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.

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

$(x - a)h(x) = g(x) + (x - a)g'(x) \tag{6}$

$g(x) = (x - a)(h(x) - g'(x)), \tag{7}$

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,