Division algorithm for polynomials

Long division with numbers never really left you: $17 \div 5$ gives quotient $3$ and remainder $2$ , and $17 = 5\times 3 + 2$ . The division algorithm for polynomials is exactly the same bookkeeping applied to algebraic expressions — divide $p(x)$ by a nonzero polynomial $g(x)$ and you get a quotient $q(x)$ and a remainder $r(x)$ tied together by $p(x) = g(x)q(x) + r(x)$ , with $r(x)$ either zero or of smaller degree than $g(x)$ . A note on scope: this process was part of the Polynomials chapter in earlier CBSE syllabi but has been removed from the current CBSE Class 10 syllabus (2025–26). Current exams test the geometrical meaning of zeroes and the sum/product relationships between zeroes and coefficients, not polynomial long division. This article treats the topic as background — useful if you meet it in an older book, a Class 11 preview, or just want the mechanics behind "one polynomial divides another."

The core insight

Dividing $p(x)$ by a nonzero polynomial $g(x)$ always produces a unique quotient $q(x)$ and remainder $r(x)$ with $p(x) = g(x)q(x) + r(x)$ , where $r(x)$ is either zero or has strictly smaller degree than $g(x)$ .

The core idea

The pattern starts with integers. Euclid's Division Lemma (from the Real Numbers chapter) says that dividing an integer $a$ by a positive integer $b$ gives a unique quotient $q$ and remainder $r$ with $a = bq + r$ and $0 \le r < b$ — the remainder is always smaller than the divisor. Polynomial division swaps "smaller number" for "smaller degree," but keeps the same shape. If $p(x)$ is divided by a nonzero polynomial $g(x)$ , there exist unique polynomials $q(x)$ (the quotient) and $r(x)$ (the remainder) such that

p(x) = g(x)\, q(x) + r(x)

where: $p(x)$ is the polynomial being divided (the dividend), $g(x)$ is the polynomial dividing it (the divisor, never the zero polynomial), $q(x)$ is the quotient produced by the division, and $r(x)$ is what is left over, subject to the rule that either $r(x) = 0$ or $\deg r(x) < \deg g(x)$ . The mechanical process that produces $q(x)$ and $r(x)$ is polynomial long division: at each step, divide the highest-degree term of the current remainder by the highest-degree term of $g(x)$ to get the next term of the quotient, multiply that new term by all of $g(x)$ , subtract the product from the current remainder, and repeat. The process halts the instant the remainder's degree drops below the degree of $g(x)$ — not before, and not one step after. The figure below tracks how the degree of the running remainder falls by (at least) one at every step, and why the division must stop once that degree is smaller than the divisor's degree.

Figure

A staircase diagram showing the degree of the running remainder dropping at each division step until it falls below the degree of the divisor where the process stops

A useful special case: if the remainder $r(x)$ comes out to $0$ , then $g(x)$ divides $p(x)$ exactly, so every zero of $g(x)$ is also a zero of $p(x)$ . This is the link between the division algorithm and the factor theorem — it is, in principle, how a known zero can be stripped out of a polynomial to reduce it to a lower-degree factor. That particular application (factorising a cubic once one zero is known) is also outside the current syllabus; it is mentioned only to show why this chapter valued the division algorithm historically.

Derivation AWhy p(x) = g(x)q(x) + r(x) actually holds, checked step by step on a real division

The goal is to show, on a concrete cubic-by-linear division, exactly how each step of polynomial long division builds up the quotient term by term while keeping the running remainder consistent with $p(x) = g(x)q(x) + r(x)$ at every stage, so that the final identity is not a coincidence but the accumulated result of the subtraction process.

Step 1. Take $p(x) = x^3 - 3x^2 + 5x - 3$ and $g(x) = x - 1$ . Divide the leading term of $p(x)$ , which is $x^3$ , by the leading term of $g(x)$ , which is $x$ , giving $x^2$ as the first term of $q(x)$ . Multiply $x^2(x-1) = x^3 - x^2$ and subtract it from $p(x)$ : $(x^3-3x^2+5x-3) - (x^3-x^2) = -2x^2+5x-3$ . At this point $p(x) = g(x)\cdot x^2 + (-2x^2+5x-3)$ holds exactly, by construction of the subtraction.

Step 2. The new remainder $-2x^2+5x-3$ still has degree $2$ , which is not smaller than $\deg g(x) = 1$ , so the division continues. Divide $-2x^2$ by $x$ to get $-2x$ , the next quotient term. Multiply $-2x(x-1) = -2x^2+2x$ and subtract: $(-2x^2+5x-3)-(-2x^2+2x) = 3x-3$ . Now $p(x) = g(x)(x^2-2x) + (3x-3)$ , again exact because each subtraction only ever removes a multiple of $g(x)$ from the running remainder.

