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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

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  • $\begingroup$ 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 $\endgroup$
    – user17762
    Feb 8, 2011 at 20:09
  • $\begingroup$ @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:( $\endgroup$ Feb 8, 2011 at 20:11
  • $\begingroup$ @Petar: Yes, I know; yet another pet peeve. I make up my lack of animal companions by having lots of peeves. $\endgroup$ Feb 8, 2011 at 20:24
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    $\begingroup$ 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. $\endgroup$ Feb 8, 2011 at 20:26
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    $\begingroup$ 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) $\endgroup$
    – zar
    Feb 8, 2011 at 20:57

2 Answers 2

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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. Plot of the function using grapher

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  • $\begingroup$ typos: delete the equal signs. $\endgroup$ Feb 8, 2011 at 21:41
  • $\begingroup$ ... $x^2-3x+2>0,x>2,x<1$ $\endgroup$ Feb 8, 2011 at 21:45
  • $\begingroup$ @Americo: oh ok. Done. Thanks for pointing that out. $\endgroup$
    – user17762
    Feb 8, 2011 at 21:47
  • $\begingroup$ Great! Thanks for your effort! $\endgroup$ Feb 9, 2011 at 6:38
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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 $\rm\ z\ \approx\: -2.5$ enter image description here

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  • $\begingroup$ But how will I find the original roots? $\endgroup$ Feb 8, 2011 at 20:31
  • $\begingroup$ @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. $\endgroup$ Feb 8, 2011 at 20:32
  • $\begingroup$ Arturo - The polynomial I originally posted is the second derivative:) $\endgroup$ Feb 8, 2011 at 20:33
  • $\begingroup$ 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). $\endgroup$ Feb 8, 2011 at 20:33
  • $\begingroup$ I have posted the original task, as an edit of my question. $\endgroup$ Feb 8, 2011 at 20:34

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