Exercise · ZW-POLYNOMIAL-DIVISION-ALGORITHM-007 · ★★☆ Intermediate

In \(\mathbb{Z}[x]\), we want to divide \(x^2\) by \(2x\) with \(r=0\) or \(\deg r<1\), and \(q,r\in\mathbb{Z}[x]\). What happens?

Choose one option.

Show answer
  • D. No such \(q\) and \(r\) exist

Explanation

If \(x^2=2x(ax+b)+r\) with \(a,b\in\mathbb{Z}\), the \(x^2\) coefficient would give \(2a=1\), impossible. The algorithm needs the leading coefficient of the divisor to be invertible, as in a field.

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Review the concept: Polynomial division algorithm

Given polynomials \(f\) and \(g\neq 0\) over a field \(F\), there exist a unique quotient \(q\) and a unique remainder \(r\) satisfying the equality below, with \(r\) zero or of degree less than that of \(g\).

\[f=gq+r,\qquad r=0\ \text{ or }\ \deg r<\deg g\]

More practice problems: polynomial division algorithm

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