(adsbygoogle = window.adsbygoogle || []).push({}); factoring "puzzle"

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

[tex] \frac {a^2 - 2x -3}{x^2 - 4x + a} [/tex]

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

[tex] x = \frac {-(-2) \pm \sqrt{(-2)^2 - 4 * 1 * (-3)} } {2 * 1} [/tex]

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

[tex] (x-3)(x+1) [/tex]

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

I start off by rewriting the denominator as an equation

[tex] x = \frac {4 \pm \sqrt{16 - 4 * 1 * a} } {2} [/tex]

At this point I can tell that there are only solutions for [tex] a < 5 [/tex]

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.

[tex]

\begin{align*}

\frac{4 + R}{2} = 3\\

4 + R = 6\\

R = 2\\

\end{align*}

[/tex]

and

[tex]

\begin{align*}

\frac{4 + R}{2} = -1\\

4 + R = -2\\

R = -6\\

\end{align*}

[/tex]

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

[tex]

\begin{align*}

\sqrt{16 - 4 * 1 * a} = 2\\

16 - 4a = 4\\

a = 3\\

\end{align*}

[/tex]

Finally there. I factor the polynomial using [tex] a = 3 [/tex]

[tex] x = \frac {4 \pm \sqrt{16 - 4 * 3} } {2} [/tex]

The roots are 3 and 1 which gives the factors

[tex] (x-3)(x-1) [/tex]

And I can now factor the original rational expression

[tex]

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

[/tex]

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

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# Homework Help: Factoring puzzle

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