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## 5.2. The division algorithm.

We say that a polynomial divides a polynomial (written ) if there is a polynomial such that . Notice that if then either or . Notice that if then either or (or both). This means that we can (only) divide by nonzero polynomials.

Problem 5.3. Let be two polynomials. Prove that if and then .

In fact, there is an analog of long division for polynomials, as there is for integers.

Theorem 5.4. Let be polynomials with . Then there exist unique polynomials such that and .

Proof. We ﬁrst prove existence of . If then we set and . Otherwise, . Let

and

Deﬁne the integer . We will use induction in .

Base step: Let , then . We set and . Notice that is well-deﬁned because and that the coefficient of in vanishes, whence .

Induction step: Now we assume that this is true whenever and let , so that . Let . Notice that , whence by induction there exist with and . Therefore

whence deﬁne and the result is true for .

Finally we prove uniqueness. Suppose that , with and . Rearranging, we have , whence divides . Since , this can only happen if , whence . In this case, , which since implies that or, equivalently, that .□

The polynomials and in the statement of the theorem are called the quotient and remainder, respectively.

Given we can ﬁnd by long division, as the following example shows.

Example 5.5. Let and . Then we let . Similarly let , whose degree is less than that of . Therefore,

whence and .

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