### All PSAT Math Resources

## Example Questions

### Example Question #4 : Factoring Equations

Solve for .

**Possible Answers:**

**Correct answer:**

Find all factors of 24

1, 2, 3,4, 6, 8, 12, 24

Now find two factors that add up to and multiply to ; and are the two factors.

By factoring, you can set the equation to be

If you FOIL it out, it gives you .

Set each part of the equation equal to 0, and solve for .

and

and

### Example Question #11 : How To Factor An Equation

Assume that and are integers and that . The value of must be divisible by all of the following EXCEPT:

**Possible Answers:**

**Correct answer:**

The numbers by which *x*^{6} – *y*^{6 }is divisible will be all of its factors. In other words, we need to find all of the factors of *x*^{6} – *y*^{6} , which essentially means we must factor *x*^{6} – *y*^{6 }as much as we can.

First, we will want to apply the difference of squares rule, which states that, in general, *a*^{2} – *b*^{2} = (*a* – *b*)(*a* + *b*). Notice that *a* and *b* are the square roots of the values of *a*^{2} and *b*^{2}, because √*a*^{2} = *a*, and √*b*^{2} = *b* (assuming *a* and *b* are positive). In other words, we can apply the difference of squares formula to *x*^{6} – *y*^{6} if we simply find the square roots of *x*^{6} and *y*^{6}.

Remember that taking the square root of a quantity is the same as raising it to the one-half power. Remember also that, in general, (*a ^{b}*)

*=*

^{c}*a*.

^{bc}√*x*^{6} = (*x*^{6})^{(1/2)} = *x*^{(6(1/2))} = *x*^{3}

Similarly, √*y*^{6} = *y*^{3}.

Let's now apply the difference of squares factoring rule.

*x*^{6} – *y*^{6 }= (*x*^{3} – *y*^{3})(*x*^{3} + *y*^{3})

Because we can express *x*^{6} – *y*^{6} as the product of (*x*^{3} – *y*^{3}) and (*x*^{3} + *y*^{3}), both (*x*^{3} – *y*^{3}) and (*x*^{3} + *y*^{3}) are factors of *x*^{6} – *y*^{6 }. Thus, we can eliminate *x*^{3} – *y*^{3 }from the answer choices.

Let's continue to factor (*x*^{3} – *y*^{3})(*x*^{3} + *y*^{3}). We must now apply the sum of cubes and differences of cubes formulas, which are given below:

In general, *a*^{3} + *b*^{3} = (*a *+ *b*)(*a*^{2} – *ab* + *b*^{2}). Also, *a*^{3} – *b*^{3} = (*a –* *b*)(*a*^{2} + *ab* + *b*^{2})

Thus, we have the following:

(*x*^{3} – *y*^{3})(*x*^{3} + *y*^{3}) = (*x –* *y*)(*x*^{2} + *xy* + *y*^{2})(*x* + *y*)(*x*^{2} – *xy* + *y*^{2})

This means that *x –* *y* and *x* + *y* are both factors of *x*^{6} – *y*^{6 }, so we can eliminate both of those answer choices.

We can rearrange the factorization (*x –* *y*)(*x*^{2} + *xy* + *y*^{2})(*x* + *y*)(*x*^{2} – *xy* + *y*^{2}) as follows:

(*x –* *y*)(*x *+ *y*)(*x*^{2} + *xy* + *y*^{2})(*x*^{2} – *xy* + *y*^{2})

Notice that (*x –* *y*)(*x *+ *y*) is merely the factorization of difference of squares. Therefore, (*x –* *y*)(*x *+ *y*) = *x*^{2 }– *y*^{2}.

(*x –* *y*)(*x *+ *y*)(*x*^{2} + *xy* +*y*^{2})(*x*^{2} – *xy* + *y*^{2}) = (*x*^{2} – *y*^{2})(*x*^{2} + *xy* +*y*^{2})(*x*^{2} – *xy* + *y*^{2})

This means that *x*^{2} – *y*^{2} is also a factor of *x*^{6} – *y*^{6}.

By process of elimination, *x*^{2} +* y*^{2} is not necessarily a factor of *x*^{6} – *y*^{6 }.

The answer is *x*^{2} + *y*^{2}^{ }.

### Example Question #11 : How To Factor An Equation

Factor .

**Possible Answers:**

Cannot be factored

**Correct answer:**

First pull out any common terms: 4*x*^{3} – 16*x* = 4*x*(*x*^{2} – 4)

*x*^{2} – 4 is a difference of squares, so we can also factor that further. The difference of squares formula is *a*^{2} – *b*^{2} = (*a* – *b*)(*a* + *b*). Here *a* = *x* and *b* = 2. So *x*^{2} – 4 = (*x* – 2)(*x* + 2).

Putting everything together, 4*x*^{3} – 16*x* = 4*x*(*x *+ 2)(*x *– 2).

### Example Question #11 : Equations / Solution Sets

Factor the following:

**Possible Answers:**

**Correct answer:**

Start by looking at your last term. Since this term is negative, you will need to have a positive group and a negative group:

Now, since the middle term is positive, you can guess that the positive group will contain the larger number. Likewise, since the coefficient is only , you can guess that the factors will be close. Two such factors of are and .

Therefore, your groups will be:

### Example Question #12 : Equations / Solution Sets

Factor the following:

**Possible Answers:**

**Correct answer:**

Begin by looking at the last element. Since it is positive, you know that your groups will contain either two additions or two subtractions. Since the middle term is negative (), your groups will be two subtractions:

Now, the factors of are and , and , and and .

Clearly, the last is the one that works, for when you FOIL , you get your original equation!

### Example Question #211 : Algebra

Factor

**Possible Answers:**

**Correct answer:**

We can factor out a , leaving .

From there we can factor again to

.

### Example Question #1 : How To Find A Solution Set

Solve for :

**Possible Answers:**

No solution

can be any real number

**Correct answer:**

can be any real number

The original statement is equivalent to a statement that is identically true regardless of the value of ; therefore, so is the original statement itself. The solution set is the set of all real numbers.

### Example Question #2 : How To Find A Solution Set

If , what is the value of:

**Possible Answers:**

**Correct answer:**

To solve this equation, simply plug 12 in for in the equation.

### Example Question #1 : How To Find A Solution Set

Which of the following equations has more than one solution?

**Possible Answers:**

All of the other responses gives a correct answer.

**Correct answer:**

The question is equivalent to asking the following:

For what value of does the equation

have more than one solution?

The equation simplifies as follows:

If the absolute values of two expressions equal, then either the expressions themselves are equal or they are each other's opposite.

Taking the latter case:

Regardless of the value of , exactly one solution is yielded this way.

The question becomes as follows: for which value of does the *other* way yield a solution?

Set:

If this is a false statement, then this yields no solutions.

If this is a true statement, then this automatically yields the set of all real numbers as the solution set. We solve for :

As a result, the statement

has infinitely many solutions, and the other three statements have exactly one.

### Example Question #762 : Psat Mathematics

Which of the following is true of the solution set of the equation ?

**Possible Answers:**

The solution set comprises two irrational numbers.

The solution set comprises one irrational number.

The solution set comprises two imaginary numbers.

The solution set comprises one rational number.

The solution set comprises two rational numbers.

**Correct answer:**

The solution set comprises two rational numbers.

First, since the equation is quadratic, put it in standard form

as follows:

To determine the nature of the solution set, evaluate discriminant for :

The discriminant is positive and a perfect square, so the solution set comprises two rational numbers.

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