All ACT Math Resources
Example Questions
Example Question #1 : How To Find The Value Of The Coefficient
What is the value of the coefficient in front of the term that includes in the expansion of ?
Using the binomial theorem, the term containing the x2 y7 will be equal to
(2x)2(–y)7
=36(–4x2 y7)= -144x2y7
Example Question #2 : Coefficients
A function of the form passes through the points and . What is the value of ?
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
.
Example Question #2 : Binomials
is equivalent to which of the following?
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 .
Example Question #1 : How To Find The Value Of The Coefficient
Give the coefficient of in the product
.
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 .
Example Question #2 : Binomials
Give the coefficient of in the product
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 .
Example Question #1 : How To Find The Value Of The Coefficient
Give the coefficient of in the binomial expansion of .
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
:
Example Question #2 : How To Find The Value Of The Coefficient
Give the coefficient of in the binomial expansion of .
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
:
Example Question #3 : How To Find The Value Of The Coefficient
Give the coefficient of in the binomial expansion of .
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
Example Question #2 : How To Find The Value Of The Coefficient
Give the coefficient of in the product
.
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.
Example Question #24 : Polynomials
Solve for .
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.