Q.A zero of $x^3+64$ is

A polynomial of degree $n$ in $x$ is $P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$ with $a_n\ne0$; the corresponding polynomial equation is $P(x)=0$. A number $c$ with $P(c)=0$ is called a root (or zero) — the two words describe exactly the same thing. The leading coefficient is $a_n$, the leading term is $a_nx^n$, and a polynomial with $a_n=1$ is monic. A polynomial's exponents must be non-negative integers, though its coefficients may be any real or complex number — this is exactly why $3x^{-1}+2$, $5x^{1/2}+1$, and trigonometric expressions like $\cos x-\sin x$ are not polynomials, however polynomial-looking they seem.

For the familiar quadratic $ax^2+bx+c=0$ ($a\ne0$), the discriminant $\Delta=b^2-4ac$ governs the roots via $x=\dfrac{-b\pm\sqrt\Delta}{2a}$: with real $a,b,c$, $\Delta>0$ gives real distinct roots, $\Delta=0$ gives equal real roots, and $\Delta<0$ gives no real roots (a non-real conjugate pair — see Complex Conjugate Root Theorem).

Translating a word problem into a polynomial equation. Many real-world conditions — a box's dimensions and volume, an age or rate relationship — translate directly into a polynomial equation once the unknown is named. E.g. a box with breadth $x$, length $x+6$, height $x+3$ has volume $x(x+6)(x+3)$; requiring this volume to equal a fixed number gives a cubic equation in $x$, and solving it (checking which root is a physically valid, positive length) answers the real question. The same principle handles factoring shortcuts too — e.g. $x^3+64=x^3+4^3=(x+4)(x^2-4x+16)$ shows directly that $x=-4$ is a zero of $x^3+64$, without any trial-and-error. It also underlies a basic but easily-confused fact: if $f,g$ are polynomials of degree $m,n$ respectively, the composition $h(x)=(f\circ g)(x)=f(g(x))$ has degree $mn$ (multiplied, not added) — substituting a degree-$n$ expression into every power up to $x^m$ of $f$ produces a top term of degree $m\times n$.