However, in common usage and in this tutorial, the word “trinomial” is used to refer more specifically to a second-degree trinomial in one variable. And of course, having three terms.

In this tutorial, we’ll learn about factoring trinomials. But before we get to the meat of it, it’s important to address a few basic but important concepts.

Technically, a trinomial is any polynomial with three terms.

Here's an example.

$x^2 + 2x + 1$

So, as you can see, there's only one variable, $\hspace{0.2em} x \hspace{0.2em}$. The highest power of the variable is $\hspace{0.2em} 2 \hspace{0.2em}$ (hence second degree) and, of course, there are exactly three terms.

A few more examples.

- $\hspace{0.2em} x^2 + 5x + 1 \hspace{0.2em}$
- $\hspace{0.2em} 3y^2 - y + 9 \hspace{0.2em}$
- $\hspace{0.2em} 4x^2 - 2x - 7 \hspace{0.2em}$

When we arrange the terms of a polynomial such that the term with a higher degree comes first, we have the trinomial in its standard form.

For example —

$x^2 - 7x + 2$

So basically, in the standard form, you have the second-degree term, followed by the first-degree term, and then finally, the constant (zeroth degree term).

An example of a trinomial not in standard form would be $\hspace{0.2em} 5x - x^2 +3 \hspace{0.2em}$.

The leading coefficient in a polynomial is the coefficient in the second–degree term.

For example, the leading coefficient in the trinomial $\hspace{0.2em} 9x + 4x^2 - 1 \hspace{0.2em}$ is $\hspace{0.2em} 4 \hspace{0.2em}$. And that in $\hspace{0.2em} x^2 + 3x - 4 \hspace{0.2em}$ is $\hspace{0.2em} 1 \hspace{0.2em}$.

If you multiply two linear binomials (binomials of degree $\hspace{0.2em} 1$), you get a trinomial. For example —

$(x + 2)(x - 6) \hspace{0.2em} = \hspace{0.2em} x^2 - 4x - 12$

When factoring, we want to reverse this and go back from a trinomial to two linear binomials. Let's see how we can do that.

Note — This method works only for trinomials with a leading coefficient of $\hspace{0.2em} 1 \hspace{0.2em}$. That's because only such trinomials can be factored to the form $\hspace{0.2em} (x + a)(x + b)$.

Alright, say we want to factor the trinomial $\hspace{0.2em} x^2 + 3x + 2 \hspace{0.2em}$.

Here are the steps to follow.

Step 1. Make sure the trinomial's leading coefficient is $\hspace{0.2em} 1 \hspace{0.2em}$. It is, so we can move to the next step.

Step 2. Write the trinomial in its factored form as shown below.

Based on the what we saw a moment ago, we know that in the factored form the trinomial would something like —

$x^2 + 3x + 2 = (x + \rule{1em}{0.05em})(x + \rule{1em}{0.05em})$

Step 3. Find what goes into those two blanks.

And for that, we'll use the coefficient of $\hspace{0.2em} x \hspace{0.2em}$, which is $\hspace{0.2em} {\color{Red} 3} \hspace{0.2em}$, and the constant term, $\hspace{0.2em} {\color{Teal} 2} \hspace{0.2em}$. We need two numbers whose sum is $\hspace{0.2em} {\color{Red} 3} \hspace{0.2em}$ and product is $\hspace{0.2em} {\color{Teal} 2} \hspace{0.2em}$.

So what are those numbers? Well, $\hspace{0.2em} 1 \hspace{0.2em}$ and $\hspace{0.2em} 2 \hspace{0.2em}$ ($\hspace{0.2em} 1 + 2 = {\color{Red} 3} \hspace{0.2em}$ and $\hspace{0.2em} 1 \times 2 = {\color{Teal} 2} \hspace{0.2em}$).^{*}

Step 4. Plug those numbers into the blank spaces. So,

$x^2 + 3x + 2 = (x + 1)(x + 2)$

That's it.

If you want to be good at factoring trinomials, there’s one thing you need to master. You’ll be using it all the time. Here’s an example.

Can you think of two numbers whose sum is $\hspace{0.2em} -7 \hspace{0.2em}$ and whose product is $\hspace{0.2em} 12 \hspace{0.2em}$?

Well, the numbers are $\hspace{0.2em} -3 \hspace{0.2em}$ and $\hspace{0.2em} -4 \hspace{0.2em}$. You may have got them on your own.

