Question

find the integral zeros of the polynomial :
p (x)=x^3+6x^2+11x+6​

Answer 1

Putting x =1

(1)^3 -6(1)^2+11(1)-6=0

one root is (x-1)

divide whole eqn by (x-1)

(x-1) | x^3 -6x^2 + 11x -6 | X^2 -5x +6

x^3 -x^2

=======

-5x^2 +11x-6

-5x^2+5x

========

6x-6

6x-6

=====

0

so eqn is x^2-5x+6=0

x^2-2x-3x+6=0

(x-2)(x-3)=0

so the roots are 1 ,2,3

hope it helps!!!

Mark as brainliest!!!