Multiple root of a polynomial is also a root of the derivative Suppose $a\in\Bbb R$ is a root of $f(x)$ in $\Bbb R[x]$. Show that $a$ is a multiple root of $f(x)$ if and only if $f'(a) = 0$, if and only if the graph of $y = f(x)$ is tangent to the $x$-axis at $x=a$. 
So for a I know that if $f'(a) = 0$, then $f(a)$ is a local extremum so $a$ is a root but I do not see why a would be a multiple root. 
 A: Actually the statement, "$a$ is a multiple root of $f(x)$ if and only if $f′(a)=0$" is not quite true as it stands; indeed, if $a$ is a local extremum of $f(x)$ with $f(a) \ne 0$, then clearly $f'(a) = 0$ but $a$ is not a root of $f(x)$, multiple or otherwise; for an example take $f(x) = x^2 - 1$:  $f'(x) = 2x$; $f'(0) = 0$, but the roots of $f(x)$ are $\pm 1$; indeed, $f(0) = -1$.  What is true is the modified assertion,  "$a$ is a multiple root of $f(x)$ if and only if $f(a) = f′(a)=0$," and that is how I will take the question here.
If, then, $a$ is a multiple root of $f(x)$, we have
$f(x) = (x -a)^kg(x) \tag{1}$
for some integer $k \ge 2$ and some $g(x) \in R[x]$.  Differentiating (1) we obtain
$f'(x) = k(x - a)^{k - 1}g(x) + (x - a)^kg'(x), \tag{2}$
from which $f'(a) = 0$ obviously immediately follows.  If now we suppose that $f'(a) = f(a) = 0$, then the curve $y = f(x)$ clearly intersects the $x$-axis at $a$, and the slope of the tangent line at $(a, f(a)) = (a, 0)$ is clearly $0$, the same as the slope of the $x$-axis itself; the curve $y = f(x)$ is thus manifestly tangent to the axis in this case.
And if the curve $y = f(x)$ is tangent to the $x$-axis at $a$, we must have $f(a) = 0$, the condition for intersection, and $f'(a) = 0$, the additional condition for tangency.  We have now established that $y = f(x)$ is tangent to the $x$-axis at $a$ if and only if $f'(a) = f(a) = 0$, and that $f(x)$ has a multiple root at $a$ only if $f'(a) = f(a) = 0$; all that remains is to show that $f'(a) = f(a) = 0$ forces $(x - a)^k \mid f(x)$ with $k \ge 2$.  But if $f(a) = 0$, the we have
$f(x) = (x - a)g(x) \tag{3}$
for some $g(x) \in R[x]$; furthermore, since $f'(a) = 0$ we also have
$f'(x) = (x - a)h(x) \tag{4}$
for some $h(x) \in R[x]$.  From (3),
$f'(x) = g(x) + (x - a)g'(x), \tag{5}$
where $g'(x) \in R[x]$, and by (4) we obtain
$(x - a)h(x) =  g(x) + (x - a)g'(x) \tag{6}$
or
$g(x) = (x - a)(h(x) - g'(x)), \tag{7}$
and substituting (7) in (3) we see that
$f(x) = (x - a)^2 (h(x) - g'(x)), \tag{8}$
which shows $a$ is a multiple root of $f(x)$.  QED
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
A: If $a$ is a root of $f$, then you can write
$$f(x) = (x - a) g(x)$$
with $g$ a polynomial of lower degree than $f$. Now by the product rule,
$$f'(x) = g(x) + (x - a) g'(x)$$
Therefore,
$$f'(a) = g(a)$$
Can you use this to finish the proof?
A: Use the taylor expansion. If $a$ is a root and $f^{\prime}(a)=0$, the first two terms in taylor expansion around $a$ must vanish and the first nonzero term is $\frac{f^{\prime \prime}(a)}{2}(x-a)^{2}$. Then you can factor out $(x-a)^{2}$ from the taylor expansion to write
$$f(x)=(x-a)^{2}g(x)$$
where $g$ is the rest of the taylor polynomial. 
A: The Taylor expansion idea is nice, and probably the quickest.
However, if one wants to use more elementary techniques, then one can also use integration for the more difficult direction. Namely, assume that $f(a) = f'(a) = 0$ for a polynomial $f$ with degree at least $2$ (degree $1$ and $0$ cases are trivial). We would like to prove that $a$ is a zero of $f$ with multiplicity at least $2$.
Then
$$ f'(x) = (x-a)p(x) $$
with some polynomial $p(x)$ of degree at least $1$. Since $f(a)=0$, we have
$$ f(x) = \int_a^x (t-a)p(t) dt = \int_0^{x-a} s \cdot (c_k s^k + \dots + c_0) dt, $$
and this latter expression is a linear combination of $(x-a)^{j+1}$ ($j+1 \geq 2$). Hence $a$ is a zero with multiplicity at least $2$.
