若f(x)=2x^3-3x^2+ax+b除以x+1所得的余数为7,除以x-1所得的余数为5,试求a,b的值
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若f(x)=2x^3-3x^2+ax+b除以x+1所得的余数为7,除以x-1所得的余数为5,试求a,b的值
若f(x)=2x^3-3x^2+ax+b除以x+1所得的余数为7,除以x-1所得的余数为5,试求a,b的值
若f(x)=2x^3-3x^2+ax+b除以x+1所得的余数为7,除以x-1所得的余数为5,试求a,b的值
f(x)=2x^3-3x^2+ax+b
=2x^3+2x^2-5x^2+ax+b
=2x^2*(x+1)-5x^2-5x+5x+ax+b
=2x^2*(x+1)-5x(x+1)+(a+5)x+(a+5)-a-5+b
=2x^2*(x+1)-5x(x+1)+(a+5)(x+1)+(b-a-5)
所以:b-a-5=7 --------------(1)
f(x)=2x^3-3x^2+ax+b
=2x^3-2x^2-x^2+ax+b
=2x^2*(x-1)-x^2+x+(a-1)x+b
=2x^2*(x-1)-x(x-1)+(a-1)(x-1)+(b+a-1)
所以:a+b-1=5 -------------- (2)
联立(1),(2)得:
a=-3,b=9
若f(x)=2x^3-3x^2+ax+b除以x+1所得的余数为7,除以x-1所得的余数为5
则f(1)=5=2-3+a+b=a+b-1
f(-1)=7=-2-3-a+b=-a+b-5
所以a=-3
b=9
由f(x)除以x+1所得的余数为7可得:
f(x)=2x^2(x+1)-5x(x+1)+(5+a)x+(b-7)+7
5+a=b-7 ①
由f(x)除以x-1所得的余数为5可得:
f(x)=2x^2(x-1)-x(x-1)+(a-1)x+(b-5)+5
a-1=5-b ②
联立①、②得
a=-3,b=9。