若代数式x3+y3+3x2y+axy2含有因式x-y,则a=_,在实数范围内将这个代数式分解因式,得x3+y3+3x2y+axy2=_.
若代数式x3+y3+3x2y+axy2含有因式x-y,则a=______,在实数范围内将这个代数式分解因式,得x3+y3+3x2y+axy2=______.
数学人气:429 ℃时间:2019-08-19 14:49:52
优质解答
∵代数式x
3+y
3+3x
2y+axy
2含有因式x-y,
∴当x=y时,x
3+y
3+3x
2y+axy
2=0,
∴令x=y,即x
3+x
3+3x
3+ax
3=0,
则有5+a=0,解得a=-5.
将a=-5代入x
3+y
3+3x
2y+axy
2,得
x
3+y
3+3x
2y-5xy
2
=x
3-x
2y+4x
2y-5xy
2+y
3=(x-y)x
2+y(x-y)(4x-y)
=(x-y)(x
2+4xy-y
2)
=
(x−y)(x+2y+y)(x+2y−y).
故答案为:
(x−y)(x+2y+y)(x+2y−y).
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