Prove If a polynomial has a double zero, then after multiplying the coefficients by an arithmetic sequence it still is a zero Theorem: if $r$ is a double zero of $$a_0x^n+...+a_{n-1}x+a_n$$ then it is also a zero of $$a_{0}b_{0}x^n+...+a_{n-1}b_{n-1}x+a_n b_n$$ where $b_0,b_1,...,b_n$ for an arithmetic sequence
I try to prove this by strong  induction that if $r$ is a double zero of a polynomial then it is also a zero of the same polynomial multiplyed by an arithmetic sequence,base $(BI)$ for $n=2$
$$ (x-r)^2=x^2-2r\ x+r^2=0   $$
then i multiply by $b_n=b+(n)d$ to get  $$(b)x^2-2r(b+d)\ x+(b+2d)r^2=0 $$ $$ b(x-r)^2+2dr(x-r)=0$$
same for $n=3$ $$(x-r)^2(x-c)=0$$ $$x^3-x^2(2r+c)+x(2rc+r^2)-cr^2=0 $$
then i multiply by $b_n=b+(n)d$ to get
$$ b*(x-r)^2(x-c)+d[-x^2(2r+c)+2x(2rc+r^2)-3cr^2]=0$$
and by plugging in $x=r$ for $[-r^2(2r+c)+2r(2rc+r^2)-3cr^2]=0$
so we proved the base case and now we do the induction given the $(IH)$ works for polynomials up to  $deg P(x)=n-1$ so for $P(x)=a_0x^n+...+a_n$ which we by assume is divided by $(x-r)^2$ ie $r$ is a double zero we prove  $$a_0bx^n+...+a_n(b+(n)d)=b*P(x) + d*[a_{n-1}x^{n-1}+...+(n)a_n]$$ we know $P(r)=0$ and  by strong $(IH)$ the polynomial should be  $d*[a_{n-1}x^{n-1}+...+(n)a_n]=0$ for $x=r$ but i cannot prove this
This was listed as Hade's rule.
I was wondering if there is more on explaining why this topic is the way it is? Looks like a scalar product to me and was wondering if there is some algebra that explains this more deeply?
 A: If $b_k = b_0 + kd$ is an arithmetic sequence ($0 \le k \le n$) then
$$
a_{0}b_{0}x^n+\cdots+a_{n-1}b_{n-1}x+a_n b_n \\=
a_0 (b_n - nd)x^n + \cdots + a_{n-1}(b_n - d) x + a_n b_n \\
= b_np(x)  - dx p'(x)
$$
which confirms your conjecture, since $p(r) = p'(r) = 0$ at a double zero $r$ of $p$.
(You were actually quite close. The “trick” was to use $b_k = b_n - (n-k)d$, and not $b_k = b_0+dk$.)

Here is another way to look at it: Let $\Bbb P$ be the vector space of all polynomials (over $\Bbb R$ or $\Bbb C$). Consider the linear operator $T : \Bbb P \to \Bbb P$ defined by
$$
 T(p)(x) = c_0p(x) + c_1 x p'(x)
$$
for some constants $c_0, c_1$. Then $T$ “preserves double zeros” (in the sense of your question), and
$$
 T(x^k) = (c_0 + k c_1) x^k \, ,
$$
i.e. the action of $T$ on a polynomial is to multiply its coefficients by an arithmetic sequence.
Conversely, any such action can be described by such an operator $T$ with suitably chosen $c_0, c_1$.
This can be used to prove generalizations: If $b_k$ is a quadratic sequence then the action of multiplying the coefficients of a polynomial with $b_k$ preserves triple zeros, and similarly for higher order sequences.
