If $(x-a)^2$ is a root of a polynomial, then the graph will be tangent to the $x$ axis at $x=a$ but why? I know this is always the case for real double roots however I do not know the explanation for this behavior. Does this behavior apply to all real polynomial roots with even index multiplicities?


2 Answers 2


First a comment

Saying that $(x-a)^2$ is a root of a polynomial $p(x)$ is an improper wording. You should say that $(x-a)^2$ divides $p(x)$ or that $a$ is a root of $p(x)$ of multiplicity at least equal to $2$.

Regarding your question

So you suppose that $p(x) = (x-a)^2q(x)$. Then taking the derivative you get

$$p^\prime(x) = 2(x-a)q(x) + (x-a)^2 q^\prime(x).$$

Therefore you have $p(a)=p^\prime(a)=0$. Proving that the graph of the function $x \ \mapsto p(x)$ is tangent to the $x$-axis at $x=a$.


You can prove this easily using calculus, but I'll give an even more elementary argument.

The graph of $f(x)=(x-a)^{2}$ is an upward facing parabola. In particular, it is the graph of $f(x)=x^{2}$ shifted to the right by $a$ units (if a is positive).

Note that, being a square, $(x-a)^{2}$ is always greater than or equal to $0$. And it is only equal to $0$ at $x=a$.

Combining the above two observations, the graph of $f(x)$ is tangent to the $x$-axis at $x=a$.

  • 2
    $\begingroup$ The OP is asking about any polynomial that has $(x-a)^2$ as a factor, not just $(x-a)^2$. $\endgroup$
    – amd
    Jan 16, 2019 at 18:24

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