题型一 二次系数为1型
十字相乘法分解因式：
【典例1】(1)$x^2 + 3x + 2$；(2)$x^2 - 3x + 2$.
变式.(1)$x^2 + 2x - 3$；(2)$x^2 - 2x - 3$.

【典例1】解:(1)$(x+1)(x+2)$；十字分解：$\begin{array}{r} x\ \ \ \ 1\\ x\ \ \ \ 2\end{array} \diagdown \diagup$
(2)$(x-1)(x-2)$；十字分解：$\begin{array}{r} x\ \ \ -1\\ x\ \ \ -2\end{array} \diagdown \diagup$
变式.解:(1)$(x+3)(x-1)$；十字分解：$\begin{array}{r} x\ \ \ \ 3\\ x\ \ \ -1\end{array} \diagdown \diagup$
(2)$(x-3)(x+1)$；十字分解：$\begin{array}{r} x\ \ \ -3\\ x\ \ \ \ 1\end{array} \diagdown \diagup$

【典例2】(1)$2x^2 - x - 1$；(2)$3x^2 + 5x + 2$.
变式.(1)$4x^2 - 5x + 1$；(2)$3x^2 + 7x + 4$.

【典例2】解:(1)$(2x+1)(x-1)$；十字分解：$\begin{array}{r} 2x\ \ \ \ 1\\ x\ \ \ -1\end{array} \diagdown \diagup$
(2)$(3x+2)(x+1)$；十字分解：$\begin{array}{r} 3x\ \ \ \ 2\\ x\ \ \ \ 1\end{array} \diagdown \diagup$
变式.解:(1)$(4x-1)(x-1)$；十字分解：$\begin{array}{r} 4x\ \ \ -1\\ x\ \ \ -1\end{array} \diagdown \diagup$
(2)$(3x+4)(x+1)$；十字分解：$\begin{array}{r} 3x\ \ \ \ 4\\ x\ \ \ \ 1\end{array} \diagdown \diagup$

【典例3】(1)$x^2 - 3xy - 4y^2$；(2)$2x^2 -5xy +3y^2$.
变式.(1)$x^2 -(k+2)x +2k$；(2)$x^2 +2nx -m^2 +n^2$.

【典例3】解:(1)原式=$(x+y)(x-4y)$；十字分解：$\begin{array}{r} x\ \ \ \ y\\ x\ \ -4y\end{array} \diagdown \diagup$
(2)原式=$(2x-3y)(x-y).$ 十字分解：$\begin{array}{r} 2x\ \ -3y\\ x\ \ \ -y\end{array} \diagdown \diagup$
变式.解:(1)原式=$(x-2)(x-k)$；十字分解：$\begin{array}{r} x\ \ \ \mathrm{}\ \ \ -2\\ x\ \ \ \ \ \ \ \ \ \ -k\end{array}$
(2)原式=$(x-m+n)(x+m+n).$ 十字分解：$\begin{array}{r} x\ \ \ \mathrm{}\ \ \ -(m-n)\\ x\ \ \ \ \ \ \ \ \ \ m+n\end{array}$