Given $p(x)=x^5+x^2+1$ have roots $r_1,\,r_2,\,r_3,\,r_4,\,r_5$. Let $q(x)=x^2-2$. Determine the product of $q(r_1)\cdot q(r_2)\cdot q(r_3)\cdot q(r_4)\cdot q(r_5)$.

If we attempt at the problem using the Vieta's formulas, then we might have to work a bit harder and longer to determine the value of the required product. But, it is a workable approach and let's see how and what procedures are needed in order to reach to the final result.

From the polynomial of $p(x)=x^5-0x^4+0x^3-(-1)x^2+0x-(-1)$, we have:

$r_1+r_2+r_3+r_4+r_5=0$

$r_1r_2+r_1r_3+r_1r_4+r_1r_5+r_2r_3+r_2r_4+r_2r_5+r_3r_4+r_3r_5+r_4r_5=0$

$r_1r_2r_3+r_1r_2r_4+r_1r_2r_5+r_1r_3r_4+r_1r_3r_5+r_1r_4r_5+r_2r_3r_4+r_2r_3r_5+r_2r_4r_5+r_3r_4r_5=-1$

$r_1r_2r_3r_4+r_1r_2r_3r_5+r_1r_2r_4r_5+r_1r_3r_4r_5+r_2r_3r_4r_5=0$

$r_1r_2r_3r_4r_5=-1$

Our target expression is $(r_1-2)(r_2-2)(r_3-2)(r_4-2)(r_5-2)$, which upon expanding we have:

$(r_1-2)(r_2-2)(r_3-2)(r_4-2)(r_5-2)$

$=(r_1r_2r_3r_4r_5)^2+2(r_1^2r_2^2r_3^2r_4^2+r_1^2r_2^2r_3^2r_5^2+r_1^2r_2^2r_4^2r_5^2+r_1^2r_3^2r_4^2r_5^2+r_2^2r_3^2r_4^2r_5^2)$

$\,\,\,\,\,\,+4(r_1^2r_2^2r_3^2+r_1^2r_2^2r_4^2+r_1^2r_2^2r_5^2+r_1^2r_3^2r_4^2+r_1^2r_3^2r_5^2+r_1^2r_4^2r_5^2+r_2^2r_3^2r_4^2+r_2^2r_3^2r_5^2+r_2^2r_4^2r_5^2+r_3^2r_4^2r_5^2)$

$\,\,\,\,\,+8(r_1^2r_2^2+r_1^2r_3^2+r_1^2r_4^2+r_1^2r_5^2+r_2^2r_3^2+r_2^2r_4^2+r_2^2r_5^2+r_3^2r_4^2+r_3^2r_5^2+r_4^2r_5^2)$

$\,\,\,\,\,+16(r_1^2+r_2^2+r_3^2+r_4^2+r_5^2)+32$

But we know too:

$(r_1+r_2+r_3+r_4+r_5)^2$

$=r_1^2+r_2^2+r_3^2+r_4^2+r_5^2+2(r_1r_2+r_1r_3+r_1r_4+r_1r_5+r_2r_3+r_2r_4+r_2r_5+r_3r_4+r_3r_5+r_4r_5)$

This implies $0=r_1^2+r_2^2+r_3^2+r_4^2+r_5^2+2(0)$, i.e.

[MATH]\color{yellow}\bbox[5px,purple]{r_1^2+r_2^2+r_3^2+r_4^2+r_5^2=0}[/MATH]

$(r_1^2+r_2^2+r_3^2+r_4^2+r_5^2)^2$

$=r_1^4+r_2^4+r_3^4+r_4^4+r_5^4+2(r_1^2r_2^2+r_1^2r_3^2+r_1^2r_4^2+r_1^2r_5^2+r_2^2r_3^2+r_2^2r_4^2+r_2^2r_5^2+r_3^2r_4^2+r_3^2r_5^2+r_4^2r_5^2)$

$0=r_1^4+r_2^4+r_3^4+r_4^4+r_5^4+2(r_1^2r_2^2+r_1^2r_3^2+r_1^2r_4^2+r_1^2r_5^2+r_2^2r_3^2+r_2^2r_4^2+r_2^2r_5^2+r_3^2r_4^2+r_3^2r_5^2+r_4^2r_5^2)$

$-4(r_1^4+r_2^4+r_3^4+r_4^4+r_5^4)=8(r_1^2r_2^2+r_1^2r_3^2+r_1^2r_4^2+r_1^2r_5^2+r_2^2r_3^2+r_2^2r_4^2+r_2^2r_5^2+r_3^2r_4^2+r_3^2r_5^2+r_4^2r_5^2)$

Ah, I just don't have the motivation to keep going with this most likely the fruitless approach, as it will be testing my patient to the limit to look for the appropriate expression or values for:

$r_1^2r_2^2r_3^2r_4^2+r_1^2r_2^2r_3^2r_5^2+r_1^2r_2^2r_4^2r_5^2+r_1^2r_3^2r_4^2r_5^2+r_2^2r_3^2r_4^2r_5^2$ and

$r_1^2r_2^2r_3^2+r_1^2r_2^2r_4^2+r_1^2r_2^2r_5^2+r_1^2r_3^2r_4^2+r_1^2r_3^2r_5^2+r_1^2r_4^2r_5^2+r_2^2r_3^2r_4^2+r_2^2r_3^2r_5^2+r_2^2r_4^2r_5^2+r_3^2r_4^2r_5^2$

It's high time to wise up to stop with this unproductive approach and think for alternative.

I will show you the solution provided by my dearest math mentor from the U.K. in my next blog post, in the mean time, I hope you could make the best of this opportunity to attempt at the solution, who knows, perhaps you could come up with the heuristic solution? :D

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