Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose $p$ is a polynomial with coefficients in $\mathbb{C}$ and it has degree $m$. Prove that $p$ has $m$ distinct roots if and only if $p$ and its derivative $p′$ have no roots in common.

share|improve this question
4  
What have you done so far? What do you know? –  Arkamis Mar 26 '13 at 17:35
add comment

3 Answers

Let $r$ be the highest power of $x-a$ that divides $f(x),$

we can write $f(x)=(x-a)^rg(x)$ where $1\le r\le m$ so that $g(x)$ is a polynomial of degree $m-r$ with $(x-a)\not\mid g(x)$

$f'(x)=(x-a)^rg'(x)+r(x-a)^{r-1}g(x)$

If $a$ is a root of $f'(x)=0,f'(a)=0\iff r-1\ge1$ as $(x-a)\not\mid g(x)$

$\implies r\ge 2$ using Polynomial remainder theorem.

Alternatively,

if $f(x)=(x-a)g(x)\implies f(a)=0\iff (x-a)\mid f(x)$

Now, $f'(x)=(x-a)g'(x)+g(x)$

If $a$ is a root of $f'(x)=0\iff f'(a)=0$ $\implies g(a)=0\implies (x-a)\mid g(x)\implies (x-a)^2\mid f(x)$

share|improve this answer
add comment

Let $p(x)$ be a polynomial with complex coefficients, of degree $m\ge 1$. The hard part, which is essentially the Fundamental Theorem of Algebra, is that we can express $p(x)$ in the form $$p(x)=a(x-r_1)(x-r_2)\cdots (x-r_m),$$ where the $r_i$ are complex numbers, and $a$ is a non-zero complex number. The solutions of the equation $p(x)=0$ are then $r_1,\dots,r_m$.

We show (A) If the $r_i$ are not all distinct, then $p(x)$ and $p'(x)$ have a common root and (B) If $f(x)$ and $f'(x)$ have a common root, then the $r_i$ are not all distinct.

Proof of A: Suppose the roots are not all distinct. So suppose for example that $r_1=r_2=r$. Then $p(x)=(x-r)^2q(x)$ for some $q(x)$. Differentiate, We get $p'(x)=(x-r)^2q'(x)+2(x-r)q(x)$. It follows that $r$ is a root $p'(x)$, and hence is a common root of $p(x)$ and $p'(x)$.

Proof of B: Let $r$ be a common root of $p(x)$ and $p'(x)$. We show that $(x-r)^2$ divides $p(x)$. This will show that $r$ is a root of $p(x)$ of multiplicity $\gt 1$, meaning that the roots of $p(x)$ are not all distinct.

Because $r$ is a root of $p(x)$, by the Factor Theorem we have that $x-r$ divides $p(x)$, and therefore $p(x)=(x-r)q(x)$ for some polynomial $q(x)$.

Differentiate. We get $p'(x)=(x-r)q'(x)+q(x)$. But $r$ is a root of $p'(x)$. So $p;(r)=(r-r)q'(r)+q(r)$. It follows that $q(r)=0$. So again by the Factor Theorem, $x-r$ divides $q(x)$.

Suppose that $q(x)=(x-r)s(x)$. Then $p(x)=(x-r)q(x)=(x-r)^2s(x)$, and therefore $r$ is a root of $p(x)$ of multiplitity $\gt 1$.

Remark: The Fundamental Theorem is needed for the proof of the result in the OP. We could prove a weaker result without using the Fundamental Theorem.

Without the Fundamental Theorem, we can show, as in the argument above, that if $p(x)$ has a root of multiplicity $\gt 1$, then it is a common root of $p(x0$ and $p'(x)$. One can also show that if $p(x)$ and $p'(x)$ have a common root, then $p(x)$ has a root of multiplicity $\gt 1$.

But one cannot show that there are (counting multiplicity) $m$ roots. I suspect the wording of the problem is not quite what was intended.

share|improve this answer
add comment

Hint $\ $ It follows immediately by the double root test, one proof of which is sketched below.

$$\rm\begin{eqnarray} &&\rm\!\! (x\!-\!a)^2 |\ p(x)\!\!\!\!\!\!\!\\ \iff\ &&\rm x\!-\!a\ |\ p(x)\ &\rm and\ \ &\rm x\!-\!a\ \bigg|\ \dfrac{p(x)}{x\!-\!a}\\ \\ \iff\ &&\rm p(a) = 0 &\rm and&\rm x\!-\!a\ \bigg|\ \dfrac{p(x)-p(a)}{x\!-\!a}\ \ \left[\!\iff \color{#C00}{\dfrac{p(x)-p(a)}{x\!-\!a}\Bigg|_{\large\:x\:=\:a}} \!=\: 0\ \right] \\ \\ \iff\ &&\rm p(a) = 0 &\rm and&\rm \color{#C00}{p'(a)} = 0\end{eqnarray}$$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.