The Math Forum

Ask Dr. Math - Questions and Answers from our Archives
Associated Topics || Dr. Math Home || Search Dr. Math

Factoring Trinomials

Date: 04/29/2002 at 11:32:43
From: Jessica
Subject: Factoring Trinomials

I'm very confused on how to go about solving these problems. I know 
for instance if you have a problem that you change them into two 
parentheses to figure it out. But then I get stuck. Please help me.

Date: 04/29/2002 at 14:09:50
From: Doctor Ian
Subject: Re: Factoring Trinomials

Hi Jessica,

It probably seems like magic unless you know _why_ anyone cares about 
being able to do this. 

Suppose you have an equation like 

  y = x^2 - 5x + 6

and you'd like to plot the equation. You could start putting in values 
of x, and getting corresponding values of y:

  x = 1,   y = 1^2 - 5(1) + 6 = 2

  x = 2,   y = 2^2 - 5(2) + 6 = 0

and so on. But how do you know which values to choose for x?  

It turns out that many quadratic equations can be written in a 
different form, which looks like 

  y = (x - 2)(x - 3) 

What's the _point_ of writing it this way?  Well, note that right away 
we can find the places where the graph of the equation crosses the 
x-axis. How? By setting y = 0:

  0 = (x - 2)(x - 3)

Now, the _only_ way that you can multiply two things together and get 
zero is if at least one of them is zero. Does that make sense? So we 
know that the only values of x for which this equation can be true are 
x = 2, 

  0 = (2 - 2)(2 - 3)

or x = 3,

  0 = (3 - 2)(3 - 3)

At _every_ other value for x, you have to get something besides zero. 

So this tells us the locations of the points where the graph crosses 
the x-axis:  (2,0) and (3,0).  

Now, we also know that parabolas are symmetric, so if the parabola 
crosses the line at these points, then the vertex of the parabola - 
the highest or lowest point, depending on whether it opens up or 
down - must be exactly halfway between these, at x = 2.5. 

To find the location of that point, we substitute this value into the 

  y = (2.5 - 2)(2.5 - 3)

    = (0.5)(-0.5)

    = -0.25

So now we know where the vertex is:  (2.5, -0.25). 

From these three points, it's easy to sketch the graph. So that's one 
reason we want to be able to factor equations this way. Here is 

Sometimes a problem leads to an equation like 

      x^2 - 5x + 6
  y = ------------
      x^2 + 2x - 8

Now, this is pretty messy. But note that we can factor both the 
numerator and denominator:

      (x - 2)(x - 3)
  y = --------------
      (x - 2)(x + 4)

      (x - 2)   (x - 3)
    = ------- * -------
      (x - 2)   (x + 4)

      (x - 3)
    = -------
      (x + 4)

which is a much nicer equation to deal with.

The point I'm trying to make is that factoring quadratic (and other) 
equations isn't just something that math teachers made up to torture 
math students. If you can factor a quadratic, it becomes much easier 
to deal with. You can't always do it, but if you can, it's almost 
always a good idea. 

Now, the reason I explained all that wasn't just to test your ability 
to stay awake, but to try to give you enough context about what's 
going on that if you _forget_ how to factor an equation, you'll be 
able to figure it out again from scratch. 

The basic idea is a very general one, which pops up over and over 
again in mathematics. To change something from one form to another, 
you write it in the other form using parameters; and then you figure 
out what the parameters have to be.  

In this case, we know that we want to end up with something that looks 
like (x + a)(x + b). So we go ahead and pretend that we've already 
found it:

  x^2 - x - 12 = (x + a)(x + b) 

Now, we can expand the right side, to get something in the same form 
as the left side, but using our new parameters (a and b):

  x^2 - x - 12 = (x + a)(x + b) 

               = x^2 + (a+b)x + ab

Now this is pretty interesting, because if these are really equal, 
then they have to have the same coefficients:

                   ab = -12
            |                   |
  x^2 - x - 12 = x^2 + (a+b)x + ab
       |                 |
           (a+b) = -1 

So this gives us two constraints:

  1)  a + b = -1

  2)  a * b = -12

This doesn't _look_ like much of an improvement, but actually we have 
a third constraint that turns out to be very helpful:

  3)  a and b are both integers

How does that help?  Well, it turns out that there just aren't that 
many ways to multiply two integers to get -12.  In fact, here are all 
the possibilities:

   a     b
  ---   ---
   -1    12
    1   -12
   -2     6
    2    -6
   -3     4
    3    -4

And if we check the sums, we'll find that only one pair of values 
will add up to -1:

   a     b    a+b
  ---   ---   ---
   -1    12    11
    1   -12   -11
   -2     6     4
    2    -6    -4
   -3     4     1
    3    -4    -1    <----- The winner!

So now we know that, if we did everything correctly, the factored 
version of the equation must be

  x^2 - x - 12 = (x + 3)(x + -4)

               = (x + 3)(x - 4)

To be sure, we always want to check to make sure that we didn't make a 
mistake. In this case, we do that by plugging -3 and 4 into the 
original equation:

  (-3)^2 - (-3) - 12 = 9 + 3 - 12

                     = 0

   (4^2) - (4) + 12  = 16 - 4 + 12

                     = 0

So we're done.  

When it works, the key to finding the values of a and b is to narrow 
down the possibilities using the fact that a*b has to be equal to the 
constant at the end. When you get some practice at this, you'll be 
able to do it in your head without even writing anything down.  For 
example, suppose you see

  x^2 - 6x + 8

The first thing you look at is the sign of the final term. It's 
positive, and the only way that you can get a positive term is by 
multiplying two positives, or two negatives. So the possibilities are 

    1 *  8
   -1 * -8
    2 *  4
   -2 * -4
and the only pair that adds up to -6 is the final pair. So the answer 
must be 

  (x - 2)(x - 4)

But the point is, this can look like magic if you don't understand 
_why_ you need to find the factors of the final constant. 

If the final constant is negative, e.g., 

  x^2 + 2x - 15

then the factors have to have opposite signs:

  -1 *  15 
   1 * -15
  -3 *   5
   3 *  -5

As always, only one pair - the third one - gives the right sum, so the 
answer must be 

  (x - 3)(x + 5)

So what at first seems to be this whole big messy process boils down 
to a very simple shortcut at the end. However, if you just try to 
memorize the shortcut, you're likely to forget it under pressure - for 
example, during a test. But if you understand _why_ the shortcut 
works, then you can work it out from scratch whenever you need it. 

I hope this helps. Write back if you'd like to talk more about 
this, or anything else. 

- Doctor Ian, The Math Forum 
Associated Topics:
High School Polynomials

Search the Dr. Math Library:

Find items containing (put spaces between keywords):
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

Math Forum Home || Math Library || Quick Reference || Math Forum Search

Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.