Question

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

Answer 1

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)