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### Factoring an Equation

```
Date: 11/13/95 at 14:43:6
From: Anonymous
Subject: factor polynomials

The problem is to factor 7y^3 + 14y^2 - 7y.  I factor out the 7y:

7y(y^2 + 7y - 1)

I get stuck there.  Can I factor any further?
```

```Date: 11/13/95 at 15:9:41
From: Doctors Ethan and Ian
Subject: Re: factor polynomials

Hi,

You were right to factor out the 7y, but you may have rushed through it a
little too quickly:

7y^3 + 14y^2 - 7y
-----------------
7y

7y^3   14y^2   7y
= ---- + ----- - --
7y      7y    7y

=  y^2 + 2y - 1

However, this isn't any easier than what you ended up with.

You need a pair of roots whose sum is -2 and whose product is -1, and
there aren't any integers that fit the bill.  So there are two
possibilities:  Either the roots are real, or they're complex.

To find out which is the case, we can look at the discriminant,

b^2 - 4ac            where   ax^2 + bx + c = 0

If this is negative, the roots are complex.  Otherwise, they're real.
In this case, a = 1, b = 2, and c = -1, so

b^2 - 4ac = 2^2 - 4(1)(-1)

= 4 + 4

= 8

That's positive, so the roots are real.  But what are they?  We can use
the quadratic formula to find out:

-b +/- sqrt(b^2 - 4ac)
x = ----------------------     whenever  ax^2 + bx + c = 0
2a

-2 +/- sqrt(8)
= --------------
2

-2 +/- 2sqrt(2)
= ---------------
2

= -1 + sqrt(2)    OR      -1 - sqrt(2)

Now, if these are the roots of the quadratic expression, then

0 = (y - (-1 + sqrt(2))(y - (-1 - sqrt(2))

Can that be right?  Let's multiply it out and see what we get.  (I'm going to use 'sq2' instead of 'sqrt(2)' to save space and typing.)

0 = y^2 - [(-1 + sq2) + (-1 - sq2)]y + [(-1 + sq2) * (-1 - sq2)]

= y^2 - [-1 + -1 + sq2 - sq2]y + [(-1)^2 - (sq2)^2]

= y^2 - [-2]y + [-1]

= y^2 + 2y - 1

So it ain't pretty!  But it works.

-Doctors Ethan and Ian,  The Geometry Forum
```
Associated Topics:
Middle School Factoring Expressions

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