All ACT Math Resources
Example Questions
Example Question #1 : Binomial Denominators
For the equation , what is(are) the solution(s) for ?
, can be factored to (x -7)(x-3) = 0. Therefore, x-7 = 0 and x-3 = 0. Solving for x in both cases, gives 7 and 3.
Example Question #51 : Variables
Simplify:
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 .
Example Question #641 : Algebra
Simplify:
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 .