Factoring: b Negative

We know that we can use the Distributive Law to expand polynomials like this:

$3 (4 n + 5) = 12 n + 15$

Furthermore, we know that we can use the FOIL method to multiply binomials:

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

What about going the other way though? In this article, we'll explore how to start with a trinomial in ${a x}^{2} + b x + c = 0$ format and determine the two binomials that multiply into it. Put an and going to $(n + 2) (n + 3)$ . Let's get started!

Trinomial factoring: b negative

More specifically, this article focuses on how to factor trinomials when b is a negative number but c is a positive number. If you're looking for guidelines on how to factor trinomials in ${a x}^{2} + b x + c = 0$ format when both b and c are positive, when c is negative, or when a is a value other than one, those topics have dedicated pages.

The general procedure for factoring trinomials is to find two numbers that multiply to the constant of the equation $c$ while adding to the coefficient of the middle term $b$ . The easiest way to do this is typically to list out all of the factor pairs that produce the product you need and identify one that also produces the desired sum. Since this article focuses on trinomials where $b$ is negative but $c$ is positive, we'll wind up with two negative numbers so their sum is negative but the product is positive.

An example of trinomial factoring: b negative

Let's take a look at a practice question to see how this works. For instance, let's try and factor $x^{2} - 7 x + 10 = 0$ .

We need a pair of numbers that multiply to 10 and add to -7, so we'll begin by listing all of the numbers that multiply to 10:

$(10) (1)$

$(-10) (-1)$

$(5) (2)$

$(-5) (-2)$

Next, we add each factor pair and look for one that has a sum of -7:

$10 + 1 = 11$

$-10 + (-1) = -11$

$5 + 2 = 7$

$-5 + (-2) = -7$

We have a winner! The numbers -5 and -2 produce both the sum and product we need, so the factors we're looking for are $(x - 5) (x - 2)$ .

If you're looking to streamline this process, you don't have to list positive factor pairs when $b$ is negative because you know you're looking for negative numbers. Similarly, you can stop listing factor pairs as soon as you find one that works. We can also double-check our work by FOILing our answers to make sure we get the trinomial we started with. Finally, it's important to remember that not every trinomial can be factored into 2 binomials this way. Irreducible polynomials exist, so don't assume you're doing something wrong if you can't find an answer.

Practice problems on trinomial factoring: b negative

a. Factor $x^{2} - 9 x + 1$

We need a pair of numbers with a product of 18 and a sum of -9, and we know they'll both be negative because b is negative. Our potential factor pairs for 18 are:

$(-9) (-2)$

$(-18) (-1)$

$(-6) (-3)$

The first pair adds to -11, which isn't what we're looking for. The second adds to -19. However, the third pair adds to -9, meaning we have an answer: $(x - 6) (x - 3)$ . Note that you could also write it as $(x - 3) (x - 6)$ and still be correct.

b. Factor $x^{2} - 11 x - 15$

We need a pair of numbers with a product of 15 and a sum of -11, and we know they'll both be negative since b is negative. Our potential factor pairs for 15 are:

$(-15) (-1)$ sum: $-16$

$(-5) (-3)$ sum: $-8$

Uh oh. We listed all of the negative factor pairs for 15 and didn't find one that adds to -11. That means that our trinomial is an irreducible or prime polynomial that cannot be factored into polynomials of a lower degree.

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Varsity Tutors helps with trinomial factoring: b negative

Factoring trinomials requires a little detective work, and some students might feel uncomfortable with the guesswork involved. It can also be tempting to assume a trinomial is irreducible when in reality a factor pair was missed or forgotten. Working with an experienced tutor can help students of all ability levels avoid both of these issues while deepening their understanding of the material. Reach out to Varsity Tutors today to learn more about the benefits of 1-on-1 instruction and how easy the sign-up process can be.