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:

Factor out the 7:

Take the 8 from the x-term, cut it in half to get 4, then square it to get 16.  Make this 16 equal to C/7:

Solve for C:

We can factor this trinomial using the FOIL method backwards. This method allows us to immediately infer that our answer will be two binomials, one of which begins with  and the other of which begins with . This is the only way the binomials will multiply to give us .

The next part, however, is slightly more difficult. The last part of the trinomial is , which could only happen through the multiplication of 1 and 2; since the 2 is negative, the binomials must also have opposite signs.

Finally, we look at the trinomial's middle term. For the final product to be , the 1 must be multiplied with the  and be negative, and the 2 must be multiplied with the  and be positive. This would give us , or the  that we are looking for.

In other words, our answer must be 

to properly multiply out to the trinomial given in this question.

One way to factor a trinomial like this one is to put the terms of the polynomial into a box/grid:

Notice that there are 4 boxes but only 3 terms. To fix this, we find two numbers that add to -7 [the middle coefficient] and multiply to -30 [the product of the first and last coefficients]. By examining the factors of -30, we discover that these numbers must be -10 and 3.

Now we can put the terms into the box, by splitting the -7x into -10x and 3x:

To finish factoring, determine the greatest common factor of each of the rows and columns. For instance, and have a greatest common factor of , and and -15 have a greatest common factor of 3.

Our final answer just combines the factors on the top and side into binomials. In this case, .

The first step in setting up our binomial is to figure out our possible leading terms. In this case, the only reasonable factors of  are :

Next, find the factors of . In this case, we could have  or . Either combination can potentially produce , so the signage is important here.

Note that since the last term in the ordered trinomial is positive, both factors must have the same sign. Further, since the middle term in the ordered triniomial is negative, we know the signs must both be negative.

Therefore, we have two possibilities,  or .

Let's solve for both, and check against the original triniomial.

Thus, our factors are  and .

The first step in setting up our binomial is to figure out our possible leading terms. In this case, the only reasonable factors of  are :

Next, find the factors of . In this case, we could have  or . Since the factor on our leading term is , and no additive combination of  and  can create , we know that our factors must be .

Note that since the last term in the ordered trinomial is negative, the factors must have different signs.

Therefore, we have two possibilities,  or .

Let's solve for both, and check against the original triniomial.

Thus, our factors are  and .

The first step in setting up our binomial is to figure out our possible leading terms. In this case, the only reasonable factors of  are :

Next, find the factors of . In this case, we could have  or . Since the factor on our leading term is , and no additive combination of  and  can create , we know that our factors must be .

Note that since the last term in the ordered trinomial is positive, the factors must have different signs. Since the middle term is also positive, the signs must both be positive.

Therefore, we have only one possibility, .

Let's solve, and check against the original triniomial.

Thus, our factors are  and .