来自刘决仕的问题
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=______.
1回答
2020-10-15 02:10
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=______.
∵代数式x3+y3+3x2y+axy2含有因式x-y,
∴当x=y时,x3+y3+3x2y+axy2=0,
∴令x=y,即x3+x3+3x3+ax3=0,
则有5+a=0,解得a=-5.
将a=-5代入x3+y3+3x2y+axy2,得
x3+y3+3x2y-5xy2
=x3-x2y+4x2y-5xy2+y3
=(x-y)x2+y(x-y)(4x-y)
=(x-y)(x2+4xy-y2)
=(x−y)(x+2y+
5y)(x+2y−
5y)