Given equation

We need the number of distinct real roots of $|x+1||x+3|-4|x+2|+5=0.$

Use a useful identity

Notice that $|x+1||x+3|=|(x+1)(x+3)|=|x^2+4x+3|.$

Also, $x^2+4x+3=(x+2)^2-1.$

Let $t=|x+2| \, , \quad t\ge 0.$

Then $|x+1||x+3|=|(x+2)^2-1|=|t^2-1|.$

So the equation becomes $|t^2-1|-4t+5=0, \qquad t\ge 0.$

Solve the equation in $t$ by cases

We consider the modulus $|t^2-1|$.

Case 1: $t^2-1\ge 0 \Rightarrow t\ge 1$

Then $|t^2-1|=t^2-1.$

So, $t^2-1-4t+5=0$ $t^2-4t+4=0$ $(t-2)^2=0.$

Hence, $t=2.$

This satisfies $t\ge 1$.

Case 2: $t^2-1<0 \Rightarrow 0\le t<1$

Then $|t^2-1|=1-t^2.$

So, $1-t^2-4t+5=0$ $-t^2-4t+6=0$ $t^2+4t-6=0.$

Solving, $t=\frac{-4\pm\sqrt{16+24}}{2}=\frac{-4\pm\sqrt{40}}{2}=-2\pm\sqrt{10}.$

Since $t\ge 0$, only $t=-2+\sqrt{10}$ is possible. Also, $\sqrt{10}\approx 3.16 \Rightarrow -2+\sqrt{10}\approx 1.16>1,$ which does not satisfy $0\le t<1$.

So Case 2 gives no valid solution.

Thus the only value of $t$ is

$t=2.$

Since $t=|x+2|$, we solve $|x+2|=2.$

Therefore, $x+2=2 \Rightarrow x=0,$ or $x+2=-2 \Rightarrow x=-4.$

Count distinct real roots

The distinct real roots are $x=0,\,-4.$

Hence the number of distinct real roots is $\boxed{2}.$

Comparison with stored answer

Stored correct answer: $2$

Our derived answer also is $2$, so they agree.