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By now, you've learned several ways to factor polynomials. For example, you know that you can use the distributive law to do the following:

$3\left(4n+5\right)=12n+15$

Likewise, we've shown that you can use FOIL to multiply two binomials to get a trinomial:

$\left(n+2\right)\left(n+3\right)={n}^{2}+5n+6$

However, what if you start with a trinomial and want to find its
factors? Put another way, you're given
${n}^{2}+5n+6$
and want to go back to
$\left(n+2\right)\left(n+3\right)$
. This article will show you how to factor trinomials in a
${x}^{2}+bx+c$
form when **c** is negative. Let's get started!

The trick to factoring trinomials in
$a{x}^{2}+bx+c$
form is that you need to find two numbers that produce the constant
of the term **c** when multiplied while adding to the coefficient
of the middle term **b**. The easiest way to do this is generally
listing the factor pairs for **c** and looking for one that also
gives you the sum you need. Here, we're focusing on examples where
**c** is negative, meaning that our 2 numbers will always have
opposite signs so their product is negative. If you would like to
review examples where
b and c are both positive, when
b is negative but c is positive, or when
a ≠ 1, simply follow the included links.

It's worth noting that some polynomials cannot be factored into polynomials of a lower degree because there is no factor pair producing the sum you need. These are called irreducible polynomials or prime polynomials, so don't automatically assume you're missing something if you cannot find a factor pair that works. Of course, we cannot assume that any given polynomial is an irreducible polynomial either.

Working out a couple of practice questions could help you feel more comfortable factoring trinomials. Let's consider ${x}^{2}+6x-16$ as our first example. We need to find 2 numbers with a product of -16 and a sum of 6. Here are all of the factor pairs for -16:

$-16\times 1$ , sum: -15

$-1\times 16$ , sum: 15

$-4\times 4$ , sum: 0

$-8\times 2$ , sum: -6

$-2\times 8$ , sum: 6

Only one pair produces the sum we need: -2 and 8. Therefore, the factors for our trinomial are $\left(x-2\right)\left(x+8\right)$ . Since we're only looking for one factor pair per problem, we can stop listing factor pairs as soon as we find one that works.

Let's look at another problem: ${x}^{2}-x-20$ . This time, we need a pair with a product of -20 and a sum of -1. Let's start listing:

$-10\times 2$ , sum: -8

$-2\times 10$ , sum: 8

$-4\times 5$ , sum: 1

$-5\times 4$ , sum: -1

We haven't listed all of the possible factor pairs for -20 yet, but -5 and 4 give us the sum of -1 we're looking for. Therefore, we know the factors of the trinomial are $\left(x-5\right)\left(x+4\right)$ and don't need to list any more pairs.

a. Factor ${x}^{2}-6x-27$

We need two numbers with a product of -27 and a sum of -6. Let's list the factor pairs for -27:

$-27\times 1$ , sum: -26

$27\times -1$ , sum: 26

$9\times -3$ , sum: 6

$-9\times 3$ , sum: -6

-9 and 3 are the pair we're looking for, so the trinomial factors to $\left(x-9\right)\left(x+3\right)$ .

b. Factor ${x}^{2}+7x-8$

We need two numbers with a product of -8 and a sum of 7. Here are the factor pairs for -8:

$-8\times 1$ , sum: -7

$-1\times 8$ , sum: 7

$4\times -2$ , sum: 2

$-4\times 2$ , sum: -2

-1 and 8 are the correct pair, so the trinomial factors to $\left(x-1\right)\left(x+8\right)$ .

Graphing Quadratic Equations Using Factoring

College Algebra Diagnostic Tests

If students don't understand how to work with polynomials, they'll be completely unprepared for the topics explored in more advanced classes. If your student struggles with the factor pair guessing game or consistently chooses factors with the wrong signs, a private math tutor can provide the 1-on-1 intervention they need to get back on track. A tutor can also establish a personal rapport with their pupil to help them feel more comfortable. Contact the Educational Directors at Varsity Tutors today for more information including a personalized quote.

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