### All PSAT Math Resources

## Example Questions

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

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 #761 : Psat Mathematics

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 : Equations / Inequalities

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 #101 : Algebra

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 / Inequalities

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 #12 : Equations / Inequalities

Factor

**Possible Answers:**

**Correct answer:**

We can factor out a , leaving .

From there we can factor again to

.