$t_0 = 0$, in which case $$\begin{aligned}
        h&=&\left(x_2 - \sqrt{-t_1}x_1 \right)\left(x_2 + \sqrt{-t_1} x_1\right).
        \end{aligned}$$

[*Proof.*]{} Suppose that $h$ is reducible. Then any factorization of $h$ must be of the form $$\begin{aligned}
 \label{Must-1}
h &=&(a_0 + a_1 x_1+ \cdots+a_nx_n)(b_0 + b_1 x_1+ \cdots+b_nx_n).\end{aligned}$$ We may also assume that $a_n=b_n=1,$ since $a_n b_n = 1$. Comparing coefficients of $x_nx_i$, $1 \le i \le n-1$, and coefficients of $x_n$, we see that $b_i=-a_i$ for $0 \le i \le n-1$. Therefore (\[Must-1\]) can be rewritten as $$\begin{aligned}
 \label{Must-2}
h &=&(a_0 + a_1 x_1+ \cdots+a_{n-1} x_{n-1}+ x_n)(-a_0 -a_1 x_1- \cdots -a_{n-1} x_{n-1} + x_n).\end{aligned}$$ Comparing coefficients of $x_i^2$, $1 \le i \le n-1$, and the constant terms, we see that $-t_i = a_i^2$ for $0 \le i \le n-1$. Thus $- t_i$ is a square for all $i$, $0 \le i \le n-1.$ Of course, $t_n = 1$.

If char ($k$) = 2, then $h$ factors into $$\begin{aligned}
h &=&(a_0 + a_1 x_1+ \cdots+a_{n-1} x_{n-1}+ x_n)^2.\end{aligned}$$

If char ($k$) $\ne 2$, and $n > 2$, then comparing the coefficients of $x_1x_2$ (in (\[Must-2\])), we obtain $-2 a_1 a_2 = 0$, contradicting the assumptions that $a_1^2 = -t_1 \ne 0$, $a_2^2 = -t_2 \ne 0$, 