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### 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
Signed,
Confused

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
equation:

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
another:

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

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

(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
http://mathforum.org/dr.math/
Associated Topics:
High School Polynomials

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