The polynomial (s+1)x^2 - (s-2)x + 1 is a perfect square. Find the value(s) of s. ​

let's recall ourselves on a perfect square triinomial[tex]\bf \qquad \textit{perfect square trinomial} \\\\ (a\pm b)^2\implies a^2\pm \stackrel{\stackrel{\text{\small 2}\cdot \sqrt{\textit{\small a}^2}\cdot \sqrt{\textit{\small b}^2}}{\downarrow }}{2ab} + b^2[/tex]so, let's notice, the middle term is really just the product of 2 times the square root of the other guys on the left and right side.  So let's get that product and equate it to the middle term.[tex]\bf \stackrel{\textit{product of 2 and the }\sqrt{\qquad }\textit{ of other terms}}{2(\sqrt{(s+1)x^2})(\sqrt{1})}~~=~~\stackrel{\textit{middle term of the trinomial}}{(s-2)x} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ 2\sqrt{(s+1)x^2}=(s-2)x\implies 2x\sqrt{s+1}=(s-2)x\implies 2\sqrt{s+1}=s-2[/tex][tex]\bf \stackrel{\textit{squaring both sides}}{(2\sqrt{s+1})^2=(s-2)^2}\implies 2^2(\sqrt{s+1})^2=s^2-4s+4 \\\\\\ 4(s+1)=s^2-4s+4\implies 4s+4=s^2-4s+4\implies 4s=s^2-4s \\\\\\ 0=s^2-8s\implies 0=s(s-8)\implies s= \begin{cases} 0\\ 8 \end{cases}[/tex]