Step 3. The remainder $3x-3$ still has degree $1$ , equal to $\deg g(x)$ , so one more step is needed. Divide $3x$ by $x$ to get $3$ . Multiply $3(x-1) = 3x-3$ and subtract: $(3x-3)-(3x-3) = 0$ . The remainder is now $0$ , which has no degree to compare — by definition the process stops here, since $r(x) = 0$ satisfies the stopping condition automatically.

Step 4. Collect the quotient terms built up across the three steps: $q(x) = x^2 - 2x + 3$ , and $r(x) = 0$ . Because every step only subtracted an exact multiple of $g(x)$ from the previous remainder, unwinding the three subtractions shows $p(x) = g(x)\big(x^2-2x+3\big) + 0$ exactly — which is the claimed identity $p(x) = g(x)q(x)+r(x)$ , verified directly by expanding: $(x-1)(x^2-2x+3) = x^3-2x^2+3x-x^2+2x-3 = x^3-3x^2+5x-3 = p(x)$ .

Worked example

Divide $p(x) = 2x^2 + 3x + 1$ by $g(x) = x + 1$ , state the quotient and remainder, and use the result to find all zeroes of $p(x)$ .

Step 1. Divide the leading term $2x^2$ by $x$ to get $2x$ , the first quotient term. Multiply $2x(x+1) = 2x^2+2x$ and subtract from $p(x)$ : $(2x^2+3x+1)-(2x^2+2x) = x+1$ .

Step 2. The remainder $x+1$ has degree $1$ , equal to $\deg g(x)$ , so divide again: $x \div x = 1$ , the next quotient term. Multiply $1(x+1) = x+1$ and subtract: $(x+1)-(x+1) = 0$ . The remainder is now $0$ and the process stops.

Step 3. Collect the quotient: $q(x) = 2x+1$ , with $r(x) = 0$ . Check by expanding: $(x+1)(2x+1) = 2x^2+x+2x+1 = 2x^2+3x+1 = p(x)$ , confirming the identity $p(x)=g(x)q(x)+r(x)$ .

Step 4. Since the remainder is zero, $g(x) = x+1$ divides $p(x)$ exactly, so $x=-1$ is a zero of $p(x)$ (by the factor theorem connection above). The quotient itself factorises further: $2x+1 = 0$ gives the second zero $x = -\tfrac12$ . Both zeroes of $p(x) = 2x^2+3x+1$ are $x=-1$ and $x=-\tfrac12$ , matching the full factorisation $p(x) = (x+1)(2x+1)$ .

Common traps

Not knowing this is off the current syllabus. Don't spend exam-prep time drilling polynomial long division for Class 10 — the current syllabus tests zeroes-and-coefficients relationships instead, and this process is not asked.
Sign errors while subtracting. As with numerical long division, forgetting to distribute the minus sign across every term of the subtracted product is the single most common slip.
Stopping too early or too late. The division stops exactly when the remainder's degree becomes smaller than the divisor's degree — not when the remainder "looks small."
Confusing this with Euclid's Division Lemma for integers. The two ideas share the same shape ( $p(x)=g(x)q(x)+r(x)$ mirrors $a=bq+r$ ) but belong to different chapters and use a different stopping condition — degree of remainder versus size of remainder.

Ask the tutor: if you want to see how this connects to the factor theorem or to Euclid's Division Lemma for integers, or just want another practice division worked through, ask the AI tutor.

Since this is not on my CBSE Class 10 syllabus anymore, why should I bother learning it?

You don't need it for exams, and no current Class 10 question requires it. It is worth a light read only because the same shape — dividend equals divisor times quotient plus remainder, with a stopping condition on size or degree — reappears as Euclid's Division Lemma for integers and, later, as the basis for the factor theorem. Seeing the pattern once makes both of those feel familiar rather than new. If your time is tight, skip straight to zeroes-and-coefficients practice instead.

How is this different from Euclid's Division Lemma that I studied for real numbers?

The statement looks almost identical: $a = bq+r$ for integers versus $p(x) = g(x)q(x)+r(x)$ for polynomials. The difference is the stopping condition. For integers, you stop once $0 \le r < b$ — the remainder is a smaller number than the divisor. For polynomials, you stop once the remainder is zero or has smaller degree than the divisor — size of the coefficients doesn't matter at all, only degree does. They also live in different chapters: Real Numbers versus Polynomials.

If the remainder comes out to zero, does that always mean I have found a zero of the polynomial

Not quite — it means the divisor g of x divides p of x exactly, and therefore every zero of g of x is also a zero of p of x. If your divisor is a simple linear factor like x minus 1, then yes, its zero (here x equals 1) becomes a zero of p of x too. But if g of x itself has degree 2 or more, a zero remainder tells you g of x is a factor of p of x — you would still need to solve g of x equals zero separately to extract its own zeroes.

Division algorithm for polynomials

The core idea

Worked example

Common traps

Before this

Where this leads