Square root each term (clarification on polynomials?) So I'm in Algebra 2, and right now we're learning about conic sections (circles/ellipse/etc). I thought some problems in the workbook looked weird, like this one:
$\y^2 = x^2 + 16

By my understanding, I should be able to take the square root of the "y", "x", and "16", leaving me with "y = x + 4," since I'm taking the square root of each term ("distributing"?), but my teacher said I have to solve the polynomial first.
 Back in Algebra 1, I was also confused when we started factoring & canceling polynomials, because the teacher never explained when we could/couldn't cancel terms. Could I have some clarification on how polynomials are defined? [And also when to use the principal root?]
EDIT: Thanks for the answers guys. Sorry if I wasn't specific in my question (I'm not exactly good at expressing my thoughts). I understood that the two were different (linear/hyperbolic). The question in the book was y^2 - x^2 = 16. When I moved "x^2" to the other side, I assumed that it's not directly attatched to "16." In a generic situation, what indicates a polynomial (parentheses)?
 A: As a quick example of why your algebra teacher is right, consider $(x,y)=(3,5)$: This is a solution since $5^2 = 3^2+4^2$, but nevertheless $5\neq 3+4=7$. The reason it doesn't work is that $(x+4)^2=x^2+8x+16,$ which only agrees with $x^2+16$ when the middle term disappears (i.e. $x=0$.) In general, then, $\sqrt{a+b}\neq \sqrt{a}+\sqrt{b}$ for nonzero $a,b$.
For a picture relating to the analytic geometry you're learning, note that $y^2=x^2+16$ defines a hyperbola whereas $y=x+4$ defines a line. The only place where they intersect--which is to say, the only place where this square-root would appear correct--is along the $y$-axis. In that case one is effectively solving $y^2=16$ which indeed has $y=4$ (and $y=-4$) as square roots.

A: I'm not sure if I really understand your question yet, but I'll give it a shot.

In general, a polynomial is any (finite) sum of monomials. A monomial is a product of a real number and some number of variables, possibly more than one copy of each.


*

*Examples of monomials include $5abc$, $14x^2y^3$, $16$, and $-3k$. 

*Nonexamples of monomials include $1+2p$, $\sqrt x$, and $\frac{5e}{d}$.
"Any finite sum" here also means the trivial case where there is only one summand. Therefore, every monomial is a polynomial by definition. But of course there are polynomials which are not monomials, like $x^2+16$.
"Any finite sum" also means that it is possible to consider multiple polynomials in a single expression. To use your example, $x^2+16$ "contains" three polynomials. There is the obvious one: $x^2+16$. But there are also two more from its constituent parts: $x^2$ and $16$.
Therefore, the idea that $x^2$ is somehow "free" from the $16$ is entirely correct. To put it in another way, one can consider $x^2+16$ as either $(x^2+16)$ or $(x^2)+(16)$, and there are advantages to both perspectives.

A parting thought: when we say "subtract $x^2$ from both sides", this is a shorthand for a tremendous amount of mental shifting that must occur to reinterpret the equation in a more useful form. Of course, you don't need to write it on paper but this process of shifting is one that you should get comfortable with. If you were to write it down, it would be almost painfully slow to read:
\begin{align*}
(y^2) &= (x^2 + 16) \\
(y^2) &= (x^2) + (16) \\
(y^2) - (x^2) &= (x^2) - (x^2) + (16) \\
(y^2-x^2) &= (x^2) - (x^2) + (16) \\
(y^2-x^2) &= (x^2-x^2) + (16) \\
(y^2-x^2) &= (0) + (16) \\
(y^2-x^2) &= (0+16) \\
(y^2-x^2) &= (16)
\end{align*}
Someone could accuse the above calculation of being hyperbole, and that's not wrong. But I hope that you can see that each of these steps is, conceptually, significant: the shifting of parentheses really should mean something in your mind.
