All SAT Math Resources
Example Questions
Example Question #1 : Polynomial Operations
If 3 less than 15 is equal to 2x, then 24/x must be greater than
6
5
3
4
3
Set up an equation for the sentence: 15 – 3 = 2x and solve for x. X equals 6. If you plug in 6 for x in the expression 24/x, you get 24/6 = 4. 4 is only choice greater than a.
Example Question #2 : Polynomials
Given a♦b = (a+b)/(a-b) and b♦a = (b+a)/(b-a), which of the following statement(s) is(are) true:
I. a♦b = -(b♦a)
II. (a♦b)(b♦a) = (a♦b)2
III. a♦b + b♦a = 0
II & III
I and III
I and II
I, II and III
I only
I and III
Notice that - (a-b) = b-a, so statement I & III are true after substituting the expression. Substitute the expression for statement II gives ((a+b)/(a-b))((a+b)/(b-a))=((a+b)(b+a))/((-1)(a-b)(a-b))=-1 〖(a+b)〗2/〖(a-b)〗2 =-((a+b)/(a-b))2 = -(a♦b)2 ≠ (a♦b)2
Example Question #1 : Polynomial Operations
If a positive integer a is divided by 7, the remainder is 4. What is the remainder if 3a + 5 is divided by 3?
5
4
6
3
2
2
The best way to solve this problem is to plug in an appropriate value for a. For example, plug-in 11 for a because 11 divided by 7 will give us a remainder of 4.
Then 3a + 5, where a = 11, gives us 38. Then 38 divided by 3 gives a remainder of 2.
The algebra method is as follows:
a divided by 7 gives us some positive integer b, with a remainder of 4.
Thus,
a / 7 = b 4/7
a / 7 = (7b + 4) / 7
a = (7b + 4)
then 3a + 5 = 3 (7b + 4) + 5
(3a+5)/3 = [3(7b + 4) + 5] / 3
= (7b + 4) + 5/3
The first half of this expression (7b + 4) is a positive integer, but the second half of this expression (5/3) gives us a remainder of 2.
Example Question #2 : Polynomials
36
100
42
45
38
42
Example Question #3 : How To Divide Polynomials
Simplify:
Cancel by subtracting the exponents of like terms:
Example Question #12 : Polynomials
Divide by .
It is not necessary to work a long division if you recognize as the sum of two perfect cube expressions:
A sum of cubes can be factored according to the pattern
,
so, setting ,
Therefore,
Example Question #374 : Algebra
By what expression can be multiplied to yield the product ?
Divide by by setting up a long division.
Divide the lead term of the dividend, , by that of the divisor, ; the result is
Enter that as the first term of the quotient. Multiply this by the divisor:
Subtract this from the dividend. This is shown in the figure below.
Repeat the process with the new difference:
Repeating:
The quotient - and the correct response - is .
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