Factorize polynomial I am trying to factorize $-6x^5+15x^4-30x^2+30x-13$ for hours:( Could someone help me? I tried making a system of equations from $(Ax^3 + Bx^2 + Cx + D) (Ex^2 + Fx + G)$ but it is a nightmare:(
In case you are interested, the system is:
$AE = -6$
$AF + BE = 15$
$AG + BF + CE = 0$
$BG + CF + DE = -30$
$CG + DF = 30$
$DG = -13$
Thanks in advance!
Edit: The original task is to draw the following function: $\ln \dfrac{x^2 - 3x + 2}{x^2 + 1}$. The polynomial above is the numerator of the second derivative of the function.
Best regards,
Petar
 A: First note that the domain over which the function makes sense in real variables is when $x^2 - 3x + 2 > 0$ i.e. when $(x-1)(x-2) > 0$ i.e. when $x > 2$ or $x < 1$. Now the way out is to rewrite 
$\log \frac{x^2 - 3x + 2}{x^2 + 1}$ as $\log (x-1) + \log (x-2) - \log (x^2+1)$ if $x > 2$
and
$\log \frac{x^2 - 3x + 2}{x^2 + 1}$ as $\log (1-x) + \log (2-x) - \log (x^2+1)$ if $x < 1$
The individual plots for the three terms can be drawn trivially. All you need to do now is to superpose these three together.
You could do further analysis which will help you with your plotting.
For instance, when $\frac{x^2-3x+2}{x^2+1} \geq 1$, the function will be non-negative and when $\frac{x^2-3x+2}{x^2+1} < 1$, the function will be negative.
So the function will be non-negative when $\frac{x^2-3x+2}{x^2+1} \geq 1 \Rightarrow -3x+2 \geq 1 \Rightarrow x \leq \frac{1}{3}$ and will be negative when $x > \frac{1}{3}$. The zero crossing is at $x= \frac{1}{3}$.
As mentioned previously the function is not-defined for $1 \leq x \leq 2$. And the function tends to $-\infty$ as $x \rightarrow 2^+$ or as $x \rightarrow 1^{-}$.
Further as $x \rightarrow \pm \infty$, the function tends to $0$.
The derivative when $x>2$ is $-\frac{2x}{x^2+1} + \frac{1}{x-2} + \frac{1}{x-1} > - \frac{2x}{x^2} + \frac{1}{x} + \frac{1}{x} = 0$ when $x>2$. So in the domain $(2,\infty)$ we have the function to be increasing and $f(2^+) = - \infty$ and $\displaystyle \lim_{x \rightarrow \infty}f(x) = 0$. Hence, $f(x) < 0$, $\forall x \in (2, \infty)$.
So we have $f(x)$ negative and it increases from $-\infty$ to $0$ in the domain $(2,\infty)$.
In the domain $(-\infty,1)$, we know that $\displaystyle \lim_{x \rightarrow -\infty}f(x) = 0$ and $f(1^-) = -\infty$ and we know that there is only one zero crossing, which means there has to be at least one maximum.
Setting the derivative to zero, we get a quadratic in $x$ which gives $x = \frac{1 \pm \sqrt{10}}{3}$. The positive root falls in $[1,2]$ and hence can be ruled out. The negative root is where the maximum occurs. And there is only one maximum.
Further, $x=0$ gives $f(0) = \log (2) > 0$ as expected since $f(\frac{1}{3}) = 0$.
So the summary is,
The function increases from $0$ to $f(\frac{1 - \sqrt{10}}{3})$ in the domain $(-\infty,\frac{1 - \sqrt{10}}{3})$.
The function decreases from $f(\frac{1 - \sqrt{10}}{3})$ to $-\infty$ in the domain $[f(\frac{1 - \sqrt{10}}{3}),1)$ with zero crossing at $x = \frac{1}{3}$.
The function is not defined in the domain $[1,2]$.
The function $f(x)$ negative and it increases from $-\infty$ to $0$ in the domain $(2,\infty)$.
This information should be enough to help you make a sketch of the plot.

A: HINT $\ $ If you let $\rm\ z = x-1\ $ then the 2nd derivative is $\rm - 6 \ z^5 - 15\ z^4 - 4\ $ so the roots satisfy $\rm 6\ z + 15 = -4/z^4\ $ which one easily sees has a unique real root roughly $\rm\ z\ \approx\: -2.5$

