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10) 27y^2+90yz+75z^2
= 3(9y^2+30yz+25z^2)
=3(3y+5z)(3y+5z)
=3(3y+5z)^2

11) (a+b)^2+8(a+b)+16
let a+b=c
then, (a+b)^2+8(a+b)+16 become
c^2+8c+16
=(c+4)^2
=[(a+b)+4]^2
=(a+b+4)^2

12) z^4-z^2+10z-25
=z^4-(z^2-10z+25)
=(z^2)^2-(z-5)^2
=[z^2+(z-5)][z^2-(z-5)]
=(z^2+z-5)(z^2-z+5)

10.27y^2+90yz+75z^2
=3(9y^2+10yz+25z^2)
=3(9y+5z)^2 < ---------------------(x+y)^2=x^2+2xy+y^2
11.(a+b)^2+8(a+b)+16
=(a+b+4)(a+b+4)
將(a+b)用x代替,變成x^2+8x+16,用化簡成(x+4)(x+4),再將(a+b)代返入去個答案
12.z^4-z^2+10z-25
=z^4-(z^2-10z+25)<----------------------------(x-y)^2=(x^2-2xy+y^2)
=z^4-(z-5)^2

2007-08-06 17:56:12 補充：
第12題有少少錯12.z^4-z^2 10z-25=z^4-(z^2-10z 25)<----------------------------(x-y)^2=(x^2-2xy y^2)=z^4-(z-5)^2 =(z^2-z 5)(z^2 z-5)

10. 27y(二次方)+90yz+75z(二次方)

= 3(9y^2 + 30yz +25z)

=3((3y+5z)^2)


11. (a+b)二次方+8(a+b)+16

= ((a+b+4))^2

12. z(四次方)-z(二次方)+10z-25
= z^2-(z^2 -10z+25)
= z^4 - (z-5)^2
= (z^2+z-5) (z^2-(z-5))
=(z^2+z-5) (z^2-z+5)

註：^2 = 二次方; ^4 = 四次方