(1)
(p^2-4)(p^2+4p)-48
=(p-2)(p+2)p(p+4)-48
=(p-2)p(p+2)(p+4)-48
设:q=p+1
则原式=(q-3)(q-1)(q+1)(q+3)-48
=(q^2-1)(q^2-9)-48
=q^4-10q^2+9-48
=q^4-10q^2-39
=(q^2-13)(q^2+3)
把q=p+1代入得:
原式=((p+1)^2-13)((p+1)^2+3)
=(p^2+2p+1-13)(p^2+2p+1+3)
=(p^2+2p-12)(p^2+2p+4)
(2)
x^2+xy-2y^2+ay-9
=(x+2y)(x-y)+ay-9
所以,设x^2+xy-2y^2+ay-9=(x+2y+m)(x-y+n)
=(x+2y)(x-y)+m(x-y)+n(x+2y)+mn
=(x+2y)(x-y)+(m+n)x+(2n-m)y+mn
则:
m+n=0
2n-m=a
mn=-9
解得:m=3,n=-3,a=-9
或:m=-3,n=3,a=9
即:a=±9
a=-9时,x^2+xy-2y^2+ay-9=(x+2y+3)(x-y-3)或:
a=9时,x^2+xy-2y^2+ay-9=(x+2y-3)(x-y+3)
(3)
设y^4+6y^3+my^2+ny+36=(y^2+3y+6)(y^2+ay+6)
=y^4+(3+a)y^3+(6+6+3a)y^2+(18+6a)y+36
3+a=6
a=3
m=6+6+3a=12+3*3=21
n=18+6a=18+6*3=36
(4)
（a+3b)^3-a^3-27b^3
=(a+3b-a)(a^2+6ab+9b^2+a^2+3ab+a^2)-(3b)^3
=3b(3a^2+9b^2+9ab)-(3b)^3
=3b(3a^2+9b^2+9ab-9b^2)
=3b(3a^2+9ab)
9ab(a+3b)
(5)
(x-2y)(x+3y)----k=1
(x-y)(x+6y)----k=5
(x-3y)(x+2y)----k=-1
(x+y)(x-6y)----k=-5
(6)