问题描述: x2-(p2+q2)x+pq(p+q)(p-q),x2-2xy-8y2-x-14y-6,这2道体求因式分解,(字母后为平方) 1个回答 分类:数学 2014-11-25 问题解答: 我来补答 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) 展开全文阅读