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Given: $P(x)$ is a polynomial of at least the second degree and $L(x)$ is the tangent to $P(x)$ at $x=a$.

Questions: Can one then say that $P(x)-L(x)$ has a double root at $x=a$? If so, why? If not, why not?

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Let $$g(x) = P(x) - L(x)$$

Since $P(a) = L(a)$ (because tangent touches curve at $x = a$), we have

$$g(a) = P(a) - L(a) = 0$$

Now, differentiate $g(x)$ to get: $$g'(x) = P'(x) - L'(x)$$

At $x = a$, $L'(a) = P'(a)$ (because tangent) so

$$g'(a) = P'(a) - L'(a) = 0$$

Since $a$ is (at least) a double root of $g(a)$ if and only if $g(a) = 0, g'(a) = 0$, we deduce that $a$ is indeed (at least) a double root.

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  • $\begingroup$ What if $g''(a) = 0$? Then it is a triple root, not double. So, no "iff". $\endgroup$ – Evgeny Dec 23 '15 at 11:39
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    $\begingroup$ @Evgeny (at least) a double root $\endgroup$ – Yiyuan Lee Dec 23 '15 at 11:53
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You can always rewrite your polynomial as $P(x) = P(a) + P'(a) (x-a) + \frac{P''(a)}{2}(x-a)^2 + \dots $. Using this form, you can write $L(x)$ as $L(x) = P(a) + P'(a)(x-a)$. Combining these two formulas we get that $P(x) - L(x) = (x-a)^2 \cdot (\frac{P''(a)}{2} + \dots )$, which suggests that at point $a$ function $P(x)-L(x)$ has at least double root: it is double root when $P''(a) \neq 0$. In other cases the multiplicity of root depends on which derivative has first non-zero value.

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