Dividing a polynomial $p(x)$ by $x-k$, but $x-k$ is set to be $0$?

Question

While studying, I read the following,

When a polynomial $p(x)$ is divided by $(x-k)$, if we set $(x-k)$ to be $0$, we get $x=k$ and the remainder as $p(k)$.

However, if we divide $p(x)$ by $(x-k)$, or $0$, wouldn't the answer be undefined?

In short: you seem to be confusing division by a number and the division of two polynomials — Ben Grossmann, Jul 17 '15 at 15:37
@Omnomnomnom However, if the denominator polynomial is equivalent to 0, wouldn't they be exchangeable? — RK01, Jul 17 '15 at 15:39
Right. Similarly, for any polynomial $p(x)$, the function $\frac{p(x)}{x-k}$ is only undefined when $x=k$. — Ben Grossmann, Jul 17 '15 at 15:42

score 4 · Accepted Answer · answered Jul 17 '15 at 15:42

Given a polynomial $p(x)$, one can phrase the fact (coming from the division algorithm) that $$p(x)=(x-k)\cdot q(x)+p(k)\quad$$ for some polynomial $q(x)$ as either

dividing $p(x)$ by $x-k$ leaves a remainder of $p(k)$
in the ring of polynomials $F[x]$, if we "set $x-k$ equal to $0$", i.e., quotient by the ideal $(x-k)$ to produce the ring $F[x]/(x-k)$, then we have $p(x)\equiv p(k)\bmod (x-k)$.

You're just mixing up these two phrasings of the same mathematics.

score 3 · Answer 2 · answered Jul 17 '15 at 15:43

The "Remainder Theorem" says if you divide $p(x)$ by $(x-k)$, the remainder is equal to $p(k)$. This follows since upon division, one obtains $p(x)=(x-k)q(x)+r$ (note the remainder is of lower degree than the divisor, and hence in this case constant).

From this you get $p(k)=(k-k)q(k)+r=r$.

score 2 · Answer 3 · answered Jul 17 '15 at 15:43

The division there isn't for integers but for polynomials. If for integers, the Euclidian Division states that $$\forall a,b\in\mathbb{Z},\,\exists ! \,q,r\in\mathbb{Z},\,a=bq+r\,\,\land\,\, 0\le r<|b|$$ its version for polynomials is $$\forall P,T\in\mathbb{C}[X],\,\exists !\,Q,R\in\mathbb{C}[X],\,P=TQ+R\,\,\land\,\, \deg{R}<\deg{T}$$ where $\mathbb{C}[X]$ is the set of all polynomials with complex coefficients (this set contains the set of all polynomials with real coefficients $\mathbb{R}[X]$ since $\mathbb{R}\subset\mathbb{C}$)

So in your case:$$\exists !\, Q,R\in\mathbb{C}[X],\,p(x)=(x-k)Q(x)+R(x)\,\,\land\,\,\deg{R}<\deg{(X-k)}=1$$

Since $\deg{R}<1$ therefore $R=c$ is constant and is equal to the number $c\in\mathbb{C}$ that can be $0$ or not $0$. To find it, replace $x$ by $k$ to get $p(k)=R(k)=c$ so the remainder is $p(k)$.

Dividing a polynomial $p(x)$ by $x-k$, but $x-k$ is set to be $0$?

3 Answers3