78. Arrange the polynomial $f(x)=x^{4}+2 x^{3}-3 x^{2}-4 x+1$ in powers of $x+1$.
Solution. We assume:

$$
\begin{gathered}
x^{4}+2 x^{3}-3 x^{2}-4 x+1=a_{4}(x+1)^{4}+a_{3}(x+1)^{3}+a_{2}(x+1)^{2}+a_{1}(x+1)+a_{0}, \\
\text { from which } x^{4}+2 x^{3}-3 x^{2}-4 x+1=a_{4} x^{4}+\left(4 a_{4}+a_{3}\right) x^{3}+ \\
+\left(6 a_{4}+3 a_{3}+a_{2}\right) x^{2}+\left(4 a_{1}+3 a_{3}+2 a_{2}+a_{1}\right) x+\left(a_{4}+a_{3}+a_{2}+a_{1}+a_{0}\right)
\end{gathered}
$$

and

$$
\left\{\begin{array}{l}
a_{4}=1 \\
4 a_{4}+a_{3}=2 \\
6 a_{4}+3 a_{3}+a_{2}=-3 \\
4 a_{4}+3 a_{3}+2 a_{2}+a_{1}=-4 \\
a_{4}+a_{3}+a_{2}+a_{1}+a_{0}=1
\end{array}\right.
$$

From the system, we find:

$$
a_{4}=1, a_{3}=-2, a_{2}=-3, a_{1}=4, a_{0}=1
$$

Thus,

$$
f(x)=(x+1)^{4}-2(x+1)^{3}-3(x+1)^{2}+4(x+1)+1
$$

Representation of the product in the form of a standard polynomial.