Question

find zeros of x cube + 6 X square + 11 x + 6​

Answer 1

here's your answer...

Given (x^3)-6(x^2)+11x-6=0
x^3-3x^2-3x^2+11x-6=0
x^2(x-3)-(3x^2-11x+6)=0
x^2(x-3)-(3x^2-9x-2x+6)=0
x^2(x-3)-(3x(x-3)-2(x-3))=0
x^2(x-3)-[(x-3)(3x-2)]=0
(x-3)(x^2-3x+2)=0
(x-3)(x^2-2x-x+2)=0
(x-3)(x(x-2)-1(x-2))=0
(x-3)(x-2)(x-1)=0

hope it helps u...
pls mark brainliest!!

Answer 2

you just use hit and trial method

that is (-1) is satisfied the equation..

then divide whole equation by X+1 ..

u get the answer.

go step by step