Factoring Equations - SAT Math

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.

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) = _x_2 + x – 2, because we cannot factor an x out of this polynomial. For the same reason, we can eliminate f(x) = _x_2 – x – 2.

Let's look at the function f(x) = _x_3 – x_2 + 2_x. When we factor this, we are left with x(_x_2 – 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) = _x_4 + _x_3 – 2_x_2 .

We can factor out _x_2.

_x_2 (_x_2 + x – 2)

When we factor _x_2 + 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) = _x_3 – x_2 – 2_x.

When we factor out an x, we get (_x_2 – 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) = _x_3 – x_2 – 2_x.

The numbers by which _x_6 – _y_6 is divisible will be all of its factors. In other words, we need to find all of the factors of _x_6 – _y_6 , which essentially means we must factor _x_6 – _y_6 as much as we can.

First, we will want to apply the difference of squares rule, which states that, in general, _a_2 – _b_2 = (a – b)(a + b). Notice that a and b are the square roots of the values of _a_2 and _b_2, because √_a_2 = a, and √_b_2 = b (assuming a and b are positive). In other words, we can apply the difference of squares formula to _x_6 – _y_6 if we simply find the square roots of _x_6 and _y_6.

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, (ab)c = abc.

√_x_6 = (_x_6)(1/2) = x(6(1/2)) = _x_3

Similarly, √_y_6 = _y_3.

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

_x_6 – _y_6 = (_x_3 – _y_3)(_x_3 + _y_3)

Because we can express _x_6 – _y_6 as the product of (_x_3 – _y_3) and (_x_3 + _y_3), both (_x_3 – _y_3) and (_x_3 + _y_3) are factors of _x_6 – _y_6 . Thus, we can eliminate _x_3 – _y_3 from the answer choices.

Let's continue to factor (_x_3 – _y_3)(_x_3 + _y_3). We must now apply the sum of cubes and differences of cubes formulas, which are given below:

In general, _a_3 + _b_3 = (a + b)(_a_2 – ab + _b_2). Also, _a_3 – _b_3 = (a – b)(_a_2 + ab + _b_2)

Thus, we have the following:

(_x_3 – _y_3)(_x_3 + _y_3) = (x – y)(_x_2 + xy + _y_2)(x + y)(_x_2 – xy + _y_2)

This means that x – y and x + y are both factors of _x_6 – _y_6 , so we can eliminate both of those answer choices.

We can rearrange the factorization (x – y)(_x_2 + xy + _y_2)(x + y)(_x_2 – xy + _y_2) as follows:

(x – y)(x + y)(_x_2 + xy + _y_2)(_x_2 – xy + _y_2)

Notice that (x – y)(x + y) is merely the factorization of difference of squares. Therefore, (x – y)(x + y) = _x_2 – _y_2.

(x – y)(x + y)(_x_2 + xy +_y_2)(_x_2 – xy + _y_2) = (_x_2 – _y_2)(_x_2 + xy +_y_2)(_x_2 – xy + _y_2)

This means that _x_2 – _y_2 is also a factor of _x_6 – _y_6.

By process of elimination, _x_2 + _y_2 is not necessarily a factor of _x_6 – _y_6 .

The answer is _x_2 + _y_2 .

First pull out any common terms: 4_x_3 – 16_x_ = 4_x_(_x_2 – 4)

_x_2 – 4 is a difference of squares, so we can also factor that further. The difference of squares formula is _a_2 – _b_2 = (a – b)(a + b). Here a = x and b = 2. So _x_2 – 4 = (x – 2)(x + 2).

Putting everything together, 4_x_3 – 16_x_ = 4_x_(x + 2)(x – 2).

This is a difference of squares. The difference of squares formula is a_2 – b_2 = (a + b)(a – b). In this problem, a = 6_x and b = 7_y.

So 36_x_2 – 49_y_2 = (6_x_ + 7_y_)(6_x_ – 7_y_).

Find two numbers that add to and multiply to

Factors of

You can use

Then make each factor equal 0.

and

and

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

Factoring yields giving roots of and .

The numerator can be factored into .

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

We can factor out a , leaving .

From there we can factor again to

.

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 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 .

This expression can be factored by grouping.

x2 can be factored out of the first two terms.

-4 can be factored out of the second two terms.

Now we can factor out the term (x-3).

x2 - 4 is a difference of squares. It can be factored into (x + 2)(x - 2).