Remainder Theorem

Remainder Theorem

TGM Original Questions

For a linear divisor or factor \(ax\pm b\), assume \(a\ne0\).

  • Given a polynomial \( \mathrm{f}(x)\) is divided by a linear expression \(ax - b\), then its remainder is \( \mathrm{f} \left( \frac{b}{a} \right) \)
    This follows from the division, where
    \( \mathrm{f}(x) = (ax - b) \mathrm{Q}(x) + \mathrm{R}\)
    where \( \mathrm{Q}(x)\) is the quotient and \(\mathrm{R}\) is the remainder

\( \mathrm{f} \ \left(\frac{b}{a} \right) = \mathrm{Q}\left(\frac{b}{a}\right) \times \left(a \times \frac{b}{a} - b \right) + \mathrm{R} \)
\( \mathrm{f} \left( \frac{b}{a} \right) = \mathrm{Q} \left( \frac{b}{a} \right) \times 0 + \mathrm{R} = \mathrm{R} \)

Remainder when divided by higher degree polynomial

  • When \( \mathrm{f}(x) \) is divided by a polynomial of degree \(n\), its remainder is zero or has degree less than \(n\)
  • For example when \( \mathrm{f}(x)\) is divided by a quadratic expression \(ax^2 + bx + c\), its remainder has the form \(dx+e\), which may be linear, constant or zero.
  • In summary:

Divisor

Remainder

\(ax + b\) degree \(1\)

Constant

\(ax^2 + bx + c\) degree \(2\)

Linear expression \(dx + e\) or constant or zero

\(ax^3 + bx^2 + cx + d\) degree \(3\)

Quadratic expression \(ex^2 + fx + g\) or lower degree, or zero

The displayed degree of each divisor assumes its leading coefficient is nonzero. The zero polynomial has no degree in this convention.

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