Find the values of the parameter $\alpha$ so 

that the following equation : 

$P(x)=16x^{4}-\alpha x^{3}+(2\alpha+17)x^{2}-\alpha x+16=0$

has four different real solutions, and they form a 

geometric sequence

Actually I don't know how I solve but my try as

following : 

let : $x\_{1}=b$ , $x\_{2}=bq$ , $x\_{3}=bq^{2}$ and $x\_{4}=bq^{3}$

Now : $P(x)=16(x-b)(x-bq)(x-bq^{2})(x-bq^{3})$ 

And comparing but I can't find any things! 

If any one have any idea please drop here to see 

Thanks!