To factor a quadratic trinomial, list factors of the quadratic  term and the constant (no variables) term, then combine them into binomials that when multiplied back out will give the original trinomial.

Here, the quadratic term has only one factorization: .

The constant term has factorizations of  and .

We know the constant term is negative, so the binomials have a different operation in each (adding or subtracting), since a positive times a negative will give a negative result.

Since the middle term  is negative, we need the "larger" factor to "outweigh" the "smaller" by , and be negative.

To factor a quadratic trinomial, list factors of the quadratic  term and the constant (no variables) term, then combine them into binomials that when multiplied back out will give the original trinomial.

Here, the quadratic term has only one factorization: .

The constant term has factorizations of , , ,  and .

We know the constant term is positive, so the binomials both have the same operation in them (adding or subtracting), since a positive times a positive OR a negative times a negative will both give a positive result.

But since the middle term  is positive, both binomial factors must contain addition. And .

First, factor out the greatest common factor (GCF), which here is . If you don't see the whole GCF at once, factor out what you do see (here, either the  or the ), and then check the result to see if any more factors can be pulled out.

Then, to factor a quadratic trinomial, list factors of the quadratic  term and the constant (no variables) term, then combine them into binomials that when multiplied back out will give the original trinomial.

Here, the quadratic term has only one factorization: .

The constant term has factorizations of , , , and .

We know the constant term is positive, so the binomials both have the same operation in them (adding or subtracting), since a positive times a positive OR a negative times a negative will both give a positive result.

But since the middle term  is negative, both binomial factors must contain subtraction. And .

To factor , we use the "reverse-FOIL".

, where and are two integers whose product is 24 and whose sum is 12. If we examine all of the factor pairs of 24, we note that the sum of each pair is as follows:

1 and 24: 25
2 and 12: 14
3 and 8: 11
4 and 6: 10

No factor pair of 24 has sum 12, so cannot be factored. The correct response is that none of the four binomials is correct.

The sum of two squares is in general a prime polynomial unless a greatest common factor can be distributed out. is the sum of squares, and its terms do not have a GCF, so it is the prime polynomial.

Of the remaining choices:

is equal to ; as the difference of squares, it is factorable.

is equal to ; this fits the pattern of a perfect square quadratic trinomial, and is therefore factorable. is factorable for a similar reason.

Let be the integer written in the box.

If  is factorable, then its factorization is , where and . In other words, must be the positive difference of two numbers of a factor pair of 7. 7 has only one factor pair, 1 and 7, so the only possible value of is . The correct choice is one.

can be factored by grouping, as follows:

cannot be factored further. Of the five choices, is the only factor.

A quadratic trinomial of the form can be factored by splitting the middle term into two terms. The two coefficients must have sum and product .

In each case, a factor pair of  must be examined. These pairs, along with their sums, are:

1 and 24: 25
2 and 12: 14
3 and 8: 11
4 and 6: 10

Each polynomial with one of these four integers as its linear coefficient can be factored. The odd one out is , since there is no factor pair of 24 whose sum is 20. This is the correct choice.

First, factor the numerator, which should be .  Now the left side of your equation looks like 

Second, cancel the "like" terms -  - which leaves us with .  

Third, solve for  by setting the left-over factor equal to 0, which leaves you with 

Here you have an expression with three variables. To factor, you will need to pull out the greatest common factor that each term has in common.

Only the last two terms have  so it will not be factored out. Each term has at least  and  so both of those can be factored out, outside of the parentheses. You'll fill in each term inside the parentheses with what the greatest common factor needs to be multiplied by to get the original term from the original polynomial: