# 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:

$AE = -6$

$AF + BE = 15$

$AG + BF + CE = 0$

$BG + CF + DE = -30$

$CG + DF = 30$

$DG = -13$

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

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Unfortunately there is no elegant factorization. At least thats what wolfram alpha says. wolframalpha.com/input/?i=-6*x^5+%2B+15*x^4+-30*x^2+%2B+30*x-13 – user17762 Feb 8 '11 at 20:09
@Sivaram - I already entered it to the wolfram and I wasn't happy:( It is a part from a homework - I must find when the second derivative of a function is 0. And I have this polynomial to be equal to zero:( – Petar Minchev Feb 8 '11 at 20:11
@Petar: Yes, I know; yet another pet peeve. I make up my lack of animal companions by having lots of peeves. – Arturo Magidin Feb 8 '11 at 20:24
There seems to be a single real root near -1.51662 plus two conjugate pairs(using Nroot in Alpha). It stays very flat at -4 around (0.8,1.4). You might check for an error-hw usually comes out neater than this. – Ross Millikan Feb 8 '11 at 20:26
Perhaps you aren't asked to study the second derivative exactly. Using numerical methods, you can "see" that there is a inflection point, near -1.5. (wolframalpha.com/input/?i=y%3Dln((x^2-3x%2B2)/(x^2%2B1))+inflection+‌​point) – zar Feb 8 '11 at 20:57

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.

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.

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typos: delete the equal signs. – Américo Tavares Feb 8 '11 at 21:41
... $x^2-3x+2>0,x>2,x<1$ – Américo Tavares Feb 8 '11 at 21:45
@Americo: oh ok. Done. Thanks for pointing that out. – user17762 Feb 8 '11 at 21:47
Great! Thanks for your effort! – Petar Minchev Feb 9 '11 at 6:38

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 $\rm\ z\ \approx\: -2.5$

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But how will I find the original roots? – Petar Minchev Feb 8 '11 at 20:31
@Petar: Perhaps if you post the original problem we can figure out what is and what is not being asked? Above you said you need to find where the second derivative is $0$, not where the function itself was zero. – Arturo Magidin Feb 8 '11 at 20:32
Arturo - The polynomial I originally posted is the second derivative:) – Petar Minchev Feb 8 '11 at 20:33
I'm assuming that the above polynomial is the 2nd derivative. Otherwise the problem makes no sense pedagogically (or perhaps there is an error in its statement). – Bill Dubuque Feb 8 '11 at 20:33
I have posted the original task, as an edit of my question. – Petar Minchev Feb 8 '11 at 20:34