✔ 最佳答案
1.(2x^3+x^2+1)^3除以2x^3+x^2+3的餘式為何?
Sol
(2x^3+x^2+1)^3
=>2x^3+x^2+3)-2]^3
=>(-2)^3
=>-8
2.設f(x)與g(x)為實係數多項式,以x^2-3x+2除f(x)得餘式3x-4,以x-1
除g(x)得餘式5,試求x-1除f(x)+g(x)的餘式?
Sol
f(x)=p(x)(x^2-3x+2)+3x-4
g(x)=q(x)(x-1)+5
f(1)+g(1)=(3*1-4)+5=4
3.設二次函數y=x^2+mx-1的圖型恆在直線y=x-5的上方,則實數m的範圍
為何?
Sol
(x^2+mx-1)-(x-5)>=0
x^2+(m-1)x+4>=0
D<=0
(m-1)^2-16<=0
-4<=m-1<=4
-3<=m<=5
4.m,n,s,t皆為整數,若x+t為x^2+mx-6與x^3+nx^2+sx+3的公因式,則t值
可能為何?
Sol
x+t為x^3+nx^2+sx+3的因式
t=1 or -1 or 3 or -3
(1) t=1
1+m-6=0,1+n+s+3=0
(2) t=-1
1-m-6=0,-1+n-s+3=0
(3) t=3
9-3m-6=0,27+9n+3s+3=0
(4) t=-3
9-3m-6=0,-27+9n-3s+3=0
So
t=1 or -1 or 3 or -3
5.若p,q為實數,且f(x)=x^3+px+3q,g(x)=x^3-qx-3p,若f(x)與g(x)的
最高公因式為一次式,則p-q為何?
Sol
f(x)-g(x)
=(x^3+px+3q)-(x^3-qx-3p)
=(p+q)x+3(p+q)
=(p+q)(x+3)
(1) p+q=0
f(x)=g(x) (不合)
(2) p+q<>0
最高公因式=x+3
f(-3)=-27-3p+3q=0,g(-3)=-27+3q-3p=0
p-q=9
6.方程式1/(x+1)+2/(x+2)+3/(x+3)=1的實根個數共有幾個
Sol
1/(x+1)+2/(x+2)+3/(x+3)=1
(x+2)(x+3)+2(x+1)(x+3)+3(x+1)(x+2)=(x+1)(x+2)(x+3)
(x^2+5x+6)+2(x^2+4x+3)+3(x^2+3x+2)=(x^2+3x+2)(x+3)
6x^2+22x+18=x^3+6x^2+11x+6
x^3-11x-12=0
f(x)=x^3-11x-12
f(-3)=-27+33-12=-6<0
f(-2)=-8+22-12=2>0
f(0)=-12<0
f(-3)*f(-2)=-12<0
-3 和 -2之間至少有一根
f(-2)*f(0)=-24<0
-2 和 0之間至少有一根
So
實根個數共有3個