# How one can show that this polynomial is negative [closed]

How one can show that this polynomial.

$$-2+2x^{2k+1}-x^{3k+2}+x^{k-1}$$

is negative for all integer $$k>1$$ and real $$x>2$$.

I have no idea to start.

## closed as off-topic by T. Bongers, Saad, Xander Henderson, Chris Custer, HoloOct 11 '18 at 3:09

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• @greedoid: No it is $k-1$. – DER Oct 10 '18 at 15:59
• Any reason why it should be $x>5$ specifically? desmos.com/calculator/esk3my2jfc – Mohammad Zuhair Khan Oct 10 '18 at 16:07
• That will likely become clear in the answer – HackerBoss Oct 10 '18 at 16:08
• @HackerBoss I only meant that because it seems that for $k>1,$ $-2+2x^{2k+1}-x^{3k+2}+x^{k-1}$ is negative for all values $x>1.18.$ – Mohammad Zuhair Khan Oct 10 '18 at 16:12
• @MohammadZuhairKhan: How you get this result – DER Oct 10 '18 at 16:12

For $$x>2$$ we have $$x^{k+1} >4x^{k-1}>x^{k-1}$$

so we have:

$$x^{2k+1}(2-x^{k+1})-(2-x^{k-1})0}\underbrace{(2-x^{k-1})}_{<0}$$

Well I can change all the signs and assert that the result is positive. If I multiply by $$x$$ I don't change the sign, because $$x$$ is positive and I then put $$y=x^k$$ to simplify the expression. Then I am looking at $$x^3y^3-2x^2y^2-y+2\gt 0$$

I am then guessing that $$-2$$ is irrelevant once $$x$$ and $$k$$ are large, so I drop it and divide through by $$y$$ (positive) and I expect to find $$x^3y^2-2x^2y-1\gt 0$$ perhaps with some extra cases to consider. Now I drop the $$-1$$ as small in relation to the other terms and divide through by $$x^2y$$ (positive) and consider $$xy-2$$

Now with $$k\gt 1$$ and $$x\gt 5$$ we have also $$y\gt 5$$ so that $$xy-2\gt23$$

Now we see how to reverse the process with crude estimates.

Multiply through by $$x^2y^\gt 1$$ (very crude estimate) and subtract $$1$$ to obtain $$x^3y^2-2x^2y-1\gt 23x^2y-1\gt 22$$

Multiply by $$y\gt 1$$and add $$2$$ to obtain $$x^3y^3-2x^2y^2-y+2\gt22y+2\gt 24\gt 0$$

which is equivalent to the original inequality. As you see, only very crude estimates are required. I used $$y$$ to see if there were any obvious squares hiding in he background.

• This is a beautiful answer. Thanks. – Surb Oct 10 '18 at 20:01

Note that $$-2+2x^{2k+1}-x^{3k+2}+x^{k-1}=-2+\left(2x^{k+2}-x^{2k+3}+1\right)x^{k-1}$$ and $$2x^{k+2}-x^{2k+3}+1=1+\left(2-x^{k+1}\right)x^{k+2}$$ If $$x > 5$$, and $$k>1$$, then $$2-x^{k+1}<-1$$, $$x^{k+2}>1$$, and $$x^{k-1}>1$$, so that $$-2+\left(1+\left(2-x^{k+1}\right)x^{k+2}\right)x^{k-1}<0$$ Informally, $$-2+(1+(<-1)(>1))(>1)\rightarrow-2+(1+(<-1))(>1)\rightarrow-2+(<0)(>1)\rightarrow-2+(<0)<0$$