Instead of performing long division step by step, we can write the quotient and remainder in their most general forms with unknown coefficients, then determine those coefficients by comparing both sides of the equation term by term. This algebraic approach often reveals the structure of the division more clearly than the mechanical long-division procedure.
Quick Reference
| Step | Action |
|---|---|
| 1 | Determine $\deg(q) = \deg(f) - \deg(d)$ and $\deg(r) \leq \deg(d) - 1$ |
| 2 | Write $q(x) = a_k x^k + \cdots + a_0$ and $r(x) = b_{m-1}x^{m-1} + \cdots + b_0$ with unknown coefficients |
| 3 | Substitute into $f(x) = q(x)\,d(x) + r(x)$ and expand |
| 4 | Match coefficients of each power of $x$ |
| 5 | Solve the resulting linear system |
The Method
Suppose we want to divide $f(x)$ (degree $n$) by $d(x)$ (degree $m$, $m \leq n$). By the division algorithm, there exist unique polynomials $q(x)$ and $r(x)$ such that
$f(x) = q(x)\,d(x) + r(x),$where $\deg(q) = n - m$ and $\deg(r) \leq m - 1$.
We write the most general forms:
$q(x) = a_k x^k + a_{k-1}x^{k-1} + \cdots + a_0 \quad (k = n - m),$$r(x) = b_{m-1}x^{m-1} + b_{m-2}x^{m-2} + \cdots + b_0.$Substituting into $f(x) = q(x)\,d(x) + r(x)$ and expanding the right side, we compare the coefficients of every power of $x$. This gives a system of $n + 1$ equations in $n + 1$ unknowns ($n - m + 1$ coefficients for $q$ plus $m$ coefficients for $r$). Solving the system yields $q$ and $r$.
For the equation $f(x) = q(x)d(x) + r(x)$ to hold for all values of $x$, the coefficients of every power of $x$ on both sides must be equal. This matching of coefficients is the key step.
Worked Examples
Divide $f(x) = 2x^4 + 3x^3 + 4x^2 + x - 2$ by $d(x) = x^2 - x + 1$.
Solution
Since $\deg(f) = 4$ and $\deg(d) = 2$, the quotient has degree $4 - 2 = 2$ and the remainder has degree at most 1. Set: $q(x) = a_2 x^2 + a_1 x + a_0, \qquad r(x) = b_1 x + b_0.$ We need: $2x^4 + 3x^3 + 4x^2 + x - 2 = (a_2 x^2 + a_1 x + a_0)(x^2 - x + 1) + (b_1 x + b_0).$ Expanding $(a_2 x^2 + a_1 x + a_0)(x^2 - x + 1)$: $a_2 x^4 + (-a_2 + a_1)x^3 + (a_2 - a_1 + a_0)x^2 + (a_1 - a_0 + b_1)x + (a_0 + b_0).$ Matching coefficients of each power of $x$: $ \begin{aligned} x^4: & \quad a_2 = 2 \\ x^3: & \quad -a_2 + a_1 = 3 \\ x^2: & \quad a_2 - a_1 + a_0 = 4 \\ x^1: & \quad a_1 - a_0 + b_1 = 1 \\ x^0: & \quad a_0 + b_0 = -2 \end{aligned} $ Solving sequentially: $a_2 = 2$ $-2 + a_1 = 3 \implies a_1 = 5$ $2 - 5 + a_0 = 4 \implies a_0 = 7$ $5 - 7 + b_1 = 1 \implies b_1 = 3$ $7 + b_0 = -2 \implies b_0 = -9$ Therefore $q(x) = 2x^2 + 5x + 7$ and $r(x) = 3x - 9$.Verification: $ \begin{aligned} (2x^2 &+ 5x + 7)(x^2 - x + 1) + (3x - 9)\\ &= 2x^4 - 2x^3 + 2x^2 + 5x^3 - 5x^2 + 5x + 7x^2 - 7x + 7 + 3x - 9\\ & = 2x^4 + 3x^3 + 4x^2 + x - 2.\quad \checkmark \end{aligned} $
Divide $f(x) = x^3 - 6x^2 + 11x - 6$ by $d(x) = x - 2$.
Solution
Since $\deg(f) = 3$ and $\deg(d) = 1$, the quotient has degree $3 - 1 = 2$ and the remainder has degree at most $0$, i.e., $r(x)$ is a constant. Set: $q(x) = a_2 x^2 + a_1 x + a_0, \qquad r(x) = b_0.$ We need: $x^3 - 6x^2 + 11x - 6 = (a_2 x^2 + a_1 x + a_0)(x - 2) + b_0.$ Expanding $(a_2 x^2 + a_1 x + a_0)(x - 2)$: $a_2 x^3 + (-2a_2 + a_1)x^2 + (-2a_1 + a_0)x + (-2a_0 + b_0).$ Matching coefficients of each power of $x$: $ \begin{aligned} x^3: & \quad a_2 = 1 \\ x^2: & \quad -2a_2 + a_1 = -6 \\ x^1: & \quad -2a_1 + a_0 = 11 \\ x^0: & \quad -2a_0 + b_0 = -6 \end{aligned} $ Solving sequentially: $a_2 = 1$ $-2(1) + a_1 = -6 \implies a_1 = -4$ $-2(-4) + a_0 = 11 \implies a_0 = 3$ $-2(3) + b_0 = -6 \implies b_0 = 0$ Therefore $q(x) = x^2 - 4x + 3$ and $r(x) = 0$. Since the remainder is zero, $d(x) = x - 2$ divides $f(x)$ exactly, and we have the factorization: $\boxed{x^3 - 6x^2 + 11x - 6 = (x - 2)(x^2 - 4x + 3).}$Verification: $ \begin{aligned} (x - 2)(x^2 - 4x + 3) &= x^3 - 4x^2 + 3x - 2x^2 + 8x - 6 \\ &= x^3 - 6x^2 + 11x - 6.\quad \checkmark \end{aligned}$
Comparison with Long Division
Both methods produce the same result. Their relative merits depend on context:
- Long division is mechanical and systematic; it is easy to execute once you know the steps.
- The method of undetermined coefficients is more algebraic; it makes the structure of the problem explicit and is useful when you want to work symbolically or when the divisor has a special form.
- For numerical problems with specific coefficients, long division (or synthetic division discussed later for linear divisors) is usually faster.
- For abstract proofs or when generalizing to multiple variables, undetermined coefficients is often cleaner.
