jmc

Expanding, we get $$[p(x){]}^2=4x^4+20x^3+21x^2-10x+1.$$Then $p(x)$ is quadratic, with leading term $2x^2.$ Let $$p(x)=2x^2+bx+c.$$Then $$[p(x){]}^2=4x^4+4b{x}^3+(b^2+4c)x^2+2bcx+c^2.$$Matching coefficients, we get $$\begin{array}{rl}4b& =20,\\ b^2+4c& =21,\\ 2bc& =-10,\\ c^2& =1.\end{array}$$From $4b=20,$ $b=5.$ Then from $2bc=-10,$ $c=-1.$ Hence, $p(x)=\boxed{2x^2+5x-1}.$