Question

find the zero of polynomial
$x^{3}-6x^{2}+11x-6$
​

Answer 1

Given polynomial is $x^{3} -6x^{2} +11x-6$

By using Trial and error method.

Put $x=1$

$1^{3} -6(1)^{2} +11(1)-6=0\\1-6+11-6=0\\12-12=0\\0=0$

$1$ is the root of equation. So $(x-1)$ is a factor of $x^{3} -6x^{2} +11x-6$.

put $x=2$

$2^{3} -6(2)^{2} +11(2)-6=0\\8-6(4)+22-6=0\\8-24+22-6=0\\30-30=0\\0=0$

$2$ is a root of equation. So, $(x-2)$ is a factor of $x^{3} -6x^{2} +11x-6$.

Put $x=3$

$3^{3} -6(3)^{2} +11(3)-6=0\\27-6(9)+33-6=0\\27-54+33-6=0\\60-60=0\\0=0$

$3$ is a root of equation. So, $(x-3)$ is a factor of $x^{3} -6x^{2} +11x-6$.

Therefore, $1,2,3$ are the zeroes of the polynomial $x^{3} -6x^{2} +11x-6$.

Answer 2

Given,

$x^{3}-6x^{2} +11x-6=0$

Substitute $x=1$ in the given equation,

$1^{3}-6.1^{2}+11.1-6\\1-6+11-6\\12-12=0\\0=0$

$x-1$ is the one of the factor of $x^{3}-6x^{2} +11x-6$

Now,

Divide the given equation with $(x-1)$

We, get quotient $x^{2} -5x+6$.

By using factorization splitting method to find the factors

$x^{2} -5x+6=0$

$x^{2} -3x-2x+6=0\\x(x-3)-2(x-3)=0\\(x-3)(x-2)=0\\x=2,3$

Therefore, $1,2,3$ are the zero polynomial of $x^{3}-6x^{2}+11x-6$.