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– y³using 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 + y²using the formula for perfect squares, we obtain the following:

x²– 2xy + y²= (x – y)².

Since we know that (x – y) = 5, (x – y)²must 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.

Whena= 9, thenx²+ 2ax+ 81 = 0 becomes

x²+ 18x+ 81 = 0.

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

Therefore whena= 9,x= –9.

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

(x –(–1)) =x+ 1

(x– 0) =x

and (x– 2).

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

We can immediately eliminate the functionf(x) =x²+x– 2, because we cannot factor anxout of this polynomial. For the same reason, we can eliminatef(x) =x²–x– 2.

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

Next, let's examinef(x) =x⁴+x³– 2x².

We can factor outx².

x²(x²+x– 2)

When we factorx²+x– 2, we will get (x+ 2)(x –1). These factors are not the same asx– 2 andx+ 1.

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

When we factor out anx, 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 isf(x) =x³–x²– 2x.

This is a difference of squares. The difference of squares formula isa²–b²= (a+b)(a–b). In this problem,a= 6xandb= 7y.

So 36x²– 49y²= (6x+ 7y)(6x– 7y).

Find two numbers that add toand multiply to

Factors of

You can use

Then make each factor equal 0.

and

and

Factoring yieldsgiving roots ofand.

The numerator can be factored into.

Therefore, it can cancel with the denominator. Soimples.

Find all factors of 24

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

Now find two factors that add up toand multiply to;andare 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