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=______.
题型:填空题难度:一般来源:不详
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=______. |
答案
∵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+y)(x+2y-y). 故答案为:(x-y)(x+2y+y)(x+2y-y). |
举一反三
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