2x∧3-x∧2+3x-5=a(x-2)∧3+b(x-2)∧2+c(x-2)+d,求a,b,c,d

1个回答

  • 首先,原式右边x的3次方项系数是a,所以 a=2

    原式等号右边

    = 2(x^2-4x+4)(x-2) +b(x^2-4x+4) +cx -c +d

    = 2(x^3 -4x^2+4x -2x^2+8x-8) +bx^2-4bx+4b+cx-c+d

    = 2x^3 -8x^2+8x-4x^2+16x-16 +bx^2-4bx+4b+cx-c+d

    = 2x^3 -(12-b)x^2+(24-4b+c)x -16+4b-c+d

    对照等号左边,可得

    12-b = 1

    24-4b+c = 3

    -16+4b-c+d = -5

    解之,得 b=11,c=23,d = -10

    所以 a =2,b =11,c =2,d = -10

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