Do stationary points of inflections occur only if the derived equation factorised has a square or above? E.g 3x(x+3)^2 with the -3 being the x-value of the inflection? Will this work with ^3,^4 etc?

Sorry if that didn't really make sense.

Also, is there a simple way to factorise a cubic without using a calculator that you cannot group?

There are several methods to factorise a cubic without using a calculator:

1. The first is to use long division of polynomials. First of all we should become familiar with the factor theorem. This states that for a polynomial, if P(a) = 0, then (x-a) is a factor of P(x). So for e.g. for the polynomial x

^{3} + - 3x

^{2} + 6x -4, P(1) [that is 1

^{3} - 3*1

^{2} + 6*1 - 4] is equal to 0. Thus (x-1) is a factor of the polynomial x

^{3} + - 3x

^{2} + 6x -4. Anyway, using the factor theorem we can find one factor of the cubic polynomial, and then we can divide the cubic with the factor. For example, (x

^{3} + - 3x

^{2} + 6x -4)/(x-1) is equal to x

^{2} -2x +7 with a remainder of 3. Take away the 3 from the result of the divison and then you can factorise the quadratic equation.

2. Now this is short divison. Using the same example as above, if we were to find one factor using the factor theorem we get (x-1). Actually, we know more than this. The resulting equation of the short division will have to be x^2 (since x^2 * x is x^3) and the constant will have to be 4, (since 4 *-1 is -4). So then we have (x^2 + ?x + 4). If we were to multiply the negative 1 of (x-1) with the ?x, we get -x. Then if we multiply x from (x-1) with the 4 from (x^2 + ?x + 4), we get 4x. (Basically I am expanding the equation (x-1)(x

^{2} + [something]x + 4). The coefficient we receive from expanding, 3, is what "?" is equal to.

3.This is through a process called "finite differences" and requires a table of x and y values of the equation. But idk if it is still in the VCE study design, (I can show you the process if u wish though).

Sorry, it is kind of difficult to explain without Latex

,

hope that helps...

Edit: Oh sorry, I didn't see your question before that.. Anyway, I think what you mean is can there be stationary points of inflection with polynomials with degrees higher than that of cubics? Well if that is what you are asking, yes there can be inflection points with fourth degree polynomials (not too sure about higher than that though...)