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