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.