# How to Factor a Quadratic

This post about how to factor a quadratic is part of a series of posts to help you prepare for the Advanced Algebra and Functions part of the Accuplacer test.

## Question

Factor completely:

$\inline&space;\fn_jvn&space;{\color{DarkBlue}&space;2x^2-4x-6}$

## Solution

The first step in factoring a polynomial (that’s an expression, like this one, that has multiples of powers of x added together) is to look for a common factor. Notice that every one of the terms (the pieces that get added together) is even. That means each term has a factor of 2. Factor that out:

$\fn_jvn {\color{DarkBlue} 2x^2-4x-6=2(x^2-2x-3)}$

Are we done? That depends on whether $\inline \fn_jvn x^2-2x-3$  can be factored.

Can it?

### Sidebar: Finding the Magic Numbers

How do you factor a quadratic trinomial whose leading coefficient is 1 — that is, whose $\inline \fn_jvn x^2$ term is not multiplied by a number (just 1, implicitly)?

Let’s reverse the process to see what’s going on. Think about how two first-degree binomials — that is, terms that look like $\fn_phv x+$ a number — multiply into a trinomial. To do that, consider two binomials, $\inline \fn_jvn x+m$ and $\inline \fn_jvn x+n$. Multiply those together to get

$\fn_jvn {\color{DarkBlue} (x+m)(x+n)=x^2+(m+n)x+mn}$

So how can you go in the other direction? That is, how can you get something like  $\fn_cm \left ( x+m \right )\left ( x+n \right )$ from something like $\fn_cm x^2+bx+c$?

Your clues to m and n are that when you multiply them together you get the constant term, the last term, of the expression getting factored, and that when you add them together you get the multiplier of $\fn_phv x$ in the middle term.

Now let’s take an example. Factor $\inline \fn_jvn x^2+2x-8$.

$\fn_jvn {\color{DarkBlue}x^2+(m+n)x+mn}$

$\fn_jvn {\color{DarkBlue}x^2+2x-8}$

So $\inline \fn_jvn m+n=2$ and $\inline \fn_jvn mn=-8$. Can you find two numbers like that? Start with the multiplication. Integer factor pairs of -8, and their sums, are:

 m n m + n 1 -8 -7 –1 8 7 2 -4 -2 –2 4 2

It looks like -2, 4 is the ticket, because those two numbers multiply to -8 and add to 2. Then the factoring is:

$\fn_jvn {\color{DarkBlue}x^2+2x-8=(x-2)(x+4)}$

### Back to the Problem

Let’s try that out on $\inline \fn_jvn x^2-2x-3$. See if you can factor the last term, -3, into two numbers that add up to the multiplier of x in the middle term, -2.

$\fn_jvn {\color{DarkBlue}x^2+(m+n)x+mn}$

$\fn_jvn {\color{DarkBlue}x^2-2x-3 }$

-3 has two pairs of integer factors: One is -1 and 3; the other is 1 and -3.

$\fn_jvn {\color{DarkBlue}-3=-3 \cdot 1}$

$\fn_jvn {\color{DarkBlue}-3+1=-2}$

So yes, there are two numbers that multiply to -3 and add to -2. Those numbers are -3 and 1. $\inline \fn_jvn x^2-2x-3$  factors into two binomials, of which one ends with -3 and one ends with 1:

$\fn_jvn {\color{DarkBlue}x^2-2x-3=(x-3)(x+1)}$

To check, you can multiply back.

Finally, don’t forget that 2 is a factor in the original expression. (I just started ignoring it, because it was not relevant to what we were doing at the moment.) So the final answer is

$\fn_jvn {\color{DarkBlue}2(x-3)(x+1)}$

The Accuplacer sample problem to which this is similar is multiple choice: You are given four possible sets of factors of a trinomial. If you don’t want to factor, you can multiply the factors in each answer choice until you find one that works.

I called this thing you factor a quadratic. That means it’s a polynomial whose highest-power term is second power. Many quadratics, like the ones considered above, have three terms; these may also be called trinomials, and you could call the problem above an example of how to factor trinomials.

Factoring is often the first step in solving a quadratic. For more about that, see “Solving a Quadratic.”

This question is similar to #7 in the sample questions for the Accuplacer Advanced Algebra and Functions test.

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