we can apply the difference of squares formula to x⁶ – y⁶ if we simply find the square roots of x⁶ and y⁶.

Remember that taking the square root of a quantity is the same as raising it to the one-half power. Remember also that, in general, (a^b)^c = a^bc.

√x⁶ = (x⁶)^(1/2) = x^(6(1/2)) = x³

Similarly, √y⁶ = y³.

Let's now apply the difference of squares factoring rule.

x⁶ – y⁶ = (x³ – y³)(x³ + y³)

Because we can express x⁶ – y⁶ as the product of (x³ – y³) and (x³ + y³), both (x³ – y³) and (x³ + y³) are factors of x⁶ – y⁶ . Thus, we can eliminate x³ – y³ from the answer choices.

Let's continue to factor (x³ – y³)(x³ + y³). We must now apply the sum of cubes and differences of cubes formulas, which are given below:

In general, a³ + b³ = (a + b)(a² – ab + b²). Also, a³ – b³ = (a – b)(a² + ab + b²)

Thus, we have the following:

(x³ – y³)(x³ + y³) = (x – y)(x² + xy + y²)(x + y)(x² – xy + y²)

This means that x – y and x + y are both factors of x⁶ – y⁶ , so we can eliminate both of those answer choices.

We can rearrange the factorization (x – y)(x² + xy + y²)(x + y)(x² – xy + y²) as follows:

(x – y)(x + y)(x² + xy + y²)(x² – xy + y²)

Notice that (x – y)(x + y) is merely the factorization of difference of squares. Therefore, (x – y)(x + y) = x² – y².

(x – y)(x + y)(x² + xy +y²)(x² – xy + y²) = (x² – y²)(x² + xy +y²)(x² – xy + y²)

This means that x² – y² is also a factor of x⁶ – y⁶.

By process of elimination, x² + y² is not necessarily a factor of  x⁶ – y⁶ .

The answer is  x² + y² .

First pull out any common terms: 4x³ – 16x = 4x(x² – 4)

x² – 4 is a difference of squares, so we can also factor that further. The difference of squares formula is a² – b² = (a – b)(a + b). Here a = x and b = 2. So x² – 4 = (x – 2)(x + 2).

Putting everything together, 4x³ – 16x = 4x(x + 2)(x – 2).

Start by looking at your last term. Since this term is negative, you will need to have a positive group and a negative group:

Now, since the middle term is positive, you can guess that the positive group will contain the larger number. Likewise, since the coefficient is only , you can guess that the factors will be close.  Two such factors of  are  and .

Therefore, your groups will be:

Begin by looking at the last element. Since it is positive, you know that your groups will contain either two additions or two subtractions. Since the middle term is negative (), your groups will be two subtractions:

Now, the factors of  are  and ,  and , and  and .

Clearly, the last is the one that works, for when you FOIL , you get your original equation!

We can factor out a , leaving .

From there we can factor again to 

.

The original statement is equivalent to a statement that is identically true regardless of the value of ; therefore, so is the original statement itself. The solution set is the set of all real numbers.

To solve this equation, simply plug 12 in for  in the equation. 

The question is equivalent to asking the following:

For what value of  does the equation 

have more than one solution?

The equation simplifies as follows:

If the absolute values of two expressions equal, then either the expressions themselves are equal or they are each other's opposite. 

Taking the latter case:

Regardless of the value of , exactly one solution is yielded this way.

The question becomes as follows: for which value of  does the other way yield a solution?

Set:

If this is a false statement, then this yields no solutions.

If this is a true statement, then this automatically yields the set of all real numbers as the solution set. We solve for :

As a result, the statement

has infinitely many solutions, and the other three statements have exactly one.

First, since the equation is quadratic, put it in standard form 

as follows:

To determine the nature of the solution set, evaluate discriminant  for :

The discriminant is positive and a perfect square, so the solution set comprises two rational numbers.