2x^3-3x^2-7x+2009
=2x^3-6x^2+3x^2-7x+2009
=2（x^3-3x^2+x）-2x+3x^2-7x+2009
=2x（x^2-3x+1）+3x^2-9x+2009
=2x（x^2-3x+1）+3(x^2-3x+1)-3+2009
=0+0+2006=2006
已知x^2-3x+1=0,求2x^3-3x^2-7x+2009
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