Linear Algebra: different determinant answers 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?

No. What you have done is $$(x+3) \cdot (x+2)=2$$ does not imply $x+3=2$ or $x+2=2$. For example, you can have $x+3 = \frac{1}{2}$ and $x+2 = 4$ which on multiplying gives $2$. Or you can have $x+3 = \frac{1}{8}$ and $x+2 =16$ which on multiplying gives $2$. So there are infinitely many possibilities which you can choose. So this is not the correct way to do.
You have the determinant as $(x+3) \cdot (x+2) -2 = x^{2} +5x+4$. This can be written as $(x+1)(x+4)=0$ which says that $x=-1,-4$.
If you have an equation of the form $f(x) \cdot g(x) =0$, then only either $f(x)=0$ or $g(x)=0$. I think this where you have got confused. But if $f(x) \cdot g(x)= K$, for $K \in \mathbb{R}$, then you cant have $f(x)=K$, or $g(x)=K$, because their product when multiplied gives $K^{2}$. You can have $f(x)= \frac{1}{K}$ and $g(x)=K^{2}$, but then this would keep on continuing. And you can't get all values by doing this. 
