已知x^2+3x-6=0,求x^3+x^2-12x+10的值。(详细过程)
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解:x^3+x^2-12x+10
=x^3+3x^2-2x^2-6x-6x+10
=x^3+3x^2-6x-2x^2-6x+10
=x(x^2+3x-6)-2x^2-6x+10
=-2x^2-6x+10
=-2(x^2+3x-6)-12+10
=-12+10
=-2
=x^3+3x^2-2x^2-6x-6x+10
=x^3+3x^2-6x-2x^2-6x+10
=x(x^2+3x-6)-2x^2-6x+10
=-2x^2-6x+10
=-2(x^2+3x-6)-12+10
=-12+10
=-2
追答
解:x^3+x^2-12x+10
=x^3+3x^2-2x^2-6x-6x+10
=x^3+3x^2-6x-2x^2-6x+10
=x(x^2+3x-6)-2x^2-6x+10
=-2x^2-6x+10
=-2(x^2+3x-6)-12+10
=-12+10
=-2
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