How many real roots of $f(x)=x^{12}-x^9+x^4-x+1$ between $0$ and $1$? How can we know if $f(x) = x^{12} - x^9 + x^4 - x + 1$ has How many real roots between 0 and 1. ?
Okay so at both 1 and 0 the value of function is 1. So there can be no or even roots between
 them so how can we tell can we use calculus ?
 A: If $0<x<1$ then
$$f(x)=1-x+x^4-x^9+x^{12}=1-x+x^4(1-x^5)+x^{12}>0$$
A: $\begin{cases}
f(x)=x^{12}-x^9+x^4-x+1 \\
f'(x)=12x^{11}-9x^8+4x^3-1 \\
f''(x)=132x^{10}-72x^7+12x^2=12x^2(11x^8-6x^5+1)=12x^2g(x) \\
g'(x)=88x^7-30x^4=x^4(88x^3-30)
\end{cases}$
And we can stop here, because we are only interested in the signs of the derivatives.
$a=\sqrt[3]{\frac{30}{88}}$ 
$\begin{array}{|c|ccccc|}
\hline 
x & 0 && a && 1\\
\hline
g'& 0 & - & 0 & + & 58 \\
\hline
g & 1 & \searrow & \simeq 0.62 & \nearrow & 6 \\
\hline
f'' & 0 && + && 72\\
\hline
f' & -1 && \nearrow && 6 \\
\hline\end{array}$
So there exists $b\in]0,a[$ such that $f'(b)=0$.
$\begin{array}{|c|ccccc|}
\hline 
x & 0 && b && 1\\
\hline
f'& -1 & - & 0 & + & 6 \\
\hline
f & 1 & \searrow & f(b) & \nearrow & 1 \\
\hline
\hline\end{array}$
From there you can calculate $b$ by dichotomy $b\simeq 0.6746$ and then compute $f(b)\simeq 0.51245>0$.
We are dealing with polynomial here, so continuity is assured and abrupt variations excluded so we can be quite confident that $f>0$ on $[0,1]$.
Now, if you want to pursue the algebraic proof, you'll have to substitute $b$ : we know that $bf'(b)=12b^{12}-9b^9+4b^4-b=0$ 
So $12f(b)=(9b^9-4b^4+b)+12(-b^9+b^4-b+1)=-3b^9+8b^4-11b+12$
And then carry on with this lesser degree equation until you get $f(b)>0$.
You wanted a general method without tricks, you got one.
