# Why can't I divide this cubic through by $x$? (I see why, but what's the rule?) [duplicate]

I had this problem at school (we're just doing polynomial equations):

$$2 x^3 - x^2 - 3x = 0$$

I can see that $x = 0$ is a solution to this. But I divided left and right side by $x$, factorised, and didn't get the $0$ solution.

I can see I was wrong. But why is there this exception to the rule that "you can always do the same thing to the left and right side of an equation"?

Thanks, Sibby

## marked as duplicate by Najib Idrissi, Takumi Murayama, Strants, Alex M., hardmathApr 30 '16 at 21:16

• What if, instead, you factorize and look for the zero's of each term ? – Claude Leibovici Apr 30 '16 at 17:56
• Why do you think that root $x=0$ should remain after you cancel $x$? – Bill Dubuque Apr 30 '16 at 18:06

Dividing by $x$ you have to consider that it is non-zero.

You have the following:

$$x(2x^2-x-3)=0 \Rightarrow x=0 \text{ or } 2x^2-x-3=0$$

"You can always do the same thing to the left and right hand side of the equation". Yes, as long as what you do is mathematically correct. You can't divide through by $x$ unless $x \neq 0$.

When you divide by $x$, the result you get is only well defined when $x \neq 0$.

So, that's why we first need to ensure that $x \neq 0$. We examine the case $x = 0$. We see it's a root of multiplicity one, so we report "$0$ is a root!" and henceforth assume that $x \neq 0$.

Then we may divide by $x$ to find the other zeroes.

You can also see the process as doing something like this:

We have $2x^3 - x^2 - 3x = 0$. We factorize it as $x(2x^2 - x - 3)$. From here we see oh, $x=0$ is a root! Instead of dividing by $x$, we can then just say that the other roots must be when $2x^2 - x - 3 = 0$, so we only have to worry about this term equalling zero now. See how we get to the same end result as when we decided to divide by zero?

• Thanks. This really helped me understand. – sibby May 2 '16 at 9:50