I started doing this by saying
\((x-a)(x-b)(x-c)(x-d)(x-e)=x^5+7x^4-2\)
a+b+c+d+e= -7 and abcde=2
ab +ac +ad +ae +bc +bd +be +cd +ce +de = 0
abc+abd+abe+acd+ace+ade+ bcd+bce+bde+cde = 0
abcd + abce + abde +acde + bcde =0
Maybe you can use that.
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So next I used calculus to find the turning points of \(y=x^5+7x^4-2\)
I found that there are only 2 turning points. One at x=0 and the other at x=-5.6
Since there are only 2 points there can be a maximum of 3 distinct real roots.
So the other 2 roots are not real but they are the conjugates of one another. ie p+qi and p-qi
If you let a=p+qi and b=p-qi then ab=p^2+q^2
I am just thinking in print.
Maybe you can get some ideas from this.