Solving by Factoring - GMAT Quantitative

is a difference of squares. The difference of squares formula is

So .

Then, .

Let's factor the expression: .

We need to look at the behavior of the function to the left and right of 1 and 5. To the left of ,

You can check this by plugging in any value smaller than 1. For example, if ,

,

which is greater than 0.

When takes values in between 1 and 5, . Again we can check this by plugging in a number between 1 and 5.

, which is less than 0, so no numbers between 1 and 5 satisfy the inequality.

When takes values greater than 5, .

To check, let's try . Then:

so numbers greater than 5 also satisfy the inequality.

Therefore .

First let's factor:

x < -8: Let's try -10. (-10 + 8)(-10 - 1) = 22, so values less than -8 don't satisfy the inequality.

-8 < x < 1: Let's try 0. (0 + 8)(0 - 1) = -8, so values in between -8 and 1 satisfy the inequality.

x > 1: Let's try 2. (2 + 8)(2 - 1) = 10, so values greater than 1 don't satisfy the inequality.

Therefore the answer is -8 < x < 1.

This questions tests the formula: .

Therefore, we have . So

Factor the equation and set it equal to zero. . So the funtion will cross the -axis when

This expression can be rewritten:

As the difference of squares, this can be factored as follows:

As the sum of squares with relatively prime terms, the first factor is a prime polynomial. The second factor can be rewritten as the difference of two squares and factored:

Similarly, the middle polynomial is prime; the third factor can be rewritten as the difference of two squares and factored:

This is as far was we can factor, so this is the complete factorization.

can be grouped as follows:

The first three terms form a perfect square trinomial, since

, so

Now use the dfference of squares pattern:

can be grouped as follows:

is a perfect square trinomial, since

Now use the difference of squares pattern:

Group the first three terms and the last three terms, then factor out a GCF from each grouping:

We try to factor as a sum of cubes; however, 5 is not a perfect cube, so the binomial is a prime.

To factor out , we try to factor it into , replacing the question marks with two integers whose product is 2 and whose sum is 3. These integers are 1 and 2, so

The original polynomial has as its factorization.

Group the first three terms and the last three terms, then factor out a GCF from each grouping:

is the sum of cubes and can be factored using this pattern:

We try to factor out the quadratic trinomial as , replacing the question marks with integers whose product is 1 and whose sum is . These integers do not exist, so the trinomial is prime.

The factorization is therefore

and

and

and

and

The expression is a perfect square trinomial, as the three terms have the following relationship:

We can factor this expression by substituting into the following pattern:

We can factor further by noting that , the difference of squares, and subsequently, factoring this as the product of a sum and a difference.

or

Note that

Therefore is equivalent to:

The correct answer is . Our work proceeds as follows:

(Factor an out of the numerator and denominator)

(Factor the quadratic polynomials)

(Cancel common terms)

(Multiply by to both sides)

(Distribute the )

(Simplify and solve)

In order to solve for we must first factor:

The zero product property states that if then or (or both).

Our two equations are then:

Solving for in each leaves us with:

and

The roots of a function are the points at which it crosses the x axis, so at these points the value of y, or f(x), is 0. This gives us:

So we will have to factor the polynomial in order to solve for the x values at which the function is equal to 0. We need two factors whose product is -18 and whose sum is -3. If we think about our options, 2 and 9 have a product of -18 if one is negative, but there's no way of making these two numbers add up to -3. Next we consider 3 and 6. These numbers have a product of -18 if one is negative, and their sum can also be -3 if the 3 is positive and the 6 is negative. This allows us to write out the following factorization:

We could solve this question a variety of ways. The simplest would be graphing with a calculator, but we will use factoring.

To begin, set our function equal to . We want to find where this function crosses the -axis—in other words, where .

Next, we need to factor the function into two binomial terms. Remember FOIL/box method? We are essentially doing the reverse here. We are looking for something in the form of .

Recalling a few details will make this easier.

1. must equal positive

2. and must both be negative, because we get positive when we multiply them and when we add them.

3. and must be factors of that add up to . List factors of : . The only pair of those that will add up to are and , so our factored form looks like this:

Then, due to the zero product property, we know that if or one side of the equation will equal , and therefore our answers are positive and positive .