Chapter 3.5 — Exercise 3.4 — Division Algorithm

The Governing Formula

For any dividend p(x) and non-zero divisor g(x), there exist a unique quotient q(x) and remainder r(x) satisfying the relationship below — the polynomial equivalent of the familiar numerical statement "dividend = divisor × quotient + remainder":

Three consequences follow directly from this one rule. If g(x) has degree 1, then degree of p(x) = 1 + degree of q(x). If p(x) is divided by (x − a), the remainder is simply p(a) — the Remainder Theorem, a genuine shortcut that skips the entire long-division process whenever the divisor is linear. And if r(x) = 0, then g(x) divides p(x) exactly, meaning g(x) and q(x) are both genuine factors of p(x).

A Full Worked Division

Dividing 3x³ + x² + 2x + 5 by x² + 2x + 1, using the same leading-term-by-leading-term process as numerical long division:

• 3x³ ÷ x² = 3x — the first quotient term. Multiplying back: 3x(x² + 2x + 1) = 3x³ + 6x² + 3x. Subtracting leaves −5x² − x + 5.
• −5x² ÷ x² = −5 — the next quotient term. Multiplying back: −5(x² + 2x + 1) = −5x² − 10x − 5. Subtracting leaves 9x + 10.
• Since 9x + 10 has degree 1, smaller than the divisor's degree 2, the division stops here.

So the quotient is 3x − 5 and the remainder is 9x + 10. Verifying against the formula: (x² + 2x + 1)(3x − 5) + (9x + 10) = 3x³ + x² + 2x + 5 — exactly the original dividend, confirming the division algorithm holds.

Three More Divisions

Running the same process on three further pairs produces a quotient and remainder each time, always satisfying the same degree condition:

The second row's remainder, 8, is worth noting — a plain number, with degree 0, which is still validly "less than" the divisor's degree 2. A remainder doesn't need to contain x at all; it only needs a smaller degree than the divisor, and a constant has the smallest degree there is (other than the zero polynomial itself).

Is the First Polynomial a Factor of the Second?

Dividing one polynomial by another and checking the remainder answers a natural follow-up question: does the first one divide the second exactly?

The third case makes the factor condition concrete: a remainder of 2, however small, is still non-zero — and by the rule stated earlier, only a zero remainder confirms that the divisor is a genuine factor of the dividend, no matter how close to zero the remainder happens to be.

Finding Every Zero When You Already Know Two

For 3x⁴ + 6x³ − 2x² − 10x − 5, two of its four zeroes are given as √(5/3) and −√(5/3). Since both are zeroes, (x − √(5/3)) and (x + √(5/3)) are both factors, so their product is too:

Dividing the original degree-4 polynomial by this degree-2 factor gives a quotient of 3x² + 6x + 3, with remainder 0 — confirming x² − 5/3 genuinely divides it. Factorising the quotient further: 3x² + 6x + 3 = 3(x² + 2x + 1) = 3(x + 1)². So the remaining two zeroes are both x = −1 — a repeated zero, hiding inside what looked like two separate unknowns. This whole technique — multiplying known zeroes into a factor, dividing it out, then factorising whatever's left — is the general strategy for any polynomial where only some of its zeroes are handed to you upfront.

Working Backward From Quotient and Remainder

A different kind of question gives the quotient and remainder and asks for the divisor. Dividing x³ − 3x² + x + 2 by an unknown g(x) gives quotient x − 2 and remainder −2x + 4. Rearranging the division formula solves directly for g(x):

Carrying out that final division gives g(x) = x² − x + 1. This "backward" version of the algorithm — solving for the divisor rather than the quotient or remainder — relies on the exact same formula as every forward division in this exercise, just rearranged to isolate a different unknown.

Constructing Your Own Examples

The closing question flips the exercise around entirely: instead of dividing a given polynomial, construct examples of p(x), g(x), q(x), and r(x) that satisfy the division algorithm under specific constraints.

• deg p(x) = deg q(x): p(x) = 2x² − 1, g(x) = 2 → q(x) = x², r(x) = −1. Dividing by a constant leaves the quotient the same degree as the dividend.
• deg q(x) = deg r(x): p(x) = x³ + 3x, g(x) = x² + 1 → q(x) = x, r(x) = 2x. Both quotient and remainder land at degree 1.
• deg r(x) = 0: p(x) = x² + 3, g(x) = x² + 1 → q(x) = 1, r(x) = 2. The remainder is a plain constant, the smallest degree a non-zero remainder can have.

Each example is a reminder that the division algorithm's degree condition — r(x) = 0, or degree of r(x) less than degree of g(x) — leaves plenty of room for the quotient and remainder to relate to each other and to the dividend in very different ways. There's no single "typical" relationship between the degrees of p(x), g(x), q(x), and r(x) beyond that one constraint, which is exactly why the exercise asks for three separate constructed examples rather than accepting one as proof of the general pattern.

The Chapter, Complete

This exercise closes Polynomials by tying together everything earlier in the chapter: zeroes found by substitution (Exercise 3.1), read off a graph (Exercise 3.2), and connected to coefficients through sum/product formulas (Exercise 3.3) — division algebra is what recovers the remaining zeroes once only a few are known, as the √(5/3) example above shows directly. Polynomial arithmetic carries forward immediately into Chapter 4's Pair of Linear Equations, where the same factoring instincts apply to systems of two variables instead of one.