Connection between number of roots for a given polynomial and its degree

Question

Why do we get $2$ solutions for a quadratic equation and $3$ solutions for a cubic equation and $4$ for biquadratic equation and so forth?

Answer 1

Why do we get $2$ solutions for a quadratic equation and $3$ solutions for a cubic equation and $4$ for biquadratic equation and so forth?

This is only true if you allow the solutions to be complex. This is given by The Fundamental Theorem of Algebra, which states (quoted from the Wiki) that

"...every non-zero, single-variable, degree $n$ polynomial with complex coefficients has, counted with multiplicity, exactly $n$ roots."

Answer 2

This is because of the factor theorem: if $r$ is a root of polynomial $P(x)$, then $P(x)$ must factor as $$P(x) = (x-r)Q(x),$$ where the degree of $Q$ is one less than the degree of $P$.

Then we have $P(x) =0$ if and only if $$(x-r)Q(r) =0,$$ which occurs when $x=r$ or when $x$ is a root of $Q$. So $P$ has at most one more root than $Q$, or exactly one more if you are willing to count multiple roots multiple times.

Since a 1st-degree polynomial evidently has 1 root, an $n$th-degree polynomial has $n$ roots (or "at most $n$" if you don't count the multiple roots separately.)