# Factoring puzzle

1. Aug 12, 2008

### kenewbie

factoring "puzzle"

For what values of "a" can the following be factored and reduced:

$$\frac {a^2 - 2x -3}{x^2 - 4x + a}$$

Ok, so the first thing on the list is to factor the top term.

$$x = \frac {-(-2) \pm \sqrt{(-2)^2 - 4 * 1 * (-3)} } {2 * 1}$$

The roots are 3 & -1, and the factors then become

$$(x-3)(x+1)$$

So, I want the denominator to become either both, or one of those factors.

I start off by rewriting the denominator as an equation

$$x = \frac {4 \pm \sqrt{16 - 4 * 1 * a} } {2}$$

At this point I can tell that there are only solutions for $$a < 5$$

I know that I want the roots the polynomial to be either 3 or -1 (or both), so I substitute everything inside the square root with R, and check which values I need from the square root.

\begin{align*} \frac{4 + R}{2} = 3\\ 4 + R = 6\\ R = 2\\ \end{align*}

and

\begin{align*} \frac{4 + R}{2} = -1\\ 4 + R = -2\\ R = -6\\ \end{align*}

Since I would be hard pressed to get the square root back as negative 6, I conclude that I want the result of the square root to be 2.

In other words

\begin{align*} \sqrt{16 - 4 * 1 * a} = 2\\ 16 - 4a = 4\\ a = 3\\ \end{align*}

Finally there. I factor the polynomial using $$a = 3$$

$$x = \frac {4 \pm \sqrt{16 - 4 * 3} } {2}$$

The roots are 3 and 1 which gives the factors

$$(x-3)(x-1)$$

And I can now factor the original rational expression

$$\frac {a^2 - 2x -3}{x^2 - 4x + a} = \frac {(x-3)(x+1)} {(x-3)(x-1)} = \frac {x + 1}{ x -1}$$

However, there are two things that bug me to no end. First off the are two solutions listed, 3 and -5. I only found one of them, but I cant see where I went wrong and "missed" the second one?

The other thing is that I found my approach rather long winded, and I wondered if there was a better way to do this kind of problems?

Thanks for any feedback.

k

2. Aug 12, 2008

### tiny-tim

Hi kenewbie!
(I assume you mean $$\frac {x^2 - 2x -3}{x^2 - 4x + a}$$) ?

erm … yes!

Two methods (basically the same):

i] Just use long division, first with (x - 3) and then with (x + 1).

The remainder will be a + something, which you want to be zero.

ii] You want 3 or 1 to be a root of the denominator, so just plug 3 or 1 into it, again you get the remainder!

3. Aug 12, 2008

### kenewbie

Re: factoring "puzzle"

I'm afraid I don't quite get what you mean. Can I bother you to type out examples how how you would do it?

k

4. Aug 12, 2008

### kenewbie

Re: factoring "puzzle"

ooh, I think I see why I missed the -5 solution.

I used $$4 + R$$ instead of $$4 \pm R$$, which would allow me to use the -6 as just 6, which would have come out with -5 as a root.

k

5. Aug 12, 2008

### tiny-tim

Well, for example, if you want x2 - 4x + a to be divisible by (x - b),

then b will be a root,

and so b2 - 4b + a must be zero ("Remainder Theorem").

6. Aug 12, 2008

### kenewbie

Re: factoring "puzzle"

Aha. I haven't learned "polynomial long division" or "polynomial remainder theorem" yet, but I see they both have neat little articles on wikipedia, so I'll look into them.

Thanks

k