Factoring "Puzzle": Find a for \frac {a^2 - 2x -3}{x^2 - 4x + a}

[kenewbie]

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 can't 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

 

[tiny-tim]

Hi kenewbie!

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?

(I assume you mean [tex] \frac {x^2 - 2x -3}{x^2 - 4x + a} [/tex]) ?

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!

 

[kenewbie]

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

 

[kenewbie]

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

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

k

 

[tiny-tim]

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?

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").

 

[kenewbie]

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