✔ 最佳答案
1.f(x)=(x²+x+1)Q(x)+(x+2),x²f(x)=(x²+x+1)P(x)+R(x),R(x)=? 答:-2x-1x²+x+1=0 => x²=-x-1x²f(x)=x²(x²+x+1)Q(x)+(x+2)x²R(x)=(x+2)x²=-(x+2)(x+1)=-x²-3x-2=(x+1)-3x-2=-2x-1......ans
2.f(x)=x²-2x-3,g(x)=x函數,f[g(x)]=4x⁴+12x³-15x²-36x+32,g(1)=?答:0或2Ans:Set g(x)=2x^2+ax+bf[g(x)]=(cx^2+ax+b)²-2(2x^2+ax+b)-3=c²x⁴+2cax³+2(a²+bc-ca)x²+2a(b-c)x+(b²-b-3)=> c=+-2, a=+-3, b=-5 or 7g(x)=2x²+3x-5 or g(x)=-2x²-3x+7=> g(1)=2+3-5=0 or g(1)=-2-3+7=2.....ans
3.a,b=Real,f(x)=ax^9+bx^8+1=Q(x)(x-1)²+2,則數對(a,b)=?答:-8,90=(x-1)²=x²-2x+1 => x²=2x-1(2x-1)²=4x²-4x+1=4(2x-1)-4x+1=4x-3(2x-1)⁴=(4x-3)²=16x²-24x+9=16(2x-1)-24x+9=8x-7 f(1)=a+b+1=2 => a+b=1......(1)f(x²)=ax(2x-1)⁴+b(2x-1)⁴+1=ax(8x-7)+b(8x-7)+1=8ax²-7ax+8bx-7b+1=8a(2x-1)+(8b-7a)x+(1-7b)=(9a+8b)x+(1-7b-8a)=(a+8)x-(a+6)......by Eq.(1)=0x+2=> a=-8, b=1-a=9.....ans
4.多項式f(x)=(x³+1)Q(x)+(2x²+3x-1)(1) f(x)除以(x+1)的餘試為?R=f(-1)=0+2-3-1=-2 (2) f(x)除以(x²-x+1)的餘試為?f(x)=(x³+1)Q(x)+(2x²+3x-1)=(x+1)(x²-x+1)Q(x)+(2x²+3x-1)
x²-x+1=0 => x²=x-1R=f(x²)=0+2(x-1)+3x-1=2x-2+3x-1=5x-3......ans 5.p,q=正整數,f(x)=x⁵-2px⁴+x³-qx²+x-2有正整數解,p,q=? 答:1,2f(x)=(x⁵-2px⁴)+(x³-qx²)+(x-2)=(x-2)(x⁴+x²+1)=> p=1, q=2.......ans 6.若f(x)為四次多項式.則f(x²-1)的次數為? 答:8 If f(x)=ax⁴+bx³+cx²+dx+ethen f(x²-1)=a(x²-1)⁴+b(x²-1)³+c(x²-1)²+d(x²-1)+e=8次方
2014-08-10 05:10:04 補充：
修正g(x)=cx^2+ax+b
