All ACT Math Resources
Example Questions
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 #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 #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.
Example Question #21 : Polynomials
Choose the answer below that is one of the solutions to the following equation:
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 .
Example Question #21 : Polynomials
Choose the answer that is a potential solution to the binomial equation below:
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
Example Question #25 : Polynomials
Choose the answer below that is a solution to the following binomial equation:
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
Example Question #31 : Polynomials
Choose the answer which is a solution to the following binomial equation:
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
Example Question #32 : Polynomials
Choose the answer below that is a possible solution to the following binomial 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.
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