如果x^2-3x+1=0,求2x^5-5X^4+2x^3-8X^2+3X

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如果x^2-3x+1=0,求2x^5-5X^4+2x^3-8X^2+3X

如果x^2-3x+1=0,求2x^5-5X^4+2x^3-8X^2+3X
如果x^2-3x+1=0,求2x^5-5X^4+2x^3-8X^2+3X

如果x^2-3x+1=0,求2x^5-5X^4+2x^3-8X^2+3X
2x^5-5x^4+2x^3-8x^2+3x
=2x^3(x^2-3x+1)+x^4-8x^2+3x
=x^4-8x^2+3x
=x^4-7x^2-(x^2-3x)
=x^4-7x^2-1
=x^2(x^2-3x+1)+3x^3-8x^2-1
=3x^3-8x^2-1
=3x(x^2-3x+1)+x^2-3x-1
=x^2-3x-1
=-1-1
=-2

用2x^5-5X^4+2x^3-8X^2+3X除以x^2-3x+1=0,即多项式的除法。

x^2-3x+1=0
x^2-3x=-1
2x^5-5X^4+2x^3-8X^2+3X
=x^3(2x^2-5X)+2x(x^2-4X)+3X
=x^3(-2+x)+2x(-1-x)+3x
=x^4-2x^3-2x^2+x
=x^2(x^2-2x)-2x^2+x
=(3x-1)(-1+x)-2(3x-1)+x
=3x^2-4x+1-6x+2+x
=3x^2-9x+3
=-3+3
=0

不一定要分解因式:
(2x^3)(x^2-3x+1)+x^4-8x^2+3x=0+(4x^2)(x^2-3x+1)+2x^2+3x=0+0+(-1)=-1

如果x²-3x+1=0,x=(3±√5)/2,x²=3x-1代入2x^5-5X^4+2x^3-8X^2+3X得
2x^5-5X^4+2x^3-8X^2+3X
=x³(2x²-5x+2)-8x²+3x
=x³[2(3x-1)-5x+2]-8(3x-1)+3x
=x^4-21x+8
=x(x³...

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如果x²-3x+1=0,x=(3±√5)/2,x²=3x-1代入2x^5-5X^4+2x^3-8X^2+3X得
2x^5-5X^4+2x^3-8X^2+3X
=x³(2x²-5x+2)-8x²+3x
=x³[2(3x-1)-5x+2]-8(3x-1)+3x
=x^4-21x+8
=x(x³+27)-48x+8
=x(x+3)(x²-3x+9)-48x+8
=(x²+3x)(x²-3x+9)-48x+8
=(3x-1+3x)(-1+9)-48*(3±√5)/2+8
=-48*(3±√5)/2
=-24(3±√5)

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