Why can't a non-zero polynomial satisfy some equations? I'm having a hard time visually picturing/understanding how to explain why a non-zero polynomial function cannot satisfy the equation:
$f''(x)$ = $-f(x)$
So is it basically asking to explain why a polynomial function must contain a 0 coefficient?
I understand that taking derivatives decreases the degree of the functions.
I need some help please. Thank you!
 A: Let's look at the degree two case: $f(x) = a x^2 + b x + c$. Suppose it satisfied those equations. Then we have
$$
2a = f''(x) = -f(x) = -ax^2 - bx-c.
$$
Comparing like terms on both sides, we get that $a = 0$, $b = 0$, and $-c = 2a = 0$ (so $c = 0$). That is, $f(x) = 0$. The same idea can be applied for higher/lower degrees. It all relies on comparing like terms from computing $f''(x) = -f(x)$.
A: $$f''(x) = -f(x)$$
This says that whatever $f(x)$ is, if you take the derivative twice, you get something equal to $0-f(x)$.
If $f(x)$ is a polynomial, then when you take the derivative twice, you'll get a polynomial that's two degrees lower — and therefore not equal.
The only way a polynomial could solve this equation is if it was an infinite-degree polynomial (since $\infty-2=\infty$). We don't call that a polynomial, we call it a power series. And, incidentally, that's one way to define the sine and cosine functions.

Perhaps it's even easier to understand with this example:
$$f''(x)=f(x)$$
This says that if we take the derivative twice, we get back the function we started with. Again, that can't be a polynomial. If you take a degree-$n$ polynomial, it's 2nd derivative will have a degree of $n-2$, which doesn't equal $n$. (Unless $n=\infty$, in which case, again, we have a power series, not a polynomial.)
A: Lemma: 
A polynomial is identically zero (i.e. $p(x)=0$ for all $x$) if and only if all its coefficients are null.
Indeed, if all coefficients are $0$, $p(x)$ always evaluates to $0$. And conversely if $p(x)\ne0$ for some $x$, then some terms are nonzero and so is their coefficient.
QED.
Now, two polynomials are equal iff their difference is identically zero, and by the lemma, iff all their coefficients are equal.
For example, $3x^2+2x-1$ and $3x^2-x-1$ aren't equal as their difference is $3x\ne0$.
When the polynomials are of different degrees, equality does not occur.
For example, $3x^2+2x-1$ and its derivative $6x+2$ are obviously unequal.

Functions that satisfy equations such as $$f''(x)=-f(x)$$ can usually be expressed as polynomials of infinite degree. They are called entire functions. The "trick" is that after derivation, the degree is still infinite.
Take the function
$$f(x)=1-\frac12x^2+\frac1{4!}x^4-\frac1{6!}x^6+\frac1{8!}x^8\cdots$$
and derive it twice. After simplification, you get
$$f''(x)=-1+\frac12x^2-\frac1{4!}x^4+\frac1{6!}x^6\cdots$$
so that all corresponding coefficients are equal (to a change of sign) and this "polynomial" satisfies the differential equation.
By the way, this entire function coincides with the familiar $\cos(x)$.
