求x²-(p²+q²)x+pq(p+q)(p-q)和x²-2xy-8y²-x-14y-6因式分解

求x²-(p²+q²)x+pq(p+q)(p-q)和x²-2xy-8y²-x-14y-6因式分解
求x²-(p²+q²)x+pq(p+q)(p-q)和x²-2xy-8y²-x-14y-6因式分解

求x²-(p²+q²)x+pq(p+q)(p-q)和x²-2xy-8y²-x-14y-6因式分解
x²-(p²+q²)x+pq(p+q)(p-q﹚＝[x－q﹙p＋q﹚][x－p﹙p－q﹚]
x²-2xy-8y²-x-14y-6＝﹙x＋2y＋2﹚﹙x－4y－3﹚
交叉相乘法.

x²-(p²+q²)x+pq(p+q)(p-q)=[x-p(p-q)][x-q(p+q)]=(x-p²+pq)(x-pq-q²)
x²-2xy-8y²-x-14y-6=(x-4y)(x+2y)-(x+14y)-6=(x-4y-2)(x+2y+3)