Polynomial division and the remainder and factor theorems

What this dot point is asking

NESA wants you to divide one polynomial by another (linear divisor at minimum, occasionally quadratic), state the result using the division algorithm, and use the remainder and factor theorems to identify roots without doing the full division when possible.

The answer

The division algorithm for polynomials

If $P(x)$ is a polynomial of degree $n$ and $D(x)$ is a polynomial of degree $k \le n$, then there exist unique polynomials $Q(x)$ (the quotient) of degree $n - k$ and $R(x)$ (the remainder) of degree less than $k$ such that

When $D(x)$ is linear (degree $1$), the remainder $R(x)$ is a constant.

The remainder theorem

If $P(x)$ is divided by $(x - a)$, the remainder is $P(a)$. In other words,

This lets you find the remainder without performing the division: just evaluate $P$ at $x = a$.

For a divisor of the form $(b x - c)$, the remainder is $P\!\left(\frac{c}{b}\right)$.

The factor theorem

The factor theorem is the special case of the remainder theorem where the remainder is zero.

$(x - a)$ is a factor of $P(x)$ if and only if $P(a) = 0$.

Equivalently, $a$ is a root of $P(x) = 0$ if and only if $(x - a)$ is a factor.

Long division of polynomials

To divide $P(x)$ by $D(x)$, set up the long-division layout and at each step divide the leading term of the current dividend by the leading term of the divisor, multiply back, and subtract.

The integer-division analogy is exact. Stop when the remainder has lower degree than the divisor.

Long division, stage by stage

Long division is procedural, and the marks come from keeping the layout aligned and each subtraction correct. Below, $x^3 - 2x^2 - 5x + 6$ is divided by $(x - 3)$ one stage at a time. (Because $x = 3$ is a root, the remainder will turn out to be $0$, confirming $(x - 3)$ is a factor.)

Stage 1, divide the leading terms. Divide the leading term of the dividend by the leading term of the divisor: $x^3 \div x = x^2$, which is the first term of the quotient. Multiply it back across the divisor, $x^2(x - 3) = x^3 - 3x^2$, write it underneath the matching columns, subtract to get $x^2$, and bring down the next term $-5x$.

Stage 2, repeat with the new leading term. Now divide the leading term of the new dividend $x^2 - 5x$ by $x$: $x^2 \div x = x$, the second quotient term. Multiply back, $x(x - 3) = x^2 - 3x$, subtract from $x^2 - 5x$ to leave $-2x$, and bring down $+6$.

Stage 3, finish and reach the remainder. Divide $-2x \div x = -2$, the last quotient term. Multiply back, $-2(x - 3) = -2x + 6$, and subtract from $-2x + 6$ to get $0$. The remainder is $0$, so division stops.

Stage 4, write the result in division-algorithm form. Collect the quotient $x^2 + x - 2$ and the remainder $0$: $x^3 - 2x^2 - 5x + 6 = (x - 3)(x^2 + x - 2)$. Because the remainder is $0$, $(x - 3)$ is a factor, and factorising the quadratic gives the full factorisation $(x - 3)(x + 2)(x - 1)$.

Strategy for factorising a cubic

To fully factorise a cubic $P(x)$ over the reals:

1. Find one rational root by trial. By the rational root theorem, any rational root $\frac{p}{q}$ (in lowest terms) has $p \mid$ constant term and $q \mid$ leading coefficient.
2. Use the factor theorem to confirm the root.
3. Divide $P(x)$ by the corresponding linear factor to get a quadratic quotient.
4. Factorise the quadratic by the standard methods.

If the quadratic factor has a negative discriminant, the cubic has one real root and two complex conjugate roots.

The rational root theorem in practice

The rational root theorem is the engine that makes factorising higher polynomials tractable. For a polynomial with integer coefficients, any rational root $\frac{p}{q}$ in lowest terms must have $p$ dividing the constant term and $q$ dividing the leading coefficient. This produces a finite, often short, list of candidates to test. For $P(x) = 2x^3 - 3x^2 - 3x + 2$, the constant is $2$ (so $p \in \{\pm 1, \pm 2\}$) and the leading coefficient is $2$ (so $q \in \{1, 2\}$), giving candidates $\pm 1, \pm 2, \pm \frac{1}{2}$. Test each with the factor theorem until one gives zero, then divide out the corresponding factor and continue. Without this theorem you would be guessing blindly.

Choosing between division methods

Two methods give the quotient after a factor is found. Long division is the general tool and works for any divisor, linear or quadratic, and is the only choice when the divisor has degree two or more. The equate-coefficients method (writing the quotient with unknown coefficients and comparing both sides) is often quicker for a linear divisor with a tidy quotient. Either is acceptable in the HSC, but show enough working that a marker can follow the steps; a bare quotient with no method risks lost marks if a sign is wrong.

Why the theorems save time

The remainder and factor theorems let you answer "what is the remainder?" or "is this a factor?" with a single substitution, no division required. This is the whole point of the topic: evaluating $P(a)$ is far faster than dividing, and for the factor theorem it converts a division question into a root-finding question. Reach for substitution first and only do the full division when you actually need the quotient.

How exam questions ask about division and the theorems

The phrasing tells you whether to substitute or to divide:

• "Find the remainder when $P(x)$ is divided by $(x - a)$." Do not divide. Apply the remainder theorem: the answer is $P(a)$. For a divisor $(bx - c)$, evaluate $P\!\left(\frac{c}{b}\right)$.
• "Find $a$ given the remainder is $r$" or "given $(x - a)$ is a factor." Set $P(a) = r$ (or $P(a) = 0$ for a factor) and solve the resulting equation for the unknown. This is the most common 2 to 3 mark version (see the HSC-style question above).
• "Show that $(x - a)$ is a factor" or "hence factorise $P(x)$." First confirm $P(a) = 0$ by the factor theorem, then divide by $(x - a)$ to get the quotient and continue factorising. The "show" step is a substitution; the "hence" step is a division.
• "The remainder when $P(x)$ is divided by $(x - a)(x - b)$ is..." The remainder has degree less than $2$, so write it as $rx + s$, then use $P(a)$ and $P(b)$ to set up two equations in $r$ and $s$. This is the standard Extension 1 quadratic-divisor remainder question.
• "Solve $P(x) = 0$" for a cubic or quartic. Find one root by testing rational-root candidates with the factor theorem, divide it out, and solve the remaining quadratic. If the quadratic has no real roots, state that the other roots are non-real.