Question about Polynomial Factor Theorem

Question

I was reading the solution to an algebra problem but got stuck at one part. Problem is here: (http://math.la.asu.edu/~ifulman/mat194/problem-solving.pdf) Example 4.2.6 -- page 140 of the PDF (the book page is numbered 129).

The solution states that $g(a_i)^2+h(a_i)^2=0, \space \therefore g(a_i) =h(a_i)=0$. By the Factor Theorem, both $g$ and $h$ must be divisible by $(x-a_i)$ I understand that.

But then it then states: "the multiplicity of these zeroes must be at least $m_i$;" therefore, $f(x)|g(x)$ and $f(x)|h(x)$.

I don't understand why the multiplicities of $g$ and $h$'s zeroes have to be at least $m_i.$ If we consider an integer analogy, $2^6 = (2^6-2)+2,$ but while the 2 on the left side has a "multiplicity" of 6, the powers of 2 in both terms of the right side is 1. To me it seems that the solution states that because both the left side and the right side are divisible by 2, $(2^6-2)$ and $2$ must have multiplicities of 6 (not a perfect analogy but you get the idea). Came someone explain to me the book's reasoning?

Answer 1

Hint $\ $ Cancel $(x-a_i)^2$ then inductively apply the same argument.

Answer 2

Because, when you divide by $(x-a_i)^2$ (which you can do), you have the same equation but with $m_i \gets m_i -1.$ So, by induction, the order of the zeros of $g, h$ is at least $m_i.$