Polynomial $p(x) = 0$ for all $x$ implies coefficients of polynomial are zero

Question

I am curious why the following is true. The text I am reading is "An Introduction to Numerical Analysis" by Atkinson, 2nd edition, page 133, line 4.

$p(x)$ is a polynomial of the form:

$$ p(x) = b_0 + b_1 x + \cdots + b_n x^n$$

If $p(x) = 0$ for all $x$, then $b_i = 0$ for $i=0,1,\ldots,n$.

Why is this true? For example, for $n=2$, I can first prove $b_0=0$, then set $x=2$ to get a linear system of two equations. Then I can prove $b_1=b_2 = 0$. Similarly, for $n=3$, I first prove $b_0=0$, then I calculate the rank of the resulting linear system of equations. That shows that $b_1=b_2=b_3=0$. But if $n$ is very large, I cannot keep manually solving systems of equations. Is there some other argument to show all the coefficients must be zero when the polynomial is always zero for all $x$?

Thanks.

Answer 1

HINT $\ $ A nonzero polynomial over a field (or domain) has no more roots than its degree, as is easily proved by induction and the Factor Theorem. In fact if every natural was a root then the polynomial would be divisible by $\rm\:(x-1)\:(x-2)\:\cdots\:(x-n)\:$ for all $\rm\:n\in \mathbb N\:,\:$ which yields a contradiction for $\rm\:n\:$ greater than the degree of the polynomial.

Note that the proof of said statement depends crucially on the hypothesis that coefficient ring is an integral domain, i.e. a ring satisfying $\rm\:ab = 0\iff a=0\ \ or\ \ b=0\:.\:$ Over non-domains such as the integers modulo $\rm\:m\:$ not prime, polynomials can have more roots than their degree. In fact if this is true then one can use such roots to factor $\rm\:m\:,\:$ see here.

Answer 2

A polynomial can be uniquely fitted by knowing its value at $d+1$ points, where $d$ is the degree of the polynomial. If it's 0 at all points, that's clearly more than $d+1$ points.

But we also know that a polynomial can be written as $(x-r_0)(x-r_1)...(x-r_k)$, where the $r_i$ are the roots of polynomial (possibly complex). Again, if there are an infinite number of roots...

Answer 3

If you are willing to accept $\displaystyle \lim_{x \rightarrow \infty}\frac{1}{x^{m}}=0$ and $\displaystyle \lim_{x \rightarrow \infty}x^m=\infty$ for all $m>0$, then you can argue as follows. Suppose you have a polynomial $f(x)=b_nx^n+ \ldots b_0$ with $b_n \neq 0$ and $n\geq 1$, then $$\lim_{x\rightarrow \infty}f(x)=\lim_{x \rightarrow \infty} x^n(b_n+ \ldots \frac{a_0}{x^n})=\lim_{x \rightarrow \infty} x^nb_n=\infty.$$

Since the constant function $0$ does not have this property, it cannot be equal to a polynomial with degree greater or equal to $1$.

Answer 4

Let $n\in \mathbb{N}$, and $$p(x)=b_0+b_1x+\ldots +b_nx^n$$ be a polynomial. If $p(x)=0$ for every $x$, then $b_i=0$ for $i=0,\, 1,\ldots\, n$.

Proof:

By induction on $n$:

If $p(x)=b_0+b_1x=0$ for every $x$, then since $p(0)=p(1)=0$ you get $b_0=b_1=0$.

Suppose that for any $$r(x)=c_0+c_1x+\ldots +c_nx^n,$$ if $r(x)=0$ for every $x$, then $c_i=0$ for $i=0,\, 1,\ldots\, n$. Let $$p(x)=b_0+b_1x+\ldots +b_{n+1}x^{n+1}.$$ Suppose that $p\equiv 0$. Since $p(0)=0$, you get that $b_0=0$. From here, I have two possible arguments. The first, that was my original one, is as follows:

We have $$p(x)=b_1x+b_2x^2+\ldots+b_{n+1}x^{n+1}=x(b_1+b_2x+\ldots+b_{n+1}x^{n}).$$ Then for all $x$, $$\begin{align*} 0&= p(x)\\ 0&= x(b_1+b_2x+\ldots+b_{n+1}x^{n}).\end{align*}$$ Then for any $x\neq 0$, $$q(x):=b_1+b_2x+\ldots+b_{n+1}x^{n}=0$$ If $q\not\equiv 0$, then $q$ is a polynomial with infinitely many roots. This can't be, therefore $q(x) = 0$ for every $x$. By the induction hypothesis, $b_i=0$ for $i=1,\,\ldots ,\, n+1$.

The second argument was inspired by @Ragib Zaman's answer. Just differentiate both sides of $p(x)=0$, then $$b_1+2b_2x+\ldots+(n+1)b_{n+1}x^n=0$$ for all $x$. By the induction hypothesis, $kb_k=0$ for $k=1,\,\ldots,\,n+1$, and this implies that $b_k=0$ for $k=1,\,\ldots,\,n+1$

Since $b_0=0$, this proves that $b_i=0$ for $i=0,\, 1,\ldots,\, n+1$.

Answer 5

Note that $p(x)=0$ implies derivatives of all orders are also $0$. Let $x=0$ into $p(x) = b_0 + b_1 x + \cdots + b_n x^n$ to obtain $b_0 = 0$.

Now differentiate both sides: $p'(x) = b_1 + 2b_2 x + 3b_3 x^2 + \cdots + nb_n x^{n-1} $. Let $x=0$ again, and we get that $b_1=0$.

If we continue to differentiate and substitution $x=0$ we will get that $b_k=0$ for $k=0,1,2,\cdots, n$. This idea can easily be made into a rigorous induction proof.

A nice corollary of this result is that two polynomials are equal if and only if they have the same degree and coefficients.