補充教材：多項式的四則運算

1、多項式的四則運算【1】下列何者為 x 的多項式？(A) x2+ V2x+1 (B) x2+ /2)4+ 3( 1+ 2n,故必3n= n+6(左端3n次項需消去，否則等式不成立)= n=3(1)令 x=0,得 8f(0)02f(0)+12=0= 6f(0)=1右 f(0) = 2(2)令 f(x) = ax3+ bx2 + cx 2(aw 0)代入原式8(ax9 + bx6+ cx3 2)x6(8a + 4bx2 + 2cx- 2) = 2(aU + b/ + cx2 2)-128bx6 + 8cx3 16 4bx8- 2cx7+ 2x6= 2a6 + 2b/+ 2cx2 16=4bx8-2c

2、x7+ (8b+ 2)x6+ 8cx3- 16= 2ax6+ 2b/+ 2cx2- 16比較兩邊系數(shù)，得4b= 0, 2c= 0, 8b+ 2 = 2a, 8c= 0, 2c= 0=b=c=0, a= 1故 f(x) = x3 2【2】設t+1 = x,試將下列二式表為x的多項式(1) t3+ 口。(2) t5+40tt3t5解答:(1) x3- 3x, (2) x5 5x3 + 5x 【詳解】：(1) t+ t1 = x= (t+t1)2：x2= t2+1 2+2 = x2n t2+1 2 = x2 2t3+1 3=(t+tT)(t2 + t 2t. t1) = x(x23);x3 3x.

3、. (t2+1 j(t3+1 3) = (x2-2)(x3- 3x)t5+1 5 +1 +1 = x5 5x3+ 6x= t5+1 5= x5-5x3+ 5x【3】f(x)為一多項式，若(x+1)f(x)除以x2+ x+ 1的余式為5x+ 3, 求f(x)除以x2+x+ 1的余式。解答:2x+ 5 【詳解】： 一2f (x) = (x x 1) Q(x) ax b一2一=(x 1) f (x) = (x 1)(x x 1) Q(x) (ax b)(x 1)=(x 1)(x2 x 1) Q(x) , a(x2 x 1) 5x 3二(ax b)(x 1) = a(x2 x 1) 5x 3= ax2

4、 +(a +b)x + b =ax2 +(a +5)x + (a +3)= b=5,a = 2n 余式 2x+ 5【4】設 f(x) = (a+ 3)x3+ (b-2)x2+(3c+ 4)x+ d,若 f(1)=1, f(2)=2, f(3)=3, f(4) = 4, 求a、b、c、d的值。解答:a= - 3, b= 2, c= 1, d= 0 【詳解】：由 f(1) = 1, f(2) = 2, f(3) = 3, f(4) = 4設 f(x) = p(x- 1)(x- 2)(x-3)+ q(x- 1)(x- 2)+r(x- 1)+ s41)=1= s= 1f(2)=2= r+s= 2n r

5、= 1f(3)=3= 2q+ 2r+ s= 3n 2q+ 2+1 = 3= q=0f(4) = 4= 6p+ 6q+3r+ s=4= 6p+ 0+ 3+1=4= p = 0 . f(x) = (x 1)+ 1 = x 與 f(x) = (a+ 3)x3+ (b-2)x2+ (3c+ 4)x+ d 比較系數(shù)得 a+ 3=0, b-2=0, 3c+ 4=1, d = 0即 a= - 3, b = 2, c= 1, d = 0【5】x的多項式f(x)滿足f(x+1) f(x) = 2x 3且f(0)=2,則最低次的f(x) =。解答:x2-4x+ 2【詳解】：_最低次的 f(x)= deg(f (x) =2=令 f(x)=ax2 +bx + 2,貝 1f (x 1) - f(x) = a(x 1)2 b(x 1) c - ax2 bx c 122ax 2ax a bx b c - ax