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).

To factor our quadratic, we are looking for two numbers that multiply to 6 and add to -5.

When we look at the factors of 6: 1 and 6, 2 and 3. We see that 2 and 3 add to 5.

To get the negative, we get -2 and -3. Thus, we have our answer. 

The equation presented in the problem is:

We know that:

Therefore we can see that the answer choice  is equivalent to .

  is equivalent to  . You can see this by first combining like terms on the right side of the equation: 

Multiplying everything by , we get back to:

We know from our previous work that this is equivalent to .

 is also equivalent  since both sides were just multiplied by . Dividing both sides by , we also get back to:

.

We know from our previous work that this is equivalent to .

 is also equivalent to  since

Only  is NOT equivalent to 

because

 is a special type of binomial. Notice that both terms are perfect squares and they are separated by a subtraction symbole; this is known as the difference of squares.

The purpose of this question is to understand the rules of algebra and recognize different forms of expressions. The correct answer is .This is because when evaluated, it equals .

When both terms of the factored form have the same coefficients with different signs, there is no number of x in the simplified version of the expression.

We can factor out a , leaving .

From there we can factor again to 

.

We need to solve this equation by factoring. 

 or 

Now, plug these values back in individually to make sure they check.

For x=-5: 

For x=4:

Both answers check. 

Since the left hand side of the problem is not set equal to 0, we cannot simply set each term equal to 0 and solve. Instead, we need to multiply out the left side, subtract the -2 over to the right side, and then re-factor. 

To double check our answers, plug in -4 and -3 into the original problem. 

For x=-4:

For x=-3:

Both answers check. 

We need two numbers that add to  and multiply to be . Trial and error will show that  and  add to  and multiply to .