First, we find two integers whose sum is 19 and whose product is . Through trial and error we find these integers are 3 and 16. We use these numbers to split the middle term, then we factor using the grouping method:

The polynomial fits the perfect square pattern:

This can be factored using the pattern

with :

Factor by grouping as follows:

Distribute the outer term through both of the inner terms.

Simplify the terms.

The answer is:

The polynomial is the difference of squares and can be factored using the pattern

where

as seen here:

The greatest common factor of the two terms is the monomial term , so factor it out:

Of the four choices, is correct.

For a quadratic trinomial with a quadratic coefficient other than 1, use the factoring by grouping method.

First, find two integers whose product is (the product of the quadratic and constant coefficients) and whose sum is 1 (the implied coefficient of ). By trial and error, we find that these are .

Split the linear term accordingly, then factor by grouping, as follows.

Factor by grouping as follows:

The first factor is the difference of squares, so further factoring can be done:

is the sum of two cubes:

As such, it can be factored using the pattern

where ;

The first factor,as the sum of squares, is a prime.

We try to factor the second by noting that it is "quadratic-style" based on . and can be written as

;

we seek to factor it as

We want a pair of integers whose product is 1 and whose sum is . These integers do not exist, so is a prime.

is the prime factorization and the correct response is .

Factor all the way to its prime factorization.

can be factored as the difference of two perfect square terms as follows:

is a factor, and, as the sum of squares, it is a prime. is also a factor, but it is not a prime factor - it can be factored as the difference of two perfect square terms. We continue:

Therefore, all of the given polynomials are factors of , but is the correct choice, as it is not a prime factor.