Divide one polynomial by another using long division, expressing the result in the form P(x) = D(x)Q(x) + R(x) where the remainder has degree less than the divisor, handling missing terms, and writing the result in the rational form P/D = Q + R/D

What this dot point is asking

This is your first time dividing one polynomial by another, and the skill is the polynomial version of the long division you learned for whole numbers. When you divide $197$ by $12$ you get a quotient of $16$ and a remainder of $5$, and you can write the whole story without fractions as $197 = 12 \times 16 + 5$. Polynomial division works the same way: you long-divide a polynomial $P(x)$ by a divisor $D(x)$, and you get a quotient $Q(x)$ and a remainder $R(x)$ that fit together as the identity $P(x) = D(x)Q(x) + R(x)$. The one new rule is when to stop: you keep going until the remainder has a smaller degree than the divisor, because once it does you can no longer divide its leading term by the leading term of the divisor.

This is the Year 11 first-contact version. We reason entirely from the long-division algorithm and the $P = DQ + R$ identity. The shortcuts that let you find a remainder without dividing at all, the remainder theorem and the factor theorem, are the very next page; the applied and exam-heavy treatment that combines division with factorising lives on the Year 12 page on polynomial division and the factor theorem. Here the goal is to carry out the division itself cleanly and to read the result off in both standard and rational form.

The answer

The division algorithm: P = DQ + R

The central result mirrors integer division exactly. Where whole-number division gives a quotient and a remainder smaller than the divisor, polynomial division gives a quotient and a remainder of lower degree than the divisor.

The degree condition $\deg R < \deg D$ is not an optional tidy-up; it is what makes the quotient and remainder unique. If you stopped early and left a remainder whose degree was still as big as the divisor's, you could divide once more, so you would not yet have the true quotient. The phrase "divide until you cannot any more" means exactly "divide until the remainder has dropped below the degree of the divisor." Two quick consequences worth knowing:

• Dividing by a linear divisor (degree $1$) always leaves a remainder of degree less than $1$, that is, a constant (possibly zero).
• Dividing by a quadratic divisor (degree $2$) leaves a remainder that is linear or constant, of the form $rx + s$.

The long-division method, step by step

The mechanics are a four-move cycle, repeated until the remainder is small enough. Lay the dividend out with its terms in descending powers, set the divisor to the left, and at each pass:

1. Divide the leading term of what remains by the leading term of the divisor. This single term goes into the quotient, above its matching power column.
2. Multiply the whole divisor by that quotient term.
3. Subtract the product from the current dividend. The leading terms cancel by design, dropping the degree by one.
4. Bring down the next term and repeat, until the running remainder has degree less than the divisor.

The reason the leading terms always cancel is the point of step 1: you chose the quotient term precisely so that, when multiplied by the divisor's leading term, it reproduces the leading term you are trying to remove. Every pass therefore kills the current highest power, and the working marches steadily down through the degrees.

Handling missing terms

A polynomial like $x^4 + 1$ has no $x^3$, $x^2$ or $x$ term, and a dividend with gaps is where divisions go wrong, because a term can silently land in the wrong column. There are two equivalent safeguards, and you should pick one and use it every time:

• Insert zero coefficients. Rewrite $x^4 + 1$ as $x^4 + 0x^3 + 0x^2 + 0x + 1$. Now every column has an entry and the subtractions line up automatically. This is the more reliable choice and the one used throughout this page.
• Leave a gap for the missing column. Write nothing in that column but keep the spacing, so later terms still fall under the right power.

The same applies to a divisor with a missing term: treat $x^2 + 1$ as $x^2 + 0x + 1$ so that, when you multiply and subtract, the middle column is handled rather than skipped.

