Binomials - ACT Math

In order to begin this kind of a problem, it's key to look at parts of the rational expression that can be simplified.

In this case, the denominator is an already-simplified binomial; however, the numerator can be factored.

The roots will be numbers that sum up to but have the product of .

The options include:

When these options are summed up:

We can negate the last three options because the first option of and fulfill the requirements. Therefore, the numerator can be factored into the following:

Because the quantity appears in the denominator, this can be "canceled out." This leaves the final answer to be the quantity .

In order to begin this kind of a problem, it's key to look at parts of the rational expression that can be simplified.

In this case, the denominator is an already-simplified binomial; however, the numerator can be factored through "factoring by grouping." This can be a helpful idea to keep in mind when you come across a polynomial with four terms and simplifying is involved.

can be simplified first by removing the common factor of from the first two terms and the common factor of from the last two terms:

This leaves two terms that are identical and their coefficients, which can be combined into another term to complete the factoring:

Consider the denominator; the quantity appears, so the in the numerator and in the denominator can be cancelled out. The simplified expression is then left as .

Even though there are three terms in the equation, there are technically only two because it's a matter of collecting like terms.

The main objective in this problem is to solve for x. In order to do so, we need to get x by itself. The first step to accomplish this is to subtract 27 from the left side of the equation and do the same to the right. This follows the idea of "what you do to one side of an equation, you must do to the other." From here, you can collect like terms between -137 and -27 because they're both constants.

The goal of getting x by itself on one side has been almost achieved. We still have that coefficient "5" that's being multiplied by x. In order to make 5x just x, 5x needs to be divided by 5 and therefore so does -164. Again, this follows the concept of mimicking actions on both sides of an equation.

The equation presented in the problem is:

To solve this type of equation, you need to factor. First, get all of the terms of the equation on one side:

Then, you need to find two factors that will give you the equation in its current form:

Therefore, and , so or .

is listed as an answer, and must therefore be correct.

The equation presented in the problem is:

First, you need to get everything on one side of the equation:

Then, you can reduce by dividing everything by :

Then, you factor:

Therefore, or .

The equation presented in the problem is:

To solve, first get all terms on the same side of the equation:

Now, you've eliminated your third term, which makes factoring difficult. If it makes things easier, you can throw a zero back in the equation:

Factor:

Therefore:

or

First, get everything on one side of the equation:

Then, to make things easier, you can reduce by dividing everything by :

Next, factor:

Therefore:

or

To solve, first pull everything over to one side of the equation:

Then, reduce by dividing by the greatest common denominator—in this case, :

Now, factor:

Therefore:

or

Using the binomial theorem, the term containing the _x_2 _y_7 will be equal to

(2_x_)2(–y)7

=36(–4_x_2 _y_7)= -144_x_2_y_7

If the expression is expanded, then by the binomial theorem, the term is

or, equivalently, the coefficient of is

Therefore, the coefficient can be determined by setting

If the expression is expanded, then by the binomial theorem, the term is

or, equivalently, the coefficient of is

Therefore, the coefficient can be determined by setting

:

While this problem can be answered by multiplying the three binomials, it is not necessary. There are three ways to multiply one term from each binomial such that two terms and one constant are multiplied; find the three products and add them, as follows:

Add: .

The correct response is .

While this problem can be answered by multiplying the three binomials, it is not necessary. There are three ways to multiply one term from each binomial such that two terms and one constant are multiplied; find the three products and add them, as follows:

Add:

The correct response is -122.

While this problem can be answered by multiplying the three binomials, it is not necessary. There are three ways to multiply one term from each binomial such that two terms and one constant are multiplied; find the three products and add them, as follows:

Add:

The correct response is .

If the expression is expanded, then by the binomial theorem, the term is

or, equivalently, the coefficient of is

Therefore, the coefficient can be determined by setting

:

The easisest way to solve for is to begin by plugging each pair of coordinates into the function.

Using our first point, we will plug in for and for . This gives us the equation

.

Squaring 0 gives us 0, and multiplying this by still gives 0, leaving only on the right side, such that

.

We now know the value of , and we can use this to help us find . Substituting our second set of coordinates into the function, we get

which simplifies to

.

However, since we know , we can substitute to get

subtracting 7 from both sides gives

and dividing by 4 gives our answer

.

To answer this problem, we need to multiply the expressions together, being mindful of how to correctly multiply like variables with exponents. To do this, we add the exponents together if the the like variables are being multiplied and subtract the exponents if the variables are being divided. So, for the presented data:

We then multiply the remaining expressions together. When we do this, we will multiply the coefficients together and combine the different variables into the final expression. Therefore:

This means our answer is .

6x (m2 +yx2 _–_3)= 6x∙m2 + 6xyx2 – 6x∙3= 6xm2 + 6yx3 -18x (Use Distributive Property)

Use the distributive property to multiply by all of the terms in :