One More Tip for Factoring Ugly Quadratics: Narrowing the Possibilities

Let’s work out an easier way to find those magic numbers in the last post. 

Again, consider the example 6x^2+x-12

As we said last time, to factor by grouping you need to find factors of 6 * -12 that add up to 1. (1 because that is the multiplier of x in the middle term.)

These are the possibilities:

Factors of 6 Factors of -12

1, 6

-1, 12

2, 3

12, -1

-2, 6

2, -6

-3, 4

3, -4


How can you combine two factors of 6 with two factors of -12 to get 1? Let’s see how we can narrow the possibilities to minimize the guesswork. We already know that the absolute values of the two magic numbers have a difference of 1. In other words, those two numbers are opposite in sign and almost the same in size. 

What else can we say?

Let’s look at the prime factors we have to work with:

6 = 2*3



Notice that our prime factors contain lots of 2s and 3s. Yet 1, the number to which the magic numbers have to add up, contains no factor of 2 or 3. If *both* magic numbers had a factor of 2, their sum would have a factor of 2. But 1 has no factor of 2. So 2 must be in just one of the magic numbers. 

Similarly for 3, if both magic numbers had a factor of 3, their sum would have a factor of 3. But 1 has no factor of 3, so 3 must be in just one of the magic numbers. Since 2s and 3s are all we’ve got, the magic numbers may be 2*2*2=8 and -3*3=-9 or -8 and 9, or -8 and 9, or -1 and 72, or 1 and -72. But we said the magic numbers must add up to a positive number, 1, so they must be -8 and 9. 

Let’s try that with another example: 


Prime factors of a, 15: 3, 5.

Prime factors of c, -16:  -1, 2, 2, 2, 2

Prime factors of 15*-16: -1, 2, 2, 2, 2, 3, 5

b for this trinomial is 8, an even number whose only prime factor is 2 (repeating). The magic numbers that you add together to get 8 must be both even or both odd, because an even plus an odd — or an odd plus an even — makes an odd sum. But they can’t both be odd, because together they must contain four factors of 2 — and a factor of 2 makes a number even. So the magic numbers must both be even. For both magic numbers to be even, each one must contain at least one factor of 2.

Here is a guess as to how those 2s fall out in the magic numbers:

Team A Team B
2 2*2*2=8

Could that work? Let’s take the divisibility reasoning one step farther.  b, 8, is divisible by 4. A number that is divisible by 4 may be the sum of either two numbers that are divisible by 4 or two that are not. That means the factors of 2 cannot be distributed with 2 on one team and 8 on the other, because 2 is not divisible by 4 but 8 is.

What possibilities remain?

What if all four 2s are on one team? That would mean 1 on Team A and 16 on Team B. That doesn’t work, because that’s one number that’s divisible by 4 and one that is not — and we need either two that are or two that are not.

Pretty much all that leaves is one 4 on each team:

Team A Team B
4 4

Then where do the 3 and the 5 go? Either they combine into 15 or they go onto opposite teams. Note that b is the difference between the absolute values of the magic numbers. b=8. That’s not a huge number. If 3 and 5 are both on the same team, then the difference between the absolute values of the magic numbers will be 15*4-4=56, way too large a number.  

Team A Team B
4*15=60 -4*1=4

The only other possibility is that one team gets 3 and the other gets 5:

Team A Team B
4*3=12 -4*5=-20

Add those together and you get -8. We need 8. So switch the -1:

Team A Team B
-1*2*2*3=-12 2*2*5=20

-12+20=8. That works. So the magic numbers are -12 and 20.

Continue and factor by grouping: 


Break up the middle term using the magic numbers:


Factor the common factors out of the first pair of terms and the second pair of terms:


Factor out 5x-4 from both pieces:

(3x+4)(5x-4). These are the factors.

Summary example:

Prime factors of 20: 2, 2, 5

Prime factors of -63: -1, 3, 3, 7

Prime factors of 20 * -63: -1, 2, 2, 3, 3, 5, 7

List the prime factors of a and c.
2, 2; 3, 3. List the repetitions
-17 Find b
b, -17, is not divisible by either repeated number. Is one of the repeated factors of ac a factor of b?
Neither 2 nor 3 can be in both magic numbers. Guess that the 2s together and the 3s together will be on separate teams:

Team A Team B
2*2=4 3*3=9
Start setting up team A and team B
Team A Team B
4*-5=-20 9*7=63
Consider the remaining factors and guess which teams they are on.
-20+63=43, not b Check. Do the numbers add up to b?
Team A Team B
4*-7=-28 9*5=45
If not, check your guesses and make new ones.
-28+45=17. Check.  Check. Do the numbers add up to b?

This method removes some of the guesswork, but not all of it. 

Will this method work every time? No, it depends on repetition of factors, something that does not always happen. But for the most part, the big coefficients in the really ugly problems factor either into many small prime numbers — of which some probably repeat — or into large prime numbers, leaving you with not that many combinations to consider. 

