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).

The correct answer is (x + 4)(x – 4) 

We neen to factor x² – 16 to solve. We know that each parenthesis will contain an x to make the x². We know that the root of 16 is 4 and since it is negative and no value of x is present we can tell that one 4 must be positive and the other negative. If we work it from the multiple choice answers we will see that when multiplying it out we get x² + 4x – 4x – 16. 4x – 4x cancels out and we are left with our answer. 

First, let's factor x³ – y³ using the formula for difference of cubes.

x³ – y³ = (x – y)(x² + xy + y²)

We are told that x² + xy + y² = 6. Thus, we can substitute 6 into the above equation and solve for x – y.

(x - y)(6) = 30.

Divide both sides by 6.

x – y = 5.

The original questions asks us to find x² – 2xy + y². Notice that if we factor x² – 2xy + y² using the formula for perfect squares, we obtain the following:

x² – 2xy + y² = (x – y)².

Since we know that (x – y) = 5, (x – y)² must equal 5², or 25.

Thus, x² – 2xy + y² = 25.

The answer is 25.

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 .