When factoring trinomials with larger coefficients, rather than guess and check, there is a 4 step process that will always yield a correct answer.

Step 1 - Multiply the first and third terms

Step 2 - Find two factors that multiply to equal the result of step 1 () and add up to equal the second/middle term.

The product  and 

Their sum is      This works!!

So,  and  are our new factors

Step 3 - Replace our middle term () with our two new factors.

Step 4 - Factor by grouping to find your answer

   Answer!!!

To check your work, FOIL, and see if you get our original trinomial 

This can be solved by looking at all of the answers and multiplying them and comparing to the answer. However this is time consuming. You can start by noting that the term  can be a result of , where one of the terms is negative, so one answer can be eliminated. It is also clear that  must be the result of  multiplied by , so two additional answers can be eliminated. Looking at the last two answers and multiplying through, the correct answer can be determined.

The polynomial factors into (x + 3) (3x - 2).

3x² + 7x – 6 = (a + b)(c + d)

There must be a 3x term to get a 3x² term.

3x² + 7x – 6 = (3x + b)(x + d)

The other two numbers must multiply to –6 and add to +7 when one is multiplied by 3.

b * d = –6 and 3d + b = 7

b = –2 and d = 3

3x² + 7x – 6 = (3x – 2)(x + 3)

(x + 3) is the correct answer.

The product of the last two numbers should be 6, while the sum of the products of the inner and outer numbers should be 5x. Factors of six include 1 and 6, and 2 and 3. In this case, our sum is five so the correct choices are 2 and 3. Then, our factored expression is (x + 2)(x + 3). You can check your answer by using FOIL.

y = x² + 5x + 6

2 * 3 = 6 and 2 + 3 = 5

(x + 2)(x + 3) = x² + 5x + 6

You can set the two numbers to equal variables, so that you can set up the algebra in this problem. The first odd number can be defined as  and the second odd number, since the two numbers are consecutive, will be .

This allows you to set up the following equation to include the given product of 195:

Next you can subtract 195 to the left and set the equation equal to 0, which allows you to solve for :

You can factor this quadratic equation by determining which factors of 195 add up to 2. Keep in mind they will need to have opposite signs to result in a product of negative 195:

Set each binomial equal to 0 and solve for . For the purpose of this problem, you'll only make use of the positive value for :

Now that you have solved for , you know the two consecutive odd numbers are 13 and 15. You solve for the answer by finding the sum of these two numbers:

The common factor here is . Pull this out of both terms to simplify:

LCM of

LCM of 

and since

The LCM 

We are looking to factor this quadratic trinomial into two factors, where the question marks are to be replaced by two integers whose product is  and whose sum is . 

We need to look at the factor pairs of  in which the negative number has the greater absolute value and the sum is :

None of these pairs have the desired sum, so the polynomial is prime.

We are looking to factor this quadratic trinomial into two factors, where the question marks are to be replaced by two integers whose product is  and whose sum is . 

We need to look at the factor pairs of  in which the negative number has the greater absolute value, and see which one has sum :

None of these pairs have the desired sum, so the polynomial is prime.

Rewrite this as 

Use the -method by splitting the middle term into two terms, finding two integers whose sum is 1 and whose product is ; these integers are , so rewrite this trinomial as follows:

Now, use grouping to factor this: