Question

求x^3-3x^2-9x-5=0;因式分解的过程

主要是求解的过程!

Answer 1

x^3-3x^2-9x-5=0
x^3-3x^2-4x-5x-5=0
x(x^2-3x-4)-5(x+1)=0
x(x+1)(x-4)-5(x+1)=0
(x+1)[x(x-4)-5]=0
(x+1)(x^2-4x-5)=0
(x+1)(x+1)(x-5)=0
(x-5)(x+1)^2=0
∵x=5 x=-1

Answer 2

x^3-3x^2-9x-5=x^3+1-3(x^2+3x+2)
=(x+1)(x^2+1-x)-3(x+2)(x+1)
=(x+1)(x^2+4x-5)
=(x+1)(x+1)(x-5)