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I need help to factor the trimonial of 2 problems.

1. 6c^2+7c+2

2.  10n^2-26n+12

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Leah A. | Biology/Chemistry TutorBiology/Chemistry Tutor
4.9 4.9 (2773 lesson ratings) (2773)

Whenever you factor a trinomial, there are two methods you can use: box method or simply listing factors.

Most students prefer the box method because it simplifies the problem, so that is what I will use.

I will use example problems so you can cross-apply my examples to ANY problem.

Before you begin the box method, look at each problem for what's called a common factor. A common factor is a number (or term) that the whole equation has in common.

For example, let's use the trinomial 4n^2+8n-12. What do 4, 8, and 12 have in common? What can be divided evenly into each of the those numbers? We can divide the whole equation by 4 (also known as factoring out 4).

4n^2/4= n^2


-12/4 = -3

Now, our equation looks like: 4(n^2+2n-3). ***Looking at the equation (right away!) to see if there are any numbers you can factor out makes everything so much easier!***

If there are no numbers you can factor out, then you have to leave the equation as is.

Let's continue with our first example (that now is simplified partially): 4(n^2+2n-3)

To do the "box" method, you have to draw a simple box with four squares. In the upper left box, place the term that is squared. In the lower right box, put your last term (the one with no variable beside it). For now, leave the upper right and lower left boxes blank.

Look at the coefficient (the number in front of the variable) for the term that is squared. If there is no number, assume it is 1. In our example, it is 1. Now look at the last term- our -3. Multiply those two numbers together off to the side of your box. 1*-3=-3. You might be asking "What was the point of that?" You just wait and see!

Think of the factors of the two numbers you just multiplied by (the coefficient of our first term and the last term). In this problem, it is very simple- just 1 and -3 OR -1 and 3. The next step is critical.

Now look at your middle term. In this case, it is +2n. NOTE: The 2n is POSITIVE. Which of the factors we listed add or subtract to positive 2n? Answer: -1n and 3n. (HINT: If you're wondering where I suddenly got my "n"s from, look at the question more carefully: what adds or subtracts to positive 2n? -1 and 3 added only gives us TWO. But NOT +2n) Place a -1n in either of the remaining boxes and a 3n in the other.

Our final step is to look at the box. Look at each column and row separately. Let's first look at the left column. For me, I have n^2 in my upper left box and the 3n in the lower left box. For each row and column, we look to see what the terms have in common. What does n^2 and 3n have in common? An n! This means at the top of our left column outside the box, write "n". This is the beginning of one our parenthesis. Now let's look at the other column. I have -1n in the upper right box and -3 in the lower left box. What does -1n and -3 have in common? It looks like only a -1! Write a -1 above the right column outside the box. We now have one parenthesis (n-1). We're halfway done!

Let's examine the rows. n^2 and -1n (my top row) have in common an n. My bottom row: 3n and -3. They have in common only a positive 3. This is my second parenthesis: (n+3).

The final answer to our example, however, has a slight trap. Remember how earlier we factored out a four? ***You must still show that you factored a 4 our the of the original equation***.

Final answer for the example: 4(n-1)(n+3).

If you found the box method confusing, you can try other method of simply listing factors. It takes more time if you're not very good with listing factors quickly, but it still works.

I'll use the example: 7n^2-10n-8.

What are the factors of 7? Easy: 7 and 1. These are the coefficients in our parenthesis: (7n____) (n_____)

Next, what are the factors of -8? -8 and +1, +8 and -1, +4 and -2, and -4 and +2. That's a lot of factors. The next part can be time consuming: test out factors. For instance, try plugging in the factors like this:
(7n+1)(n-8). FOIL the equation by multiplying the first terms together (7n *n), then the "outside" terms (7n*-8), then the inside terms (+1*n), and lastly the "last" terms (+1*-8). Immediately, you see that 7n*-8= -56n. There's no way that you'll get from -56n to -10n.
So let's try a different combination. How about trying +4 and -2? (7n+4)(n-2). 7n*n=n^2. 7n*-2= -14n. 4*n=4n. and 4*-2=-8. the -14n+4n=-10n. That's the middle term of our equation!

Now we know that our answer is: (7n+1)(n-8).

I hope that these methods help you answer your problems! Good luck and have a great evening!

Gene G. | You can do it! I'll show you how.You can do it! I'll show you how.
5.0 5.0 (257 lesson ratings) (257)

ax2 + bx + c

When the first coefficient is not 1, it's easiest to use what's called AC factoring.

1. Multiply a by c.   ac = (6 * 2) = 12
2. Find two factors of ac that add up to b.
    You'll have to pay attention to the signs.  If ac is positive, both factors will have the same sign, + or -.  If ac is negative, one factor has to positive and the other negative.  They have to add up algebraically to equal b.
3. Split the middle term's coefficient b into these two factors.  (See the steps below.)
4. Rearrange the terms and use the distributive rule in reverse to finish finding the factors.

6c2 + 7c + 2
ac = 6*2 = 12
factors of 12 can be 3 and 4, which add up to 7
Split the 7c term into 3c + 4c and group the terms like this:
(6c2 + 3c) + (4c + 2)
You can factor 3c out of the first group, and 2 out of the other:
3c(2c + 1) + 2(2c + 1)
It's no coincidence that (2c - 1) appears twice!
Use the distributive rule to finish the job.
(3c + 2)(2c + 1)

Note: When you split that center term, it doesn't matter which order you put the pieces in.  You'll get the same result by a slightly different route.

10n2 - 26n + 12

One extra step this time because there's a common factor to take out first:

(2)(5n2 -13n + 6)
ac = 5 * 6 = 30 = (-10)*(-3)
(2)(5n2 -10n - 3n + 6)
(2)[5n(n-2) -3(n -2)]       Notice that 6/(-3) = -2, not +2

(You can Google "AC Factoring" and find a much more detailed explanation at several sites.)