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.

---------------------------------------------------------------

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.