✔ 最佳答案
(1+px+x^2)^2
= 1+px+x^2 + px + (px)^2 + px^3 + x^2 + px^3 + x^4
= 1+2px+(p^2 +2)x^2+2px^3+x^4
= x^4 + 2px^3 + (p^2+2)x^2 + 2px + 1
the coefficient of x in the expansion of (1+px+X^2)^2 is -6
Therefore
2p=-6
p=-3
the coefficient of x^2
=p^2 + 2
=9+2
=11
A)
x^3-13x+12 - (x-4)(x^2-4x+3)
=x^3-13x+12 - (x^3-4x^2+3x-4x^2+16x-12)
=8x^2-32x+24
B)
(x^3-13x+12)/(x-3) - 2(x^2-4x+3)
=(x-3)(x^2+3x-4)/(x-3) - 2(x^2-4x+3)
=(x^2+3x-4) - 2(x^2-4x+3)
= -x^2+11x-10
2x^3 +3x^2 -5x+c = (2x+A)(x+B) (x+2)
2x^3 +3x^2 -5x+c = (2x+A)[x^2+(B+2)x+2B]
2x^3 +3x^2 -5x+c = 2x^3 + 2(B+2)x^2 + 4Bx + Ax^2 + A(B+2)x + 2AB
Compare the coefficient of x^2 , x and constant, we have
2(B+2)+A=3--------------(1)
A(B+2)+4B=-5------------(2)
C=2AB--------------------(3)
By (1) ,
A=3-2(B+2) --------(4)
Put (4) into (2)
[3-2(B+2)](B+2) + 4B = -5
3B+6 - 2B^2 - 8B - 8 + 4B = -5
2B^2 + B - 3= 0
(2B+3)(B-1)=0
B=1 OR -3/2
By (4),
When B=1 , A=-3
When B=-3/2 , A=2
Therefore C=2AB= -6
2x ^3 +3x^2 -5X+c=0
2x ^3 +3x^2 -5X-6=0
(2x+2)(x-3/2)(x+2)=0
Therefore , x=-1 or 3/2 or -2