Tereza solved the equation with factorials: $$(n+2)!+16(n+1)!=(n+3)!$$
for $n\in \mathbb{N}$, as follows:
(1) She rearranged the expressions with factorials, so that she had $(n+1)!$ in each term.
$$(n+2)\cdot(n+1)!+16(n+1)!=(n+3)(n+2)(n+1)!$$
(2) On the left side of the equation, she factored out $(n+1)!$.
$$(n+1)!\cdot\left[(n+2)+16\right]=(n+3)(n+2)(n+1)!$$
(3) She divided both sides of the equation by $(n+1)!$.
$$(n+2)+16=(n+3)(n+2)$$
(4) She removed the parentheses and simplified the quadratic equation to its standard form.
$$\begin{aligned} n+2+16&=n^2+3n+2n+6\cr n^2+4n-12&=0 \end{aligned}$$
(5) She solved the quadratic equation.
$$D=b^2-4ac=16+48=64$$ $$n_1,\ n_2=\frac{-b\pm\sqrt{D}}{2a}=\frac{-4\pm 8}{2}$$ $$n_1=-6,\ n_2=2$$
(6) She wrote down the solution set for the given equation.
$$K=\left\{-6; 2 \right\}$$
Did Tereza make a mistake during the calculation? If yes, determine where.
Tereza did not make a mistake and solved the problem correctly.
Tereza made the first mistake in step (1). She should have factored the expression on the right side of the equation like this: $$(n+3)!=(n+2)(n+1)!$$
Tereza made the first mistake in step (3). She divided the equation by the expression $(n+1)!$, which becomes zero for $n=-1$. The solution set for the given equation is $K=\left\{ -6; -1; 2 \right\}$.
Tereza made the first mistake in the last step (6). The root $-6$ of the quadratic equation is not a natural number, $-6\notin\mathbb{N}$. Therefore, it is not a solution to the given equation.