Suppose a function like $f(x)=ax^3+bx^2+cx ; a,b,c\in R \text{ and }a\ne0 $ so
it has a root $x$ at which $f(x)=0$ but at that point $f'(x)=c$ and we all know that it is the slope of the tangent i.e. $\tan \theta =c\implies \theta =\tan^{-1}c$. So will this tangent pass through origin?
If i talk in general then any function $g(x)$, at $x=0$, has a tangent that will pass through the origin?
The simple answer is that if the derivative exists at the point $x=0$ for a function $f:\mathbb{R}\to \mathbb{R}$and $f(0)=0$ then yes the tangent line to the curve at $x=0$ does pass through the origin.
To prove this note that the slope of the the tangent line is $f'(0)$ and the general formula for the tangent line is $y=f'(0)x+c$. We can find the constant $c$ because we know that at $x=0$ we have the point $(0,f(0))$. So substitution leads to $f(0)=0+c$ so $c=0$ iff $f(0)=0$. Obviously a straight line only passes through the origin if $c=0$.
The above argument shows that in fact this is the only condition under which a tangent line to a differentiable function at $x=0$ can pass through the origin.