5
$\begingroup$

Find all polynomials $p(x)$ such that:

$$xp(x-1) = (x-30)p(x)$$

My solution:

We can see, if $x = 0$ then $p(0) = -0/29 = 0$ similarly, $p(29) = 0$ so our polynomial is $x(x-29)$.

I'd thought this was a simple question, but apparently the answer is:

$ax(x-1)(x-2)(x-3)\cdots(x-29)$, where a is any real number.

I have no idea how this is the answer

$\endgroup$
1
  • 3
    $\begingroup$ Okay so you have deduced that $x=0$ and $x=29$ are the roots of the polynomial. But that doesn't mean that you have the whole polynomial. So our work shows that the polynomial is of the form $$ p(x) = g(x) \cdot x (x-29) $$ for some polynomial $g(x)$. Which is of course true, but you need to continue ... $\endgroup$
    – Matti P.
    Apr 3, 2020 at 7:08

5 Answers 5

5
$\begingroup$

Your approach is almost correct. You are at half way.

You got $P(0)=P(29)=0$. Now consider, $P(1)$. From $$x\cdot P(x-1)=(x-30)P(x),$$ we have when $x=1$, $$1\cdot P(0)=(-29)P(1).$$ $$\implies P(1)=0.$$ Similarly, considering $x=2, 3, \ldots, 29$, you will get $P(2)=P(3)=\cdots=P(29)=0$. Hence your result follows.

$\endgroup$
3
  • 1
    $\begingroup$ Yes, this is the answer I was looking for. I didn't consider all cases. Just wanted to ask, if the question asked for the polynomial of lowest degree which satisfies the given condition then my solution would be correct, no? $\endgroup$
    – MNIShaurya
    Apr 3, 2020 at 7:57
  • 1
    $\begingroup$ Note that this solution is currently incomplete, since $a$ is a real number (and not a polynomial). How do you know that there are no further roots? Or that those roots are not double roots? $\endgroup$
    – Calvin Lin
    Apr 3, 2020 at 11:02
  • 1
    $\begingroup$ @MNIShaurya : No, your solution would not be correct. The minimal degree solution has degree $30$. $\endgroup$ Apr 3, 2020 at 14:10
3
$\begingroup$

Even your short answer is missing something -- you should multiply by an unknown constant because knowing the roots tells you nothing about the vertical scale of the polynomial.


The correct version of the argument you are attempting is this...

What happens when $x = 1$? Then $p(0) = -29 p(1)$. Since $p(0) = 0$, we have $p(1) = 0$.

We just got $p(1)$ on the right. To get $p(1)$ on the left, set $x = 2$. Then $2 p(1) = -28 p(2)$. Since $p(1) = 0$, we have $p(2) = 0$.

Repeating with $x = 3$, $4p(2) = 27 p(3)$ and so $p(3) = 0$.

...

Continuing, you eventually show all of $0$, $1$, ..., $29$ are roots. As in the first paragraph, you need a constant multiple since you have no way to get another value of the polynomial.

That seems to give you the answer you recite, but there is more to show. How do we know there aren't more roots? Suppose there were; for example, let $$ p(x) = a x(x-1)(x-2)\cdots (x-29) \cdot (x-100) \text{.} $$ Then the equation you start with forces $101$ is a root, which forces $102$ is a root, which forces ..., producing infinitely many roots. If you work through the details, you can show that the presence of any root other than those listed in the recited answer forces infinitely many more roots. Since no polynomial has infinitely many roots, there are no roots other than those in the recited answer.

So that only leaves repetitions of the $30$ roots we know about. Set $$ p(x) = a \prod_{k=0}^{29} (x-k)^{q_k} \text{.} $$

In $x p(x-1)$, the factors $(x-30)$ and $x$ appear with multiplicities $q_{29}$ and $1$, respectively. In $(x-30)p(x)$, with multiplicities $1$ and $q_0$, respectively. So $q_{29} = 1 = q_0$. Applying these henceforth, ...

In $x p(x-1)$, the factors $(x-29)$ and $x-1$ appear with multiplicities $q_{28}$ and $1$, respectively. In $(x-30)p(x)$, with multiplicities $1$ and $q_1$, respectively. So $q_{28} = 1 = q_1$. Applying these henceforth, ...

...

Continuing, we show all the $q_{k} = 1$, so all the roots have multiplicity one.

$\endgroup$
0
1
$\begingroup$

From $(x-30)p(x)=xp(x-1)$ and $p(0)=0$, we have $$(1-30)p(1)=1\cdotp(0)=0$$ i.e. $p(1)=0$. Also $$(2-30)p(2)=2p(1)=0$$ i.e. $p(2)=0$. Generally, if $p(k)=0$, for some positive integer $k$, then $$(k+1-30)p(k+1)=(k+1)p(k)=0$$ therefore, we have $$p(x)=x(x-1)(x-2)\cdots(x-29)g(x)$$ We claim $p(30)\ne 0$ else, if $p(30)=0$ then we will have $$p(31)=p(32)=\cdots=p(n)=\cdots=0$$ for all $n\geq30$ which implies $p\equiv0$.

$\endgroup$
1
  • 1
    $\begingroup$ Note that this solution is currently incomplete, since $a$ is a real number (and not a polynomial). How do you know that there are no further roots? Or that those roots are not double roots? $\endgroup$
    – Calvin Lin
    Apr 3, 2020 at 11:03
1
$\begingroup$

You approach is correct. Indeed $p(0)=0$. Now substitute $x=1$. We get

$p(0) = -29p(1) = 0$ $\implies p(1) = 0$

Now substitute $x=2$ and in a similar fashion you obtain $p(2) = 0$. This continues until $x=30$ when $30p(29) = 0*p(30)$. Hence $x=0$ to $29$ are all roots which yields :

$p(x) = ax(x-1)(x-2)...(x-29)$

$\endgroup$
1
  • 1
    $\begingroup$ Note that this solution is currently incomplete, since $a$ is a real number (and not a polynomial). How do you know that there are no further roots? Or that those roots are not double roots? $\endgroup$
    – Calvin Lin
    Apr 3, 2020 at 11:04
1
$\begingroup$

As you can judge from my comment on the solutions, they are all incomplete. What the analysis on $ p (0) = p(1) = \ldots p (29) = 0 $ indicates is only that

$$ p(x) = A(x) \times x (x-1)(x-2) \ldots (x-29), $$

where $A(x)$ is a polynomial.

When we substitute this back into the given equation, and divide by the common factors we get is

$$ A (x-1) = A(x). $$

This implies that $ A(x-1) = A(x) = A(x+1) = A(x+2) = \ldots$.

However, the only polynomial that takes on the same values at infinitely many points, is the constant polynomial. Thus $A(x) = a$ for some real number $a$.

This completes the solution.

$\endgroup$
2
  • 1
    $\begingroup$ This is also incomplete, it doesn't contain the analysis that leads to the first equation. Also note that p(30) \neq 0 $\endgroup$
    – James K
    Apr 3, 2020 at 12:32
  • 1
    $\begingroup$ @JamesK Of course, why repeat work that was done 7 times? I'm building off of that, as my first line indicates. My point was just to complete the solution. (Edited p(30).) $\endgroup$
    – Calvin Lin
    Apr 3, 2020 at 12:53

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .