Question

Find integral roots of the polynomial x³ - 6x² + 11x - 6

Answer 1

Let p(x) = x 3 + 6x 2 + 11x + 6

Put x = – 1

p(– 1) = (– 1)3 + 6(– 1)2 + 11(– 1) + 6 = – 1 + 6 – 11 + 6 = 0

∴ (x + 1) is a factor of p(x)

So, we cam break up terms of p(x) as follows.

p(x) = x 3 + 6x 2 + 11x + 6



Therefore, the integral roots of the given equation is find out as:

p(x) = 0

⇒ (x+1)(x+2)(x+3) = 0

⇒ x = -1, -2, -3

Answer 2

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