Let $f(x) =x^5+x^2+1$ with $a, b, c, d, e$ as zeros. . .

Question

Let $f(x) =x^5+x^2+1$ with $a, b, c, d, e$ as zeros and $g(x)=x^2-2.$ Show that $$g(a)g(b)g(c)g(d)g(e)=-23. $$

I have no idea how to solve it.

Please help.

Thanks in advance.

Answer 1

The numbers $a^2,b^2,c^2,d^2,e^2$ are all zeros of $$ h(x)=x^5-x^2-2x-1. $$ This is because $$ h(x^2)=x^{10}-x^4-2x^2-1=-f(x)f(-x). $$ This implies, in turn, that $a^2-2,b^2-2,c^2-2,d^2-2,e^2-2$ are the zeros of $$ r(x):=h(x+2)=x^5+10 x^4+40 x^3+79 x^2+74 x+23. $$

We also see that $\gcd(f(x),f(-x))=1$, so consequently $r(x)$ has no other roots. Do you see why? At this point you can use the relation between the coefficients of a polynomial and the product of its roots.

Answer 2

Let $a_1,\ldots,a_5$ be the roots of $f(x)=x^5+x^2+1$. Then for every real $x$ we have:

$$f(x)=x^5+x^2+1=\prod_{i=1}^5(x-a_i).$$

So, $g(a_1)\cdot g(a_2)\cdot g(a_3)\cdot g(a_4)\cdot g(a_5)$ is equal to:

\begin{align*} \prod_{i=1}^5 g(a_i)&=\prod_{i=1}^5 (a_i^2-2)=\prod_{i=1}^5 (a_i-\sqrt{2})(a_i+\sqrt{2})\\ &=\prod_{i=1}^5 (a_i+\sqrt{2}) \cdot \prod_{i=1}^5 (a_i-\sqrt{2})\\ &=\prod_{i=1}^5 (a_i+\sqrt{2}) \cdot \prod_{i=1}^5 [(-1)(\sqrt{2}-a_i)]\\ &=\prod_{i=1}^5 (a_i+\sqrt{2}) \cdot (-1)^5 \prod_{i=1}^5 (\sqrt{2}-a_i)\\ &=\prod_{i=1}^5 [(-1)(a_i+\sqrt{2})] \cdot \prod_{i=1}^5 (\sqrt{2}-a_i)\\ &=\prod_{i=1}^5 (-\sqrt{2}-a_i) \cdot \prod_{i=1}^5 (\sqrt{2}-a_i)\\ &=f(-\sqrt{2})\cdot f(\sqrt{2})\\ &=(-4\sqrt{2}+3)(4\sqrt{2}+3)\\ &=-23 \end{align*}

Answer 3

Let $\{a,b,c,d,e\}$ be $r_i : i \in [1,5]$. The $f(x)=0$ condition then says that: $$ \sum_i r_i = \sum_{i<j} r_ir_j = \sum_{i<j<k<m} r_ir_jr_kr_m = 0 \\ \sum_{i<j<k} r_ir_jr_k = r_1r_2r_3r_4r_5 =1 \\ $$ Call this the given conditions.

And the number to be found is $$ \prod_i (r_i^2-2) = -32 + 16\sum_i r_i^2 -8\sum_{i<j} r_i^2r_j^2 \\ +4 \sum_{i<j<k} r_i^2r_j^2r_k^2 -2 \sum_{i<j<k<m} r_i^2r_j^2r_k^2r_m^2 + r_1^2r_2^2r_3^2r_4^2r_5^2 $$ We can express these terms in ways that allow us to use the given conditions. $$\sum_i r_i^2 = \left( \sum_i r_i \right)^2 -2 \sum_{i<j} r_ir_j = 0-2\cdot 0 = 0$$ $$ \sum_{i<j} r_i^2r_j^2 = \left( \sum_{i<j} r_ir_j \right)^2-2\sum_i r_i^2 \sum_{j,k\neq i:j<k} r_jr_k -6\sum_{i<j<k<m} r_ir_jr_kr_m = 2\sum_i r_i^2 \sum_{j,k\neq i:j<k} r_jr_k \\ \sum_i r_i^2 \sum_{j,k\neq i:j<k} r_jr_k = \sum_{i<j<k} r_ir_jr_k \sum_i r_i - 4 \sum_{i<j<k<m} r_ir_jr_kr_m = 1 \cdot 0 -4\cdot 0 = 0 \\ \sum_{i<j} r_i^2r_j^2 = 0 $$ In a similar way, you find that $$ \sum_{i<j<k} r_i^2r_j^2r_k^2 =1 \\ \sum_{i<j<k<m} r_i^2r_j^2r_k^2r_m^2 = -2 $$ And of course $r_1^2r_2^2r_3^2r_4^2r_5^2 = (r_1r_2r_3r_4r_5)^2 = 1$.

Putting these together, $$ \prod_i (r_i^2-2) = -32 + 16 \cdot 0 -8\cdot 0 +4 \cdot 1 -2 \cdot (-2) + 1 = -23 $$ By the way, if you started with $$x^5 + 2x^2 + 1 = 0$$ you would find $$g(a)g(b)g(c)g(d)g(e) = -32 + 16 \cdot 0 -8\cdot 0 +4 \cdot 4 -2 \cdot (-4) + 1 = -7$$ I wonder if it is a general property that this product always comes out to be an ineteger? Indeed it seems to be the case that if $P(x)$ and $Q(x)$ are polynomials with integer coefficients, and the roots of $P(x)$ are $r_i$, then $\prod_i Q(r_i)$ is always an integer.