TITLE: Linear Algebra: different determinant answers
QUESTION [5 upvotes]: I'm having a problem verifying my answer to this question:
Solve for x:
$$\left| \begin{array}{cc}
x+3 & 2 \\
1 & x+2 \end{array} \right| = 0$$
I get:
$(x+3)(x+2)-2=0$ 
$(x+3)(x+2)=2$
thus:

$x+3=2$ and    $x+2=2$

$x=-1$ and $x=0$
The book says that $x=-1$ and $x=-4$ is the correct answer.
I tried doing it a different way by expanding and got totally different answers:
$x^2+5x=-4$
$x(x+5)=-4$
$x=-4$ and $x=-9$
What is going on here?

REPLY [2 votes]: Let me point out where I think your error of reasoning arises: it comes from drawing an invalid analogy to the case where we have an equation of the form
$$(x-a)(x-b) = 0.$$
In that case, the standard next step is to say "therefore $x-a=0$ or $x-b=0$"; you are trying to do the same thing with $0$ replaced by $2$.
The reason this works for $0$ is that in the real numbers we have:

If $r$ and $s$ are real numbers, and $rs=0$, then either $r=0$ or $s=0$.

That is: the only way for a product to be equal to $0$ is if (at least) one of the factors is already equal to $0$.
This does not hold for any number other than $0$. There are many ways in which a product can equal $2$; in fact, infinitely many different ways. So you cannot simply break up the product by cases in this situation (or in any situation in which you have a product of real numbers equal to something other than $0$).