1.在整式x^2+xy-2y^2+ay-9中,把a换成一个常数后可以使原多项式因式分解,求a的值,并将该多项式因式分解.

1
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)
2,
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
3,
(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)