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# Factoring: $a\ne 1$

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

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

Likewise, we know that we can use FOIL like this:

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

In this article, we'll learn how to go the other way and take a trinomial ( $a{x}^{2}+bx+c$ format) and find its factors. Specifically, we'll be looking at cases where $a\ne 1$ , which makes the math a little bit more challenging. Let's start with a refresher on how to factor trinomials with 1 as the leading coefficient before we get to the main event.

## Refresher on factoring trinomials: $a=1$

The trinomials we'll be working with are always in form $a{x}^{2}+bx+c$ where a is the leading coefficient, b is the coefficient of the linear term, and c is the constant. When $a=1$ , factoring trinomials is like a numbers puzzle. You need to find two numbers with a sum equivalent to the coefficient of the middle term (b) and a product equal to the constant (c). The easiest way to do this is to write out all of the factor pairs for c and look for one that also adds up to the sum you need.

Let's consider ${n}^{2}+8n+15$ as an example. Our two numbers must have a product of 15, so let's list all of our options:

$15\times 1=15$

$3\times 5=15$

$15+1=16$ , which isn't what we're looking for. However, $3+5=8$ , meaning that 3 and 5 satisfy both criteria of adding to b (8) while multiplying to c (15). This information allows us to rewrite our expression like this:

${n}^{2}+3n+5n+15$

From here, we can split the expression into two parts:

$\left({n}^{2}+3n\right)+\left(5n+15\right)$

And apply the Distributive Property to get:

$n\left(n+3\right)+5\left(n+3\right)$

Both parts have $n+3$ as a factor, enabling us to factor by grouping using the Distributive Property one more time:

$\left(n+5\right)\left(n+3\right)$

Done! If no numbers have a sum of b and a product of c, you're looking at an irreducible polynomial that you cannot do anything with without getting into complex numbers or irrationals. You also need to consider whether b and c are positive or negative. If b is negative but c is positive, you'll need two negative numbers to get the sum you need. If c is negative, you'll need numbers with opposite signs to get the product you need.

## Factoring polynomials when $a\ne 1$

The standard procedure for factoring polynomials doesn't change if you have a leading coefficient besides 1, but there is an extra step. We need to multiply the leading coefficient (a) by the constant (c) and then find two numbers that multiply to that number while adding to b. Let's look at an example:

$14{x}^{2}-37x+5$

First, we have to multiply 14 and 5 to get 70. Next, we need to list all of the factor pairs for 70 looking for one that also adds to -37:

$7\times 10=70$

$7+10=17$

$-7\times \left(-10\right)=70$

$-7+\left(-10\right)=-17$

$35\times 2=70$

$35+2=37$

$-35\times \left(-2\right)=70$

$-35+\left(-2\right)=-37$

$70\times 1=70$

$70+1=71$

$-70\times \left(-1\right)=70$

$-70+\left(-1\right)=-71$

$14\times 5=70$

$14+5=19$

$-14\times \left(-5\right)=70$

$-14+\left(-5\right)=-19$

The appropriate factor pair is -35 and -2. Note that you can stop listing factor pairs once you've found one that works. Now, we can rewrite our expression:

$14{x}^{2}-2x-35x+5$

And add parentheses to split it into two parts:

$\left(14{x}^{2}-2x\right)-\left(35x-5\right)$

Note that we've changed the 5 to a -5 and the -35 to a 35. We're going to FOIL a -1 into the second part of the expression, meaning we need to change the sign to end up with a mathematically equivalent expression. The next step is applying the Distributive Property to factor common factors in pairs:

$2x\left(7x-1\right)-5\left(7x-1\right)$

And using it one final time since we have a common factor:

$\left(7x-1\right)\left(2x-5\right)$

Finished! There are a lot of steps involved, but it isn't too overwhelming as long as you take it one step at a time. Note that you're effectively doing this when a =1 as well, but multiplying the constant by 1 doesn't change its value or the product you need.

All of the caveats above continue to apply when you have a leading coefficient other than 1. For example, the signs will determine what kinds of numbers you're looking for in your factor pair and trinomials with no factor pairs adding to b are still irreducible.

## Factoring trinomials: $a\ne 1$ practice problems

a. Factor the following trinomial: $7{x}^{2}-11x-6$

We can start by listing the factor pairs of $7\times \left(-6\right)=-42$ :

-1 and 42

1 and -42

-2 and 21

2 and -21

-3 and 14

3 and -14

-6 and 7

6 and -7

The only pair that adds up to -11 is -6 and 7.

So we can rewrite the trinomial as:

$7{x}^{2}-11x-6=7{x}^{2}-14x+3x-6$ $=7x\left(x-2\right)+3\left(x-2\right)$ $=\left(7x+3\right)\left(x-2\right)$b. Factor the following trinomial: $6{x}^{2}-31x+35$

Multiply the leading coefficient (a) and the constant term (c): $ac=6\times 35=210$

Find two numbers that multiply to give 210 and add up to -31:

-21 and -10 (since $-21\times \left(-10\right)=210$ and $-21+\left(-10\right)=-31$ )

Rewrite the middle term of the trinomial using these two numbers:

$6{x}^{2}-31x+35=6{x}^{2}-21x-10x+35$

Factor by grouping:

$6{x}^{2}-21x-10x+35=3x\left(2x-7\right)-5\left(2x-7\right)$

Factor out the common factor $\left(2x-7\right)$ :

$\left(3x-5\right)\left(2x-7\right)$

c. Factor the following trinomial: $2{x}^{2}-7x+6$

$2\times 6=12$

$-3\times \left(-4\right)=12$

$-3-4=-7$

$\left(2{x}^{2}-3x\right)+\left(-4x+6\right)$

$\left(x-2\right)\left(2x-3\right)$

d. Factor the following trinomial: $4{x}^{2}-4x-3$

The coefficient of ${x}^{2}$ is 4 and the constant term is -3.

The product of 4 and -3 is -12.

The factors of -12 which sum to -4 are 2 and -6.

$4{x}^{2}-4x-3$

$4{x}^{2}-6x+2x-3$

$2\left(2x+1\right)-3\left(2x+1\right)$

$\left(2x-3\right)\left(2x+1\right)$

## Topics related to the Factoring: $a\ne 1$

Graphing Quadratic Equations Using Factoring

## Flashcards covering the Factoring: $a\ne 1$

## Practice tests covering the Factoring: $a\ne 1$

College Algebra Diagnostic Tests

## Help with factoring trinomials: a ≠1 is available with Varsity Tutors

Factoring trinomials can be a very difficult concept for students to master because there are a lot of steps involved and one mistake can derail the entire process. If your student didn't get enough classroom time to feel comfortable factoring trinomials with a leading coefficient other than 1, a private math tutor can provide additional questions while helping them identify and address any learning obstacles that arise. The Educational Directors at Varsity Tutors are currently standing by to answer any questions you may have regarding 1-on-1 tutoring and how to sign up today.

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