## Monday, August 24, 2015

### Prove that $x^7-2x^5+10x^2-1$ has no root greater than 1 (Heuristic Method)

Prove that $x^7-2x^5+10x^2-1$ has no root greater than 1.

Heuristic method:

The trick is to substitute $y+1$ for $x$ in the given function:

If $f(x)=x^7-2x^5+10x^2-1$, then after the substitution we see that we have:

$f(y+1)$

$=(y+1)^7-2(y+1)^5+10(y+1)^2-1$

$=(y^7+7 y^6+21 y^5+35 y^4+35 y^3+21 y^2+7 y+1)-2(y^5+5 y^4+10 y^3+10 y^2+5 y+1)+10(6^2+2y+1)-1$

$=y^7+7y^6+19y^5+25y^4+15y^3+11y^2+17y+8$

One thing that we could notice is how there is no sign changes in the expression $y^7+7y^6+19y^5+25y^4+15y^3+11y^2+17y+8$.

According to the Descartes' Rule of Signs(A method of determining the maximum number of positive and negative real roots of a polynomial), it says:

For positive roots, start with the sign of the coefficient of the lowest (or highest) power. Count the number of sign changes $n$ as you proceed from the lowest to the highest power (ignoring powers which do not appear). Then $n$ is the maximum number of positive roots.

In our case, $n=0$, so there is no positive roots for $x\gt 1$ and we're done!

This method is quicker and neater than the previous two, of course! :D