To factor a trinomial without using the quadratic equation, a few basic steps can be taken. The first step is always to rearrange our trinomial into  quadratic form. It isn't ideal to work with , but  in this equation cannot be reduced.

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First, create two blank binomials.

Start by factoring our first term back into the first term of each biniomial. Since the only reasonable roots of  are  and , we know that

Next, factor out our constant, ignoring the sign for now. The factors of  are  and . Note that these do not normally sum or difference to , but the presence of  as a term means one term will be doubled in value when creating . This means we either get  and  (no good), or  and  (good). Remember to add the number we want to double inside the binomial opposite , so it multiplies correctly. Now, we can add in our missing values:

One last step remains. We must check our signs. Since  is negative in our trinomial, one sign is positive and one is negative. To figure out which, check the sign of  in our trinomial. Since  is positive, the bigger of our two terms after any multipliers are completed is the positive value. Since our terms after mutiplying are  and , the  must be the positive term.

Thus, our two binomial factors are  and .

To factor a trinomial without using the quadratic equation, a few basic steps can be taken. The first step is always to rearrange our trinomial into  quadratic form, but this is already done. It isn't ideal to have , but we cannot reduce the constants further.

First, create two blank binomials.

Start by factoring our first term back into the first term of each biniomial. There are two valid factorings of . We could have  or , and there's not really a valid system for figuring out which is correct (unless you use the quadratic formula). As a loose rule of thumb, if  is near in value to , it's more likely (though by no means necessary) that the factors of  are also close in value. So, let's try .

Next, factor out our constant , ignoring the sign for now. The factors of  are either  and , or  and , but the presence of  and  as terms means one term will be doubled in value and the other will be tripled when creating . So, for  and  we either produce  and  (no good), or  and  (no good) for . If we try  and  with  and , we get either  and  (no good) or  and  (good!) Remember to add the numbers we want to use opposite the  term we want to combine them with , so it multiplies correctly. 

Now, we can add in our missing values:

One last step remains. We must check our signs. Since  is negative in our trinomial, one sign is positive and one is negative. To figure out which, check the sign of  in our trinomial. Since  is positive, the bigger of our two terms after any multipliers are completed is the positive value. Since our terms after mutiplying are  and , the number that becomes  must be the positive term.

Thus, our two binomial factors are  and .

Through factoring, the sum of the two roots must equal 2, and the product of the two roots must equal . , and 3 satisfy both of these criteria.

Given the following expression:

We need to find factors of  that add up to . 

 can be broken down into the following factors:

Of these choices, only  adds up to . Additionally, the coefficient in front of the variable is , so we do not need to worry about that when finding these values. There are no negatives in the quadratic expression, so the signs in the factored form are all positive. This gives us the final answer of

You can use the FOIL method to re-expand the expression and check your work!

To factor a quadratic equation in the form

, where , find two integers that have a sum of  and a product of .

For this equation, that would be 9 and 5.

Therefore, the solutions to this equation are  and .

Begin by setting the equation equal to 0 by subtracting 88 from both sides.

Now that the equation is in the form , find two integers that sum to  and have a product of .

For this equation, those integers are  and .

Therefore, the solutions to this equation are

Begin by setting the equation equal to zero by adding 105 to each side.

For an equation in the form , where , find two integers that have a sum of  and a product of .

For this equation, that would be 8 and 9.

Therefore, the solutions to this equation are

Rewriting the equation as , we can see there are four terms we are working with, so factor by grouping is an appropriate method. Between the first two terms, the Greatest Common Factor (GCF) is  and between the third and fourth terms, the GCF is 4. Thus, we obtain .   Setting each factor equal to zero, and solving for , we obtain  from the first factor and  from the second factor. Since the square of any real number cannot be negative, we will disregard the second solution and only accept . 

First pull out 3u from both terms.

3u⁴ – 24uv³ = 3u(u³ – 8v³) = 3u[u³ – (2v)³]

This is a difference of cubes. You will see this type of factoring if you get to the challenging questions on the GRE. They can be a pain to remember, but pat yourself on the back for getting to such hard questions! The difference of cubes formula is a³ – b³ = (a – b)(a² + ab + b²). In our problem, a = u and b = 2v:

3u⁴ – 24uv³ = 3u(u³ – 8v³) = 3u[u³ – (2v)³]

                = 3u(u – 2v)(u² + 2uv + 4v²)

This is a difference of squares. The difference of squares formula is a² – b² = (a + b)(a – b).

In this problem, a = 6x and b = 7y:

36x² – 49y² = (6x + 7y)(6x – 7y)