### All SAT Math Resources

## Example Questions

### Example Question #11 : Factoring Equations

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 : Factoring Equations

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 #13 : Factoring Equations

What do you get when you factor:

**Possible Answers:**

**Correct answer:**

To factor our quadratic, we are looking for two numbers that multiply to 6 and add to -5.

When we look at the factors of 6: 1 and 6, 2 and 3. We see that 2 and 3 add to 5.

To get the negative, we get -2 and -3. Thus, we have our answer.

### Example Question #6 : How To Factor An Equation

Which of the following equations is NOT equivalent to the following equation?

**Possible Answers:**

**Correct answer:**

The equation presented in the problem is:

We know that:

Therefore we can see that the answer choice is equivalent to .

is equivalent to . You can see this by first combining like terms on the right side of the equation:

Multiplying everything by , we get back to:

We know from our previous work that this is equivalent to .

is also equivalent since both sides were just multiplied by . Dividing both sides by , we also get back to:

.

We know from our previous work that this is equivalent to .

is also equivalent to since

Only is NOT equivalent to

because

### Example Question #11 : Factoring Equations

in factored form is equal to:

**Possible Answers:**

**Correct answer:**

is a special type of binomial. Notice that both terms are perfect squares and they are separated by a subtraction symbole; this is known as the difference of squares.

The purpose of this question is to understand the rules of algebra and recognize different forms of expressions. The correct answer is .This is because when evaluated, it equals .

When both terms of the factored form have the same coefficients with different signs, there is no number of x in the simplified version of the expression.

### Example Question #211 : Algebra

Factor

**Possible Answers:**

**Correct answer:**

We can factor out a , leaving .

From there we can factor again to

.

### Example Question #16 : Factoring Equations

Solve for x:

**Possible Answers:**

**Correct answer:**

We need to solve this equation by factoring.

or

Now, plug these values back in individually to make sure they check.

For x=-5:

For x=4:

Both answers check.

### Example Question #1991 : Sat Mathematics

Solve:

**Possible Answers:**

None of the given answers.

**Correct answer:**

Since the left hand side of the problem is not set equal to 0, we cannot simply set each term equal to 0 and solve. Instead, we need to multiply out the left side, subtract the -2 over to the right side, and then re-factor.

To double check our answers, plug in -4 and -3 into the original problem.

For x=-4:

For x=-3:

Both answers check.

### Example Question #211 : Algebra

Factor the polynomial

**Possible Answers:**

**Correct answer:**

We need two numbers that add to and multiply to be . Trial and error will show that and add to and multiply to .

### Example Question #1993 : Sat Mathematics

Solve for a.

**Possible Answers:**

No solution

**Correct answer:**

The expression can be factored.

We must find two numbers that added together equal -7, and multiplied together equal 60. Those two numbers are -12 and 5.

Now we can set both terms equal to zero and solve for a.

,

Certified Tutor