Writing the result in rational form

The identity $P(x) = D(x)Q(x) + R(x)$ can be divided through by $D(x)$ to express the original fraction as a polynomial plus a proper remainder fraction:

This is the polynomial echo of writing $\tfrac{197}{12} = 16 + \tfrac{5}{12}$ (or the mixed number $16\tfrac{5}{12}$). The quotient $Q(x)$ is the "whole" part and $\dfrac{R(x)}{D(x)}$ is the leftover proper fraction, called proper because the degree of its top is less than the degree of its bottom. This form is the one you want for sketching rational functions and, later, for integration, because it splits a hard fraction into an easy polynomial plus a small remainder. A question may ask for either form, so be ready to convert: multiply the rational form through by $D(x)$ to get back the standard identity, or divide the standard identity by $D(x)$ to get the rational form.

Verifying by reconstruction

The single best habit in this topic is to check every division by rebuilding the dividend. Because the algorithm guarantees $P = DQ + R$, you can expand $D(x) \times Q(x)$, add $R(x)$, and you must recover $P(x)$ term for term. If you do not, there is an arithmetic slip somewhere in the tableau. This check is quick, it is self-marking, and it turns a topic that is easy to fumble into one you can be certain about. Every worked example and practice solution on this page ends with this reconstruction.

A worked long-division tableau

Here is the full layout for dividing $x^3 - 2x^2 - 5x + 9$ by $x - 3$. The quotient sits above the bar in its power columns, each step subtracts a multiple of the divisor, and the final remainder is at the foot.

Reading off the tableau, the quotient is $x^2 + x - 2$ and the remainder is $3$, so

Building the tableau stage by stage

The same division, shown one cycle at a time. At each stage the newest quotient term and its subtraction are picked out in accent; earlier work is left in plain type. Build your own divisions in exactly this rhythm.

Stage 1, divide and subtract the leading terms. Divide $x^3$ by $x$ to get $x^2$, the first quotient term. Multiply the divisor: $x^2(x - 3) = x^3 - 3x^2$. Subtract this from the dividend; the $x^3$ terms cancel and you are left with $x^2 - 5x + 9$ (the $-5x$ and $+9$ brought straight down).

Stage 2, repeat on the new leading term. Divide $x^2$ by $x$ to get $x$, the next quotient term. Multiply: $x(x - 3) = x^2 - 3x$. Subtract; the $x^2$ terms cancel and the running remainder becomes $-2x + 9$.

Stage 3, the last subtraction leaves the remainder. Divide $-2x$ by $x$ to get $-2$, the final quotient term. Multiply: $-2(x - 3) = -2x + 6$. Subtract; the $x$ terms cancel and you are left with $3$. Its degree is $0$, less than $\deg(x - 3) = 1$, so the division stops. The completed quotient is $x^2 + x - 2$ with remainder $3$.

How exam questions ask about polynomial division

The wording is predictable once you map it to the algorithm.

• "Use long division to divide $P(x)$ by $D(x)$, expressing the result in the form $P(x) = D(x)Q(x) + R(x)$." Run the four-move cycle, stop at $\deg R < \deg D$, and write the identity line explicitly. State $Q(x)$ and $R(x)$ clearly.
• "Write the result in the form $\dfrac{P}{D} = Q + \dfrac{R}{D}$." Do the same division, then divide the identity through by $D(x)$. The leftover fraction $\dfrac{R}{D}$ must be proper ($\deg R < \deg D$).
• "Find the quotient and remainder when $P(x)$ is divided by $D(x)$." Just name $Q(x)$ and $R(x)$ from the tableau; you do not always need the full identity line, but writing it is a safe, mark-earning check.
• "Show that $D(x)$ is a factor of $P(x)$" or "$P(x)$ is divisible by $D(x)$." Divide and show the remainder is $0$; then $P(x) = D(x)Q(x)$, and you can quote the quotient as the cofactor.
• "Find $a$ and $b$ so that $P(x)$ is exactly divisible by $D(x)$." Either divide and set the remainder (which will contain $a$ and $b$) to zero, or write $P = D \times Q$ with an unknown quotient and equate coefficients. Both are reliable; equating coefficients is often faster for a clean divisor.
• "What are the possible degrees of the remainder on division by a cubic?" Use $\deg R < \deg D$: a cubic divisor ($\deg 3$) forces the remainder to have degree $0$, $1$ or $2$ (or be zero).