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```Date: 05/28/2002 at 16:41:09
From: Chris

Hi.

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.

-Chris
```

```
Date: 05/28/2002 at 17:54:57
From: Doctor Ian

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
patterns:

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

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

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
2a

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

-5 +/- sqrt(25 - 24)
= --------------------
2

-5 +/- 1
= --------
2

= -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
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:

http://mathforum.org/library/drmath/view/53877.html

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

I hope this helps!  Good luck with your studies, and don't
anything else.

- Doctor Ian, The Math Forum
http://mathforum.org/dr.math/
```

```
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.

-Chris
```
Associated Topics:
High School Polynomials
Middle School Factoring Expressions

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