Let $f(x)=x^5+x^2+1$ with $x_1,x_2,x_3,x_4,x_5$ as zeros and ...

Question

Let $f(x)=x^5+x^2+1$ with $x_1,x_2,x_3,x_4,x_5$ as zeros and $g(x)=x^2-2.$ Show that $$g (x_1)g (x_2)g (x_3)g (x_4)g (x_5)-30g(x_1x_2x_3x_4x_5)=7$$.

I found this question in a local question paper. And I have no idea how to solve it...

Please Help.

Answer 1

It is known that $$x^2+1+x^5=(x-x_1)(x-x_2)(x-x_3)(x-x_4)(x-x_5)\tag1$$

Comparing the constants of each side, we have $-1=x_1x_2x_3x_4x_5$.

Now, puting in $-x$ in $\text{(1)}$, we have $$x^2+1-x^5=-(x+x_1)(x+x_2)(x+x_3)(x+x_4)(x+x_5)\tag 2$$Multipling $\text{(1)}$ and $\text{(2)}$, we have $$h(x)=(x^2+1)^2-x^{10}=(x_1^2-x^2)(x_2^2-x^2)(x_3^2-x^2)(x_4^2-x^2)(x_5^2-x^2)$$ Now put in $x=\sqrt{2}$. We have $$g (x_1)g (x_2)g (x_3)g (x_4)g (x_5)-30g(x_1x_2x_3x_4x_5)=h(\sqrt{2})+30=9-32+30=7$$