Why does equating one of the brackets in $(x+1)(x+3)=0$ to zero valid? When we want to solve an equation like the one given above, we set either $(x+1)$ or $(x+3)$ equal to $0$ to get $x = -1$ or $x = -3$. However, when we put one of those values in the equation, what we end up doing is multiplying the LHS by zero. How is that a valid operation?
 A: In general, there is nothing wrong with "multiplying both sides by $0$". We don't do it because it causes a permanent loss of information since it inevitably yields $0=0$, but we could still do it if we wanted to. But in your case, we are not even doing that. The equation asks you to find values for $x$ such that when you compute $(x+1)(x+3)$, the result is $0$, no matter how you arrived at $0$. If it makes you more comfortable, you can expand this to $x^2+4x+3=0$, and plug in the solutions into this form to verify.
A: We're taking advantage of the zero-product property. The idea is that if we're given the true statement
$$a\dot b = 0$$
Then either $a=0$, $b=0$, or both $a=b=0$. One of these things is true, if the original statement is true. In your example, we probe both factors to see if this is possible. Indeed, $x=-1$ makes $x+1 = 0$ and $x=-3$ makes $x+3=0$. So both factors have zeroes.
This isn't always the case. Consider the true statement
$$\frac1x\dot(x+1)=0$$
In this case, $\frac1x=0$ is never true! It must be the case that $x+1=0$ for some $x$. And yeah, the root is $x=-1$.
A: "However, when we put one of those values in the equation, what we end up doing is multiplying the LHS by zero. How is that a valid operation?"
That is precisely the point. We want to find a value of $x$ which makes the left side equal the right. If the polynomial is degree two, there are exactly two numbers which make the equality true. This is known as the fundamental theorem of algebra.
