Let $$p(x) = 21x^4-11x^3-23x^2+11x+2.$$ It is easy to check that $p(1)=p(-1)=0$, i.e., $1$ and $-1$ are zeros of $p$. $p$ has two more real zeros. The smaller is $x=$ , and the larger is $x=$ .
Hint: Use synthetic division twice to divide by $x-1$ and $x+1$, or use long division to divide by $(x-1)(x+1)= x^2-1$. The find the zeros of the resulting quadratic quotient.