How do I factor this cubic equation? How do I rewrite this formula in plain polynomial form?
$$x(t) = a(1-t)^3 + 3b(1-t)^2t + 3c(1-t)t^2 + dt^3$$
According to what I'm reading, just fully writing out the expansion and then collecting the polynomial factors, as:
$$x(t) = (-a + 3b -3c +d)t^3 + (3a - 6b + 3c)t^2 + (-3a + 3b)t + a$$
I'm okay at factoring some quadratics but this is cubic and I want to understand this one; it's driving me crazy.  How did we get from the formula to the polynomial?  BTW, I'm doing some bezier curve learning and this is the formula I need, but I want to understand exactly how to factor this beast.
Edit: Following the suggestion by Taladris' comment I'm going to expand $(1-t)^3, (1-t)^2t, (1-t)t^2$ and group similar terms.  Here goes:
I'll use the special form: $a^3$ + $b^3$ = (a + b)($a^2$ - ab + $b^2$) for the first part:
$a(1-t)^3$ =   $(a)^3$$(-at)^3$ =  (a - at)($a^2$ + $a^2$t - $at^2$)
Next part:
$3b(1-t)^2$(t)  =   (3b-3bt)(3b-3bt)(t)  =   ($9b^2$ -18$b^2$$t + $9$b^2$$t^2$)(t)
...not sure if the above is correct.  Next part:
3c(1 - t)$t^2$ = (3c - 3ct)$(t)^2$
I can't imagine the above is correct, but next I do grouping...
 A: Welcome!
There are Three possible ways:
The first way:
Consider
\begin{align}
x(t) = a(1-t)^3 + 3b(1-t)^2t + 3c(1-t)t^2 + dt^3
\end{align}
where $a=\alpha^{1/3}$, $\qquad d=\beta^{1/3}$, $\qquad b=\alpha^{2/3}\beta^{1/3}:=a^2d$, $\qquad c=\alpha^{1/3}\beta^{2/3}:=ad^2$.
Therefore, since
\begin{align}
(au+dv)^3=a^3u^3+3a^2du^2v+3ad^2uv^2+d^3v^3
\end{align}
Setting $u=1-t$ and $v=t$, we get
\begin{align}
x(t):=(a+(d-a)t)^3&=(a(1-t)+dt)^3
\\
&=a^3(1-t)^3+3a^2d(1-t)^2t+3ad^2(1-t)t^2+d^3t^3
\\
&=\alpha(1-t)^3+3\alpha^{2/3}\beta^{1/3}(1-t)^2t+3\alpha^{1/3}\beta^{2/3}(1-t)t^2+\beta t^3,
\end{align}
and this is very special case!
The Second way:
You can read it here Cubic equation
The Third way
You can use synthetic division by assuming
\begin{align}
x(t) &= (-a + 3b -3c +d)t^3 + (3a - 6b + 3c)t^2 + (-3a + 3b)t + a
\\
&=  \alpha t^3 +\beta t^2 + \gamma t + \lambda
\\
&= (t-z_0)(z_3 t^2 + z_2 t+ z_1)
\\
[&= 
\underbrace {z_3}_{=\alpha} t^3 + \underbrace {(z_2+z_0z_3)}_{=\beta} t^2+ \underbrace {(z_1 + z_0z_2)}_{=\gamma} t+ \underbrace {z_0z_1}_{=\lambda}.]
\end{align}
Now the equation $z_3 t^2 + z_2 t+ z_1=0$; can be solved using the general law of quadratic equation (which is easier than synthetic division). Now, you need to write the $z$'s in terms of $a,b,c,d$; which are indeed very long and needs to very focus!. Hope this helps!
