Polynomial Equations Root Finding

Question

If one of the roots of the Equation :

$$2000x^6 + 100x^5 + 10x^3 + x - 2 = 0$$ ;

Is of the form $$(m + √n)/r$$. Where m is a non zero integer and n and r are relatively coprime.

We have to find $$(m+n+r)/100$$ .

I am not able to factorise this as I am not able to guess any roots.

I could only comment that there's a root between 0 and 1 which doesn't help

Using sum , products of roots seems too tough to even start...

Any hint would do ...

Thanks in advance.

Answer 1

Given $$2000x^6+100x^5+10x^3+x-2=0\Rightarrow (2000x^6-2)+(100x^5+10x^3+x)=0$$

Now Divide it by $x^3\;,$ Where $x\neq 0\;,$ We get

$$2\left[(10x)^3-\left(\frac{1}{x}\right)^3\right]+\left[(10x)^2+1+\frac{1}{x^2}\right]=0$$

So $$2\left[(10x)^3-\left(\frac{1}{x}\right)^3\right]+\frac{(10x)^3-\frac{1}{x^3}}{\left[10x-\frac{1}{x}\right]}=0$$

So $$\left[(10x)^3-\frac{1}{x^3}\right]\cdot \left[2+\frac{1}{\left(10x-\frac{1}{x}\right)}\right]=0$$

So $$\left[(10x)^2+1+\frac{1}{x^2}\right]\cdot \left(20x^2+x-2\right)=0$$

So we get $100x^4+10x^2+1=0$ or $20x^2+x-2=0$

Now $100x^4+10x^2+1>0\;\forall x\in \mathbb{R}$. So here $$\displaystyle 20x^2+x-2=0\Rightarrow x = \frac{-1\pm \sqrt{1+160}}{2\times 20} = \left(\frac{-1\pm \sqrt{161}}{40}\right)$$

So after camparing with $\displaystyle \frac{m+\sqrt{n}}{r}\;,$ We get $m+n+r=-1+61+40=200$

So $$\displaystyle \frac{m+n+r}{100} = \frac{200}{2} =2$$

Answer 2

By Descartes rule of signs, we have that the equation has exactly $1$ positive root,exactly $1$ positive root and $4$ complex roots.

As the coefficients are all real and rational, the $4$ complex roots are complex conjugates and the $2$ real roots are conjugate surds as one of the real roots has been given to be a quadratic surd.

Hence the other real root is $\frac{m-\sqrt n}{r}$.

Does this help if you now use the relation between coefficients of terms and the roots?