$x^3+2x^2+x+2\equiv 0 \mod 45$
$f(x)=(x^2+1)(x+2)$
by inspection
$\fbox{1}$$x=7$ is a possible solution $\mod 9$ ,since $45=5\times 3^2$
$\fbox{2} $ $x= 2 $ is a solution $\mod 5$ but i want to get general work on how to find other solutions...from here >>>
On
First work with modulo 9:
$$f(x) = (x^2 + 1)(x + 2) \equiv 0 \pmod 9$$
Now we have three distinct cases, because 9 is square of 3:
Case 1: $9\mid (x^2 + 1)$
This case is impossible, because $-1$ isn't quadratic residue modulo $9$.
Case 2: $3\mid (x^2 + 1)$ and $3\mid (x+2)$
The first relation is impossible, because $-1$ isn't quadratic residue modulo $3$.
Case 3: $9\mid (x+2)$
This is the only case possible and is satisfied when $x \equiv 7 \pmod 9$
So this implies that $f(x) \equiv 0 \pmod 9$ if and if $x \equiv 7 \pmod 9$
Now work modulo 5:
$$f(x) = (x^2 + 1)(x + 2) \equiv 0 \pmod 5$$
Because $5$ is prime number we have only two cases:
Case 1: $5\mid (x^2+1)$
We know that $-1$ is quadratic residue modulo $5$ and it implies that $x \equiv \pm 2 \pmod 5$
Case 1: $5\mid (x+2)$
This one implies that $x \equiv -2 \pmod 5$
So from this we obtain that $f(x) \equiv 0 \pmod 5$ if and only if $x \equiv \pm 2 \pmod 5$
Now apply Chinese Remainder Theorem for two cases:
Case 1: $x \equiv 2 \pmod 5$ and $x \equiv 7 \pmod 9$
The first conguence relation is equivalent to $x \equiv 7 \pmod 5$, so using this and the second congruence relation $x \equiv 7 \pmod {45}$
Case 2: $x \equiv -2 \pmod 5$ and $x \equiv 7 \pmod 9$
Applying CRT we get that $x \equiv -2 \equiv 43 \pmod {45}$
So we have two solutions:
$$x \equiv 7,43 \pmod {45}$$
If you want to solve it without explicitly stating the Chinese Remainder Theorem.
Hint :$ x \equiv 7 \pmod 9$ so $ x = 9k+ 7 $ for some $k$.
Now, $ 9k + 7 \equiv ? \pmod 5$.
Solve for $k$.