Let $(f\circ g)(x) =x^4+2x^3-3x^2-4x+6$ and $g(x)=x^2+x-1$. Find $f(x)$, it seem to be $f$ will have the formula $f(x)=ax^2+bx+c$. Plugging $g(x)$ in $f(x)$, we get $$ f(x^2+x-1)=a(x^2+x-1)^2+b(x^2+x-1)+c= x^4+2x^3-3x^2-4x+6$$ by comparing the coefficients in both sides we found $a=1, b=-2,c=3$. I know what got is correct. My question is
Is there another way to do it?
We know that $ (f \circ g)(x) = x^{4} + 2x^{3} - 3x^{2} - 4x + 6, \qquad g(x) = x^{2} + x - 1. $
Let $f(t)$ be a quadratic polynomial: $f(t) = at^{2} + bt + c$.
Since $f(g(x)) = f(t)$ whenever $t = g(x)$, we can evaluate $f(t)$ by choosing values of $x$ that make $g(x)$ convenient.
$g(0) = -1, \qquad f(-1) = (f \circ g)(0) = 6$.
$g(1) = 1, \qquad f(1) = (f \circ g)(1) = 1 + 2 - 3 - 4 + 6 = 2$.
$g(2) = 5, \qquad f(5) = (f \circ g)(2) = 16 + 16 - 12 - 8 + 6 = 18$.
Thus the quadratic $f(t)$ satisfies the system
\begin{cases} a(-1)^{2} - b + c = 6, \\ a(1)^{2} + b + c = 2, \\ 25a + 5b + c = 18. \end{cases}
Solving the system gives $a = 1,\qquad b = -2,\qquad c = 3$.
Therefore $f(t) = t^{2} - 2t + 3$.
You know $\deg f\circ g = \deg f \cdot\deg g$. Hence $2\deg f = 4$, i.e. $\deg f = 2$. So, $f = a(X+1)^2 +b(X+1) +c$. Now it is $f(-1) = c$, $f'(-1) = b$ and $f''(-1) = 2a$.
It is $c = f(-1) = f(g(0)) = 6$ and $-4=(f\circ g)'(0) = f'(g(0))g'(0) = f'(-1) = b$ and $-6=(f\circ g)''(0) = ((f'\circ g)g')'(0) = f''(g(0))(g'(0))^2+f'(g(0))g''(0) = f''(-1)+2f'(-1) = 2a+2b = 2a -8$ Hence $c = 6, b=-4, a = 1$. So, $f = (X+1)^2-4(X+1)+6 = X^2 -2X +3$.
