3124
數學題目-多項式-1
1. 設f(x)=3x^3-4x^2 ax-b除以x^2-x 1的餘式為-2 3
則a 2b=?A.-4 B.-2 C.6 D.02.設多項式f(x)除以x^2-x-20所得於式3x-1
則f(x)除以x 4之餘式為?A.11 B-11 C-13 D133.設x-2和x-3均為x^3 ax^2 bx 6的因式
則a b=_________.4.若x^5 1=A(x 4)^5 B(x 4)^4 C(x 4)^3 D(x 4)^2 EB(x 4) F則B D F=?A-1683 B-3124 C-242 D以上皆非5.設8x^3-10x 3=a(2x 3)^3 b(2x 3)^2 c(2x 3) d
求a b c d=?
1. f(x)=3x^3-4x^2 ax-b除以x^2-x 1的餘式為-2x 3f(x)=3x^3-4x^2 ax-b=(x^2-x 1)(ex f)-2x 3最高項係數為3
e=3二次項係數為-4
1.f+(-1).e=1.f+(-1).3=-4
f=-1f(x)=3x^3-4x^2 ax-b=(x^2-x 1)(3x-1)-2x 3一次項係數為a
a=(-1).(-1)+1.3-2=2常數項係數為-b
b=-[1.(-1)+3]=-2a 2b=2 2.(-2)=-2或者 3-4+a-b a=2
b=-2 3-3+1 ────── -1+△+○ 先推這裡△=-1
○=2
再往上推 -1+1 -1 ────── -2+32. 多項式f(x)除以x^2-x-20所得於式3x-1f(x)=q(x).(x^2-x-20)+3x-1f(x)=q(x).(x-5).(x 4)+3x-1f(-4)=3.(-4)-1=-113.x-2和x-3均為x^3 ax^2 bx 6的因式f(x)=x^3 ax^2 bx 6f(2)=2^3 a*2^2 b*2 6=0 2a b 7=0f(2)=3^3 a*3^2 b*3 6=0 3a b 11=0a=-4
b=1a b=-34. x^5 1=A(x 4)^5 B(x 4)^4 C(x 4)^3 D(x 4)^2 E(x 4) Fx=-3代入 (-3)^5 1=A(-3 4)^5 B(-3 4)^4 C(-3 4)^3 D(-3 4)^2 E(-3 4) F-242=A B C D E F---------------------(1)x=-5代入 (-5)^5 1=A(-5 4)^5 B(-5 4)^4 C(-5 4)^3 D(-5 4)^2 E(-5 4) F-3124=-A B-C D-E F--------------------(2)[(1) (2)]/2 --