Welcome to the NOVA Loudoun Math Lab Blog

Welcome to the NOVA Loudoun Math Lab Blog.

For the spring 2021 term, the NOVA Loudoun Math Lab hours are:

  • Monday and Wednesday, 9:30 am to 9 pm;
  • Tuesday and Thursday, 11 am to 9 pm; and
  • Friday, 9 am to 1 pm

Email us at or complete the tutoring request form to request tutoring.

We are here to support NOVA students in their math courses. We do that mostly through tutoring, these days all online via Zoom. We see students on a walk-in (OK, Zoom-in) basis; we do not make appointments. Our plan is to keep one Zoom meeting going all term, so after your first visit this term, you can continue to use the same link and password.

At this blog you can find resources that are available when we are closed. Please see our resource pages for MTTMTH 154, and MTH 161/167, and our blog posts cover course-related topics that we think you’ll find interesting. NOVA students are welcome to respond to posts on this blog.


Tips for Factoring Ugly Quadratics

How can you factor something like 6x^2+x-12?

To start, look for a common factor. There is none. 

You could use the ac method, but to do that you need to multiply 6 by -12 and then break that result down into factors — kind of feels like more work than you really should have to do. Is there a shortcut?

You need a pair of factors of ac. Those numbers are combinations of the factors of a and the factors of c. Instead of multiplying to create a big number, jump right into factoring: For a trinomial that looks like ax^2+bx+c, break a and c down into factor pairs. For 6x^2+x-12:

  • a=6.  Factor pairs of a are 1,6 and 2,3.
  • c=-12. Factor pairs of c are -1,12; 1, -12; -2,6; 2, -6; 3, -4; and -3,4.

To find the magic numbers that we need so we can factor by grouping, we need to match the numbers up into two cross-column pairs. For example, if we use 1 for a (from the “6” column), we can match it with, say, -1 for c (from the “-12” column); once we do that, we need

to use the other term in the same box as 1 — that’s 6 — as well as the other c term in the same box as -1. That’s 12. Then one pair of possible magic numbers we could try would be 1*-1=-1 and 6*12=72. We need those numbers to add up to b, the multiplier of x in the middle term — in this case that’s 1.

Will -1 and 72 work as the pair of magic numbers we need? As stated above, the two magic numbers need to add up to 1. -1+72=71 is way off. We need two numbers whose absolute values are close together. 

Try 1 * 12 + 6 *-1 = 6. Getting closer, but there’s still a way to go. 

Try 2, 3 for a and 3, -4 for c? 2*3+3*-4=6 – 12 = -6. Doesn’t work, so try flipping the correspondence: 2*-4 +3 * 3 = -1. Bingo. The magic numbers are 2*-4=-8 and 3*3=9.

Now break up the middle term in accordance with the magic numbers you just found and factor by grouping:



Factor a common factor from the first two pairs and another common factor from the second two pairs:


The contents of the parentheses are the same. Check. Now factor out 2x+3:


Let’s try another one, a really hard one. Factor:


What are the factor pairs of a and c?

We need to match the numbers up into two cross-column pairs that add up to -17. 

Again, we can match up all possible pairs: 1*1 and 20 * -63 (=-1259, doesn’t work); 1 * -63 and  20 *1 (=-43, doesn’t work) and work our way down the column that way. 

With some trial and error*, you can find that the magic factor pairs are 4*7=28 and +5*-9=-45. 28-45=-17. Bingo.


Break that middle term up into two pieces whose coefficients are the magic numbers you just found: 28 and -45:


Find a common factor in each pair of terms:

4x(5x+7) -9(5x+7)

The contents of the parentheses are the same. Check. Now factor out 5x+7. Then 20x^2-17x-63=(4x-9)(5x+7).

Multiply back and you’ll see that this is the correct factoring.

*There are ways to minimize that trial and error. That will be the topic of the next post.

Based on an idea from Math Lab tutor Yoel Abolnik.

From Theta to x

Once you’ve mastered the unit circle, how do you extend your knowledge to trig functions over many cycles?

The unit circle looks  like this:


The numbers around the circle represent angles, measured in radians. The angle variable is called theta.

How can you represent that angle on a coordinate plane? Imagine that you break the unit circle open at theta = 0,  open it, and straighten it into a line.

That line is an x-axis. Add a y-axis and you have a coordinate plane set up to graph trig functions.

Thanks to Asma Stanikzai, a student in MTH167, for this idea.

Video Math Resource

NOVA Manassas professor Matthew Westerhoff is working on a project, Doctrina, described in today’s Daily Flyer.  Doctrina will be a series of videos in various math topics. Prof. Westerhoff already has some videos for the project posted on YouTube, mostly in Calculus II topics: volume integration, series.

Check it out!