问题: 急!!!!!!!!!!!!因式分解!
1.(x+1)(x+2)(x+3)(x+4)+1
2.4x^2-4xy-3y^2-4x+10y-3
3.m^2+7mn-18n^2-5m+43n-24
4.若a^2+2a+5是a^4+ma^2+n的一个因式,那mn的值为多少?
5.(y^2-1)(y+3)(y+5)-9
解答:
1.(x+1)(x+2)(x+3)(x+4)+1
=(x+1)(x+4)(x+2)(x+3)+1
=(x^2+5x+4)(x^2+5x+6)+1
设:x^2+5x=y
则原式=(y+4)(y+6)+1
=y^2+10y+24+1
=(y+5)^2
所以:(x+1)(x+2)(x+3)(x+4)+1
=(x^2+5x+5)^2
2.4x^2-4xy-3y^2-4x+10y-3
=(2x-3y)(2x+y)-4x+10y-3
=(2x-3y+1)(2x+y-3)
3.m^2+7mn-18n^2-5m+43n-24
=(m-2n)(m+9n)-5m+43n-24
=(m-2n+3)(m+9n-8)
4.若a^2+2a+5是a^4+ma^2+n的一个因式,那mn的值为多少?
可以设:a^4+ma^2+n=(a^2+2a+5)(a^2+ka+p)
展开括号等式的右边=a^4+2a^3+5a^2+ka^3+2ka^2+5ka+pa^2+5pa+5p
=a^4+(k+2)a^3+(5+2k+p)a^2+(5k+5p)a+5p
根据等式的性质有:
k+2=0
5+2k+p=m
5k+5p=0
n=5p
解:k=-2,p=2,m=3,n=15
则mn=2*15=30
5.(y^2-1)(y+3)(y+5)-9
=(y-1)(y+1)(y+3)(y+5)-9
=(y-1)(y+5)(y+1)(y+3)-9
=(y^2+4y-5)(y^2+4y+3)-9
设y^2+4y=x
则原式=(x-5)(x+3)-9
=x^2-2x-15-9
=x^2-2x-24
=(x+4)(x-6)
则.(y^2-1)(y+3)(y+5)-9
=(y^2+4y+4)(y^2+4y-6)
=(y+2)^2(y^2+4y-6)
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