Show that $\lambda$ is a repeated root of $p(z)$ if and only if $p(\lambda) = p'(\lambda) = 0$ 
Suppose $p \in \mathcal{P}\mathbb{(C)}$ and $\lambda \in \mathbb{C}$. Applying the division algorithm, we know there exists $q \in \mathcal{P}\mathbb{(C)}$ and $r \in \mathbb{C}$ such that 
  $$p(z) = q(z)(z-\lambda) + r$$
  Show that $p'(\lambda) = q(\lambda)$ and use this to deduce that $\lambda$ is a repeated root of $p(z)$ if and only if $p(\lambda) = p'(\lambda) = 0$.  (Note: - A repeated root of $p$ is a root that appears at least twice in the complete factorization of $p$, which is equivalent to $(z - \lambda)^2$ is a factor of p.)

To show that $p'(\lambda) = q(\lambda)$, if I take the derivative of $p(z)$, I get $$ p'(z) = q'(z)(z-\lambda) + q(z)$$
Then plugging in $z = \lambda$, $$ p'(\lambda) = q'(\lambda)(\lambda-\lambda) + q(\lambda)$$
and $p'(\lambda) = q(\lambda)$ appears as desired.
$\rightarrow$ this direction I think is straight forward. Suppose that $\lambda$ is a repeated root of $p(z)$. Then p and p' can be represented as 
$$p(z) = (\lambda - \lambda)^2f(z)$$
$$p'(z) = 2(z-\lambda)f(z) + (z-\lambda)^2 f(z)$$
(per 4.11 theorem from textbook). Plugging in $z = \lambda$, 
$$p(\lambda) = (\lambda - \lambda)^2f(\lambda) = 0$$
$$p'(\lambda) = 2(\lambda-\lambda)f(\lambda) + (\lambda-\lambda)^2 f(\lambda) = 0$$
Thus $p(\lambda) = p'(\lambda) = 0$.
$\leftarrow$ this side is really messing me up. I start with supposing $p(\lambda) = p'(\lambda) = 0$. Considering the original $p(z) = q(z)(z-\lambda) + r$ I want to show that $p(\lambda) = p'(\lambda) = 0$ using the fact that $q(\lambda) = p'(\lambda)$, but not sure how to start this. Any help or ideas in the right direction would be appreciated. I feel like I keep going in circles looking at $p(z)$ and $p'(z)$ trying to find something when there is nothing more left to find.

 A: Let $$p(\lambda)=0\tag 1.$$
You already proved that $p'(\lambda)=q(\lambda)$ which implies $$q(\lambda)=p'(\lambda)=0\tag2.$$
For $(2)$  take $q(z)=(z-\lambda)t(z)$ where $t(\lambda)\ne 0$. On plugging this in 
$p(z)=(z-\lambda)q(z)+r$
You get $p(z)=(z-\lambda)^2t(z)+r$. Now use $(1)$ to get $r=0$ and then $p(z)=(z-\lambda)^2q(z)$ shows the result.
A: If $\lambda$ is a repeated root of $p(z)$, then
$p(z) = (x - \lambda)^2q(z), \tag 1$
from which, via differentiation,
$p'(z) = 2(z - \lambda)q(z) + (z - \lambda)^2q'(z); \tag 2$
from (1), 
$p(\lambda) = (\lambda - \lambda)^2q(\lambda) = 0, \tag 3$
whilst from (2),
$p'(\lambda) = 2(\lambda - \lambda)q(\lambda) + (\lambda - \lambda)^2 q'(\lambda) = 0 \tag 4$
as well, which shows that $\lambda$ is a multiple root of $p(z)$ implies
$p(\lambda) = p'(\lambda) = 0. \tag 5$
Going the other way:
Suppose now we are given that (5) binds; in accord with the division algorithm for polynomials we may write
$p(z) = (z - \lambda)q(z) + r, \tag 6$
where
$0 \le \deg r < \deg (z - \lambda) = 1; \tag 7$
it follows then that
$\deg r = 0 \Longrightarrow r \in \Bbb C, \tag 8$
and taking $z = \lambda$ in (6) we find
$0 = p(\lambda) = (\lambda - \lambda)q(\lambda) + r = r; \tag 9$
with $r = 0$, (6) becomes
$p(z) = (z - \lambda)q(z), \tag {10}$
whence, upon taking derivatives,
$p'(z) = q(z) + (x - \lambda)q'(z); \tag{11}$
again with $z = \lambda$,
$0 = p'(\lambda) = q(\lambda); \tag{12}$
with $q(\lambda) = 0$ we may essentially repeat the argument of (6)-(10) to infer the existence of some
$s(z) \in \Bbb C[z] \tag{13}$
with
$q(z) = (z - \lambda)s(z), \tag{14}$
and returning to (10) we find
$p(z) = (z - \lambda)q(z) = (z - \lambda)^2 s(z), \tag{15}$
so $\lambda$ is a multiple root of $p(z)$.
Show that $\lambda$ is a repeated root of $p(z)$ if and only if $p(\lambda) = p'(\lambda) = 0$
