Question

煩

求救各位大神們

設f(x)=x²+2xx²+3x+4,則以x²+x²+2x²+x+1除xf(x)+x*f(x)+xf(x)+1的餘式為

Answer 1

✨ ベストアンサー ✨

qn

x⁴+x³+2x²+x+1
=(x⁴+x³+x²)+(x²+x+1)
=x²(x²+x+1)+(x²+x+1)
=(x²+1)(x²+x+1)

令 g(x)=(x²+1)(x²+x+1)

x⁹f(x)+x⁵f(x)+xf(x)+1
=x(x⁸+x⁴+1)f(x)+1
=x(x²+x+1)(x⁶-x⁵+x³-x+1)f(x)+1
=(x²+x+1)(x⁷-x⁶+x⁴-x²+x)f(x)+1
=(x²+x+1)[(x²+1)(x⁵-x⁴-x³+2x²+x-3)+3]f(x)+1
=(x²+x+1)(x²+1)(x⁵-x⁴-x³+2x²+x-3)f(x) + (x²+x+1)·3·f(x) +1
=g(x)(x⁵-x⁴-x³+2x²+x-3)f(x) + 3(x²+x+1)(x³+2x²+3x+4) +1
=g(x)(x⁵-x⁴-x³+2x²+x-3)f(x)+3(x²+x+1)[(x²+1)(x+2)+2x+2] +1
=g(x)(x⁵-x⁴-x³+2x²+x-3)f(x)+3(x²+x+1)(x²+1)(x+2)
+3(x²+x+1)(2x+2) +1
=g(x)(x⁵-x⁴-x³+2x²+x-3)f(x)+g(x)(3x+6)+3(2x³+4x²+4x+2) +1
=g(x)[(x⁵-x⁴-x³+2x²+x-3)f(x)+(3x+6)] + 6x³+12x²+12x+7