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Factoring Quadratics Without Guessing

Date: 05/28/2002 at 16:41:09
From: Chris
Subject: Quadratic Equations


My name is Chris and I am a seventh grader, but I love math and 
like to learn on my own. I have already learned basic 
trigonometry, logs, etc... I am now learning how to solve 
quadratic equations. I found out how to, but I am still wondering 
how to do one thing. It is part of the factoring. I will use the 
example x^2 + 5x + 6 = 0.

My dad taught me that you find two factors of the 6 that add up 
to the 5 in 5x. In this case it can be 2 and 3. He told me to 
then make it into (x + 3)(x + 2) = 0. 

What I am wondering is if there is any way to find two factors of 
6 (or whatever number it is) that add up to 5 (or the other 
number) without using guess and check. If there is no way to do 
that, then how do you factor it when the two factors are 
not whole numbers? I would love to know this.


Date: 05/28/2002 at 17:54:57
From: Doctor Ian
Subject: Re: Quadratic Equasions

Hi Chris,

In fact, guess and check is the most straightforward way to 
proceed if you have reason to believe that the roots are 
integers, and what makes it reasonable is that there usually 
aren't all that many possible guesses!  This is because for any 
given integer, there are a limited number of ways to multiply two 
other integers to get it.  

For example, consider something like

  x^2 + 13x  - 68

If you break 68 into prime factors, 

  68 = 2 * 2 * 17

it becomes clear that the _only_ possible ways to factor -68 are

  -1 *  68
   1 * -68
  -2 *  34
   2 * -34
  -4 *  17
   4 * -17

and only one of these pairs of factors can add up to 13.  So it's 
guess and check, but when done correctly, it's a very _educated_ 
kind of guess and check.  

But there _is_ a way of avoiding guess and check, called 
'completing the square', which makes use of two very common 

  1.      (x + a)^2 = x^2 + 2ax + a^2

  2. (x + a)(x - a) = x^2 - a^2

You're going to be seeing a lot of these as you continue on in 
math, so you may as well take this opportunity to become friends 
with them.  :^D

How do we use them?  First, we complete the square by adding the 
square of half the coefficient of x to both sides of the 
equation.  Using your example equation,

            x^2 + 5x + 6 = 0

  x^2 + 5x + (5/2)^2 + 6 = (5/2)^2

       (x + (5/2))^2 + 6 = 25/4

Now we subtract it from both sides to get

       (x + (5/2))^2 + 6 - 25/4 = 0

    (x + (5/2))^2 + 24/4 - 25/4 = 0

            (x + (5/2))^2 - 1/4 = 0

And since 1/4 is a square (of 1/2), we can use our second pattern:

                  (x + (5/2))^2 - 1/4 = 0

   [(x + 5/2) + 1/2][(x + 5/2) - 1/2] = 0

                       [x + 3][x + 2] = 0

It has a certain elegance, doesn't it?  But as you can see, the 
numbers can get a little hairy, which is why people tend to 
prefer the guess and check method.  It offers fewer chances for 
making silly mistakes. 

But the quickest way to factor a quadratic equation is to use the 
quadratic formula.  Again, using your example, 

  x^2 + 5x + 6 = 0         

the quadratic formula says that the roots are 

      -b +/- sqrt(b^2 - 4ac)
  x = ----------------------       here a=1, b=5, c=6

      -5 +/- sqrt(5^2 - 4*1*6)
    = ------------------------

      -5 +/- sqrt(25 - 24)
    = --------------------

      -5 +/- 1
    = --------

    = -4/2, -6/2

    = -2, -3

This tells us that the original equation is zero when x = -2 or 
when x = -3; so 

  x^2 + 5x + 6 = (x + 2)(x + 3)

Note that you have to be careful with signs when using this 
method!  It's easy to get them mixed up.

The quadratic formula is by far the fastest way to factor a 
quadratic equation, and it has the advantage that it works even 
when the roots aren't integers.  

In practice, I'll usually give factoring about 5-10 seconds, and 
if I don't get it immediately, I'll go right to the quadratic 
formula.  And when the initial coefficient is anything other than 
1, e.g., 

  3x^2 - 2x + 15

I won't even _try_ factoring, because the number of possibilities 
goes up so fast that I get a headache just thinking about it. 

Note that you can use the 'discriminant', 

  sqrt(b^2 - 4ac)

to quickly predict whether you're going to end up with something 
neat, or something messy.  As we saw, for x^2 + 5x + 6, you get

  sqrt(5^2 - 4(1)(6)) = sqrt(25 - 24) = 1

which is pretty tame.  For x^2 + 13x - 68, we get

  sqrt(13^2 - 4(1)(-68)) = sqrt(169 + 272) = 21

which is also pretty tame.  For x^2 + 13x + 13, we get

  sqrt(13^2 - 4(1)(13)) = sqrt(169 - 52) = sqrt(117)

which is somewhat ugly.  And for x^2 + 13x + 68, we get

  sqrt(13^2 - 4(1)(68)) = sqrt(169 - 272) = sqrt(-103)

which means that the roots are complex instead of real:

Don't worry too much about complex numbers right now.  I just 
wanted to include them here because you've probably already 
realized that a quadratic equation describes a polynomial, and 
that the roots of the polynomial tell you where the parabola 
crosses the x-axis.  

Eventually it would dawn on you that it's possible to draw a 
parabola that doesn't cross the x-axis... which would lead you to 
wonder:  Doesn't the equation of that parabola have roots?   And 
the answer to that question is:  It has roots, but they are 
complex instead of real. 

I hope this helps!  Good luck with your studies, and don't 
hesitate to write back if you'd like to talk more about this, or 
anything else. 

- Doctor Ian, The Math Forum 

Date: 05/28/2002 at 20:55:36
From: Chris
Subject: Thank you

Thank you so much for answering my question. It helped me a 
lot. Thanks.

Associated Topics:
High School Polynomials
Middle School Factoring Expressions

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