Let $P(x)=x^5+x^2+1$ and the equation has roots $x_1,x_2,..x_5$.

Question

Let $P(x)=x^5+x^2+1$ and the equation has roots $x_1,x_2,..x_5$. If $g(x)=x^2-2$, then find the value of $g(x_1)g(x_2)....g(x_5)$

I have no idea on how to begin. Clearly, they can’t all be multiples, as it would take an eternity.

What I noticed was that the sum of roots is zero, and the product is 1, but I don’t think that’s very useful.

Answer 1

We can write

$$P(x)=\prod_{i=1}^5(x-x_i)=(-1)^5\cdot\prod_{i=1}^5(x_i-x)=-\prod_{i=1}^5(x_i-x)$$

so

$$-P(x)=\prod_{i=1}^5(x_i-x)$$

From this, setting $x=\sqrt{2}$ and $x=-\sqrt{2}$, we get that:

$$-P(\sqrt{2})=\prod_{i=1}^5(x_i-\sqrt{2})$$

and

$$-P(-\sqrt{2})=\prod_{i=1}^5(x_i+\sqrt{2})$$

Now

$$g(x_1)=(x_1-\sqrt{2})(x_1+\sqrt{2})$$

so

$$g(x_1)g(x_2)g(x_3)g(x_4)g(x_5)=\prod_{i=1}^5(x_i-\sqrt{2})\prod_{i=1}^5(x_i+\sqrt{2})=$$ $$[-P(\sqrt{2})]\cdot [-P(-\sqrt{2}]=P(\sqrt{2})P(-\sqrt{2})$$

Can you end it now?