x2-(p2+q2)x+pq(p+q)(p-q), x2-2xy-8y2-x-14y-6, 这2道体求因式分解，谢谢各位高手。（字母后为平方）
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1.x^2-(p^2+q^2)x+pq(p+q)(p-q)
=x^2-(p^2+q^2)x+pq(p^2-q^2)
=x^2-(p^2+q^2)x+pq(p-q)(p+q)
=x^2-(p^2+q^2)x+(p^2-pq)(pq+q^2)
x^2-(p^2+q^2)x+(p^2-pq)(pq+q^2)
=[x-(p^2-pq)][x-(pq+q^2)]
=(x-p^2+pq)(x-pq-q^2)
2.x2--2xy--8y2--x--14y--6
=(x-4y)(x+2y)+(2x-8y)-3x-6y-6
=(x-4y)(x+2y)+2(x-4y)-3x-6y-6
=(x-4y)(x+2y+2)-3(x+2y+2)
=(x-4y-3)(x+2y+2)
=x^2-(p^2+q^2)x+pq(p^2-q^2)
=x^2-(p^2+q^2)x+pq(p-q)(p+q)
=x^2-(p^2+q^2)x+(p^2-pq)(pq+q^2)
x^2-(p^2+q^2)x+(p^2-pq)(pq+q^2)
=[x-(p^2-pq)][x-(pq+q^2)]
=(x-p^2+pq)(x-pq-q^2)
2.x2--2xy--8y2--x--14y--6
=(x-4y)(x+2y)+(2x-8y)-3x-6y-6
=(x-4y)(x+2y)+2(x-4y)-3x-6y-6
=(x-4y)(x+2y+2)-3(x+2y+2)
=(x-4y-3)(x+2y+2)
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