However, what's important is, you want to be able to get the answers (the two numbers) quickly, and better still, in your head. So let me show how you can do that.

Step 1. Go over the factor pairs of the product (both positive and negative factors).

To limit the number of factor pairs you have to work with, here's something to keep in mind.

Quick Tip — The larger factor (in magnitude) will have the same sign as the sum.

So, here are the factor pairs of $\hspace{0.2em} 12 \hspace{0.2em}$ along with their sums. With the condition that the larger factor (again, in magnitude) is negative. Because the sum is negative $\hspace{0.2em} -7 \hspace{0.2em}$.

Factor Pair | Sum |
---|---|

$\hspace{0.2em} -1 \hspace{0.4em} \times \hspace{0.4em} -12 \hspace{0.2em}$ | $\hspace{0.2em} -13 \hspace{0.2em}$ |

$\hspace{0.2em} -2 \hspace{0.4em} \times \hspace{0.4em} -6 \hspace{0.2em}$ | $\hspace{0.2em} -8 \hspace{0.2em}$ |

$\hspace{0.2em} -3 \hspace{0.4em} \times \hspace{0.4em} -4 \hspace{0.2em}$ | $\hspace{0.2em} {\color{Red} -7} \hspace{0.2em}$ |

Notice how we ignored certain factor pairs, for example, $\hspace{0.2em} 1 \hspace{0.2em}$ and $\hspace{0.2em} 12 \hspace{0.2em}$? That's because the larger factor isn't negative as it needs to be.

Step 2. See if a factor pair gives you the required sum.

Alright, so in the present example, the pair $\hspace{0.2em} -3 \hspace{0.2em}$ and $\hspace{0.2em} -4 \hspace{0.2em}$ pass the test.

Before we move ahead, let's try one more example.

Example

Find two numbers whose sum is $\hspace{0.2em} 1 \hspace{0.2em}$ and whose product is $\hspace{0.2em} -20 \hspace{0.2em}$?

Solution

Let's go over the factor pairs of $\hspace{0.2em} -20 \hspace{0.2em}$ (the product). Now remember, because the sum is positive $\hspace{0.2em} (1) \hspace{0.2em}$, the larger factor must be positive.

Factor Pair | Sum |
---|---|

$\hspace{0.2em} -1 \hspace{0.4em} \times \hspace{0.4em} 20 \hspace{0.2em}$ | $\hspace{0.2em} 19 \hspace{0.2em}$ |

$\hspace{0.2em} -2 \hspace{0.4em} \times \hspace{0.4em} 10 \hspace{0.2em}$ | $\hspace{0.2em} -8 \hspace{0.2em}$ |

$\hspace{0.2em} -4 \hspace{0.4em} \times \hspace{0.4em} 5 \hspace{0.2em}$ | $\hspace{0.2em} {\color{Red} 1} \hspace{0.2em}$ |

That's it. The numbers we were looking for are $\hspace{0.2em} -4 \hspace{0.2em}$ and $\hspace{0.2em} 5 \hspace{0.2em}$.

Great! Now let’s look at the method that will allow you to solve pretty much any trinomial you will be working on.

It'll be easier to explain the method with an example. So, here's one.

Example

Factor the trinomial $\hspace{0.2em} 2x^2 + 5x - 3 \hspace{0.2em}$.

Solution

Step 1. Take out the common factor (if any)

Step 2. Split the middle term

Step 3. Group and extract common factors

Step 4. Again, extract the common factor to get the final answer

Example

Factor the trinomial $\hspace{0.2em} 2x^2 + 5x - 3 \hspace{0.2em}$.

Solution

Step 1. Take out the common factor (if any)

Step 2. Split the middle term

Step 3. Group and extract common factors

Step 4. Again, extract the common factor to get the final answer

Algebraic identities — definition

Here are two of the most famous identities (and relevant to our present discussion).

Now, some trinomials fit perfectly into patterns consistent with the identities above. And that means we can easily write them in their factored form.

Consider this, for example —

Check if two of the three terms are perfect squares. If you look at the two identities, both of them carry two perfect squares in the expanded form (trinomial form).

Find the square root of the two terms. Let us all the square roots a and b.

Check if the third term is equal to twice the product of a and b (from the previous step). If yes, the trinomial fits the pattern. Otherwise, it doesn’t.

If the trinomial fits the pattern, write the factored form using the appropriate identity.

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