This can be solved by substitution and factoring.

x² – y = 34 can be written as y = x² – 34 and substituted into the other equation: x – y = 4 which leads to x – x² + 34 = 4 which can be written as x² – x – 30 = 0.

x² – x – 30 = 0 can be factored to (x – 6)(x + 5) = 0 so x = 6 and –5 and because only 6 is a possible answer, it is the correct choice.

When a = 9, then x² + 2ax + 81 = 0 becomes 

x² + 18x + 81 = 0.  

This equation can be factored as (x + 9)² = 0.

Therefore when a = 9, x = –9.

In general, if a function has a root at x = r, then (x – r) must be a factor of f(x). In this problem, we are told that f(x) has roots at –1, 0 and 2. This means that the following are all factors of f(x):

(x – (–1)) = x + 1

(x – 0) = x

and (x – 2).

This means that we must look for an equation for f(x) that has the factors (x + 1), x, and (x – 2).

We can immediately eliminate the function f(x) = x² + x – 2, because we cannot factor an x out of this polynomial. For the same reason, we can eliminate f(x) = x² – x – 2.

Let's look at the function f(x) = x³ – x² + 2x. When we factor this, we are left with x(x² – x + 2). We cannot factor this polynomial any further. Thus, x + 1 and x – 2 are not factors of this function, so it can't be the answer.

Next, let's examine f(x) = x⁴ + x³ – 2x² .

We can factor out x².

x² (x² + x – 2)

When we factor x² + x – 2, we will get (x + 2)(x – 1). These factors are not the same as x – 2 and x + 1. 

The only function with the right factors is f(x) = x³ – x² – 2x.

When we factor out an x, we get (x² – x – 2), which then factors into (x – 2)(x + 1). Thus, this function has all of the factors we need.

The answer is f(x) = x³ – x² – 2x.

This is a difference of squares. The difference of squares formula is a² – b² = (a + b)(a – b). In this problem, a = 6x and b = 7y.

So 36x² – 49y² = (6x + 7y)(6x – 7y).

Find two numbers that add to  and multiply to

Factors of

You can use

Then make each factor equal 0.

 and

and

Factoring yields  giving roots of  and .

The numerator can be factored into .

Therefore, it can cancel with the denominator. So  imples .

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).