### All SAT Math Resources

## Example Questions

### Example Question #1 : Polynomials

If 3 less than 15 is equal to 2x, then 24/x must be greater than

**Possible Answers:**

5

3

4

6

**Correct answer:**

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 #1 : Polynomial Operations

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

**Possible Answers:**

I only

I and II

II & III

I, II and III

I and III

**Correct answer:**

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 : Polynomials

If a positive integer *a* is divided by 7, the remainder is 4. What is the remainder if 3*a* + 5 is divided by 3?

**Possible Answers:**

2

5

3

6

4

**Correct answer:**

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 3*a + 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 = (7*b + *4) / 7

*a = *(7*b* + 4)

then 3*a + 5 = *3 (7*b *+ 4) + 5

(3*a*+5)/3 = [3(7*b *+ 4) + 5] / 3

= (7*b *+ 4) + 5/3

The first half of this expression (7*b* + 4) is a positive integer, but the second half of this expression (5/3) gives us a remainder of 2.

### Example Question #2 : Polynomial Operations

**Possible Answers:**

100

42

45

38

36

**Correct answer:**

42

### Example Question #1 : Multiplying And Dividing Polynomials

Simplify:

**Possible Answers:**

**Correct answer:**

Cancel by subtracting the exponents of like terms:

### Example Question #12 : Polynomial Operations

Divide by .

**Possible Answers:**

**Correct answer:**

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 #13 : Polynomial Operations

By what expression can be multiplied to yield the product ?

**Possible Answers:**

**Correct answer:**

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 .