Thank you for your reply!

however, is it possible to explain to me assuming I have no knowledge of differentiation

You need to have some knowledge of differentiation to determine whether a cubic has 0, 1 or 2 turning points.

Just adding to Billuminati's response, if you want to know how many turning points a quadratic has then you can just take the discriminant of the derivative. \(f\left(x\right)=ax^3+bx^2+cx+d\) then \(f'\left(x\right)=3ax^{2}+2bx+c\)

The discriminant of the derivative is equal to, \(\Delta _{f'}=4b^2-12ac\).

\(\Delta _{f'}<0\Longrightarrow \text{0 stationary points}\)

\(\Delta _{f'}>0\Longrightarrow \text{2 stationary points}\)

\(\Delta _{f'}=0\Longrightarrow \text{1 stationary point}\)

Note: 2 stationary points implies that the stationary points are turning points, and 1 stationary point means that the stationary point is an inflection point.

So I guess what you could do, without knowledge of differentiation, is just plug in \(a,\:b,\:c\) for a cubic in the form \(ax^3+bx^2+cx+d\) into the expression \(4b^2-12ac\) to determine the amount of turning points.

This is a little formulaic, which I don't really like but I guess if you do not know differentiation, just use this.