Question

Find what straight lines are represented by the following equation and determine the angles between them.
$x^3-6x^2+11x-6=0$

Answer 1

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

Put $x=1$ in the given equation

$\therefore 1-6+11+6=0$

Clearly $x=1$ satisfies the given equation

$\therefore x-1=0$ is a root of the given equation

Dividing the given equation by $x-1$

$\frac{x^3-6x^2+11x-6}{x-1}=x^2-5x+6$

$\Rightarrow x^3-6x^2+11x-6=(x^2-5x+6)(x-1)=(x^2-3x-2x+6)(x-1)=\left(x\right(x-3)-2(x-3\left)\right)(x-1)=(x-1)(x-2)(x-3)$

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

$\Rightarrow (x-1)(x-2)(x-3)=0$

So the equation of lines are $x=1,x=2$ and $x=3$

Clearly all the lines are parallel

So the angle between them is $0$.