1. ## General Question about Cubic Polynomials

I've been studying methods for finding the roots of cubic functions. The general formulae seem overwhelmingly difficult to remember. I know it customary for students to memorize the quadratic formula, but what about the formula seen here: Cubic Formula -- from Wolfram MathWorld

?

LOL, how could anybody possilby do something like that on a test? The cubic formula seems impossible to remember, so I just made up a different method. Say I have a cubic equation $x^3-6x^2+21x-26=0$.

$f(x)=x^3-6x^2+21x-26$

$f'(x)=3x^2-12x+21$

$f''(x)=6x-12$

If you look at the graphs of cubic functions, there always seems to be a point of inflection at the x-intercept. Is this true in general? If so, then $f''(x)=0 -> x=2$ gives the root $f(2)=0$, So now I proceed with synthetic division. Is this method something I can rely on? Are there always roots at the inflection points of cubic functions?

Generally,

$f(x)=ax^3+bx^2+cx+d$

$f'(x)=3ax^2+2bx$

$f''(x)=6ax+2b$

$f''(x)=0 -> x=-\frac{b}{3a}$

Notice that $\frac{-(-6)}{3}=2$

I'm not good at proving things in math, so I don't know if I can claim that this works in general, and that all cubic functions will have at least one root of the form $(x+\frac{b}{3a})$

2. both of the following cubics have an inflection point at $x = 2$

$f(x) = x^3 - 6x^2 + 3x - 18$

$f(x) = x^3 - 6x^2 - 5x + 30$

do either have a root at $x = 2$ ?

3. Originally Posted by skeeter
both of the following cubics have an inflection point at $x = 2$

$f(x) = x^3 - 6x^2 + 3x - 18$

$f(x) = x^3 - 6x^2 - 5x + 30$

do either have a root at $x = 2$ ?

ok, so my method works mererly as a test. Set f''(x)=0, maybe you'll find a root. The $0=x^3-6x^2+3x-18$ is easy to solve by factoring, so this one wouldn't be a problem. Other cubic function are more difficult to deal with, so I'm trying to explore simplified methods of finding the roots.