Polynomial Division: Algorithm, Remainder Theorem, and Applications

Polynomial division is one of the fundamental operations in algebra. It allows us to express a polynomial as the product of another polynomial and a quotient, plus a possible remainder.

At first glance, it may seem like a purely mechanical procedure. In fact, polynomial division is a central theoretical tool: it allows us to study divisibility, identify factors, apply the Remainder Theorem, understand synthetic division, and connect the roots of a polynomial to its factorization.

The underlying idea is similar to that of integer division: given a dividend and a divisor, we look for a quotient and a remainder. In the case of polynomials, however, the quantity that governs the process is not the numerical value, but the degree.

Table of Contents

• Formal Definition
• The Meaning of the Quotient and the Remainder
• Polynomial Long Division
• Worked Example with Layout
• The Euclidean Division Theorem
• Divisibility of Polynomials
• The Remainder Theorem
• Synthetic Division
• Synthetic Division Step by Step
• Common Mistakes
• Exercises with Solutions
• Conclusion

Formal Definition

Let \(A(x)\) and \(B(x)\) be two polynomials, with \(B(x)\neq 0\). To divide \(A(x)\) by \(B(x)\) means to find two polynomials \(Q(x)\) and \(R(x)\), called the quotient and the remainder respectively, such that \(A(x)=B(x)Q(x)+R(x)\), subject to the condition:

\[ R(x)=0 \quad \text{or} \quad \deg R(x)<\deg B(x) \]

The polynomial \(A(x)\) is called the dividend, the polynomial \(B(x)\) is called the divisor, \(Q(x)\) is the quotient, and \(R(x)\) is the remainder.

The degree condition on the remainder is essential. If the remainder had degree greater than or equal to that of the divisor, it would still be possible to continue the division. The process terminates only when what remains has degree strictly less than that of the divisor — that is, when it can no longer be divided by the same procedure.

For example, \(x^2+3x+2=(x+1)(x+2)+0\). Here the dividend is \(x^2+3x+2\), the divisor is \(x+1\), the quotient is \(x+2\), and the remainder is \(0\). Since the remainder is zero, the division is exact.

The Meaning of the Quotient and the Remainder

The identity \(A(x)=B(x)Q(x)+R(x)\) states that the polynomial \(A(x)\) is decomposed into two parts: a part that is a multiple of the divisor \(B(x)\), namely \(B(x)Q(x)\), and a residual part, namely \(R(x)\).

The quotient \(Q(x)\) represents the portion of the dividend that can be obtained by multiplying the divisor by another polynomial. The remainder \(R(x)\), on the other hand, represents what is left after subtracting from the dividend every possible contribution built from the divisor.

The analogy with integer division is helpful. For example, \(17=5\cdot 3+2\). Here \(17\) is the dividend, \(5\) is the divisor, \(3\) is the quotient, and \(2\) is the remainder. The remainder must be less than the divisor.

The same structure arises in polynomial division, but the relevant condition does not concern the numerical value of the remainder — it concerns its degree:

\[ \deg R(x)<\deg B(x) \]

This is the principle that makes polynomial division a genuine Euclidean division.

Polynomial Long Division

The most general method for dividing two polynomials is long division. This is the procedure to use when the divisor has any degree, not necessarily \(1\).

Before starting, it is important to write both polynomials with their terms arranged in descending powers of \(x\). If a term is missing, it must be included with coefficient \(0\).

For example, the polynomial:

\[ x^4-3x+2 \]

must be written as:

\[ x^4+0x^3+0x^2-3x+2 \]

This does not change the polynomial, but makes the long division layout cleaner and prevents alignment errors.

The idea behind long division is to progressively eliminate the leading term of the dividend. At each step, we look at the leading term of the remaining polynomial and divide it by the leading term of the divisor.

If \(\deg A(x)<\deg B(x)\), the division is already complete: \(Q(x)=0\) and \(R(x)=A(x)\). If instead \(\deg A(x)\geq \deg B(x)\), we divide the leading term of the dividend by the leading term of the divisor. The result becomes the first term of the quotient. We then multiply the divisor by this term and subtract the product from the dividend.

Worked Example with Layout

Divide \(2x^3+3x^2-5x+1\) by \(x-2\). The dividend is already written with terms in descending order:

\[ 2x^3+3x^2-5x+1 \]

We seek two polynomials \(Q(x)\) and \(R(x)\) such that:

\[ 2x^3+3x^2-5x+1=(x-2)Q(x)+R(x) \]

First step

Divide the leading term of the dividend by the leading term of the divisor:

\[ \frac{2x^3}{x}=2x^2 \]

The first term of the quotient is \(2x^2\). Multiply the divisor by \(2x^2\):

\[ 2x^2(x-2)=2x^3-4x^2 \]

To subtract this from the dividend, change the sign of each term and add:

\[ -(2x^3-4x^2)=-2x^3+4x^2 \]

We also include the missing terms to keep the columns aligned:

\[ -2x^3+4x^2+0x+0 \]

After adding, the \(2x^3\) term cancels and the partial remainder is:

\[ 7x^2-5x+1 \]

Second step

Repeat the procedure on the partial remainder. Divide its leading term by the leading term of the divisor:

\[ \frac{7x^2}{x}=7x \]

The second term of the quotient is \(7x\), so the partial quotient becomes:

\[ 2x^2+7x \]

Multiply the divisor by \(7x\):

\[ 7x(x-2)=7x^2-14x \]

Change signs to subtract:

\[ -(7x^2-14x)=-7x^2+14x \]

Include the missing constant term:

\[ -7x^2+14x+0 \]

After adding, the new partial remainder is:

\[ 9x+1 \]

Third step

Divide the leading term of the new partial remainder by the leading term of the divisor:

\[ \frac{9x}{x}=9 \]

The third term of the quotient is \(9\), so the quotient becomes:

\[ 2x^2+7x+9 \]

Multiply the divisor by \(9\):

\[ 9(x-2)=9x-18 \]

Change signs to subtract:

\[ -(9x-18)=-9x+18 \]

The final remainder is:

\[ R(x)=19 \]

Since the divisor \(x-2\) has degree \(1\) and the remainder \(19\) has degree \(0\), we have:

\[ \deg 19=0<1=\deg(x-2) \]

The division is therefore complete.

Final result

From the layout we read the quotient:

\[ Q(x)=2x^2+7x+9 \]

and the remainder:

\[ R(x)=19 \]

Therefore:

\[ 2x^3+3x^2-5x+1=(x-2)(2x^2+7x+9)+19 \]

We verify the result by expanding the right-hand side:

\[ (x-2)(2x^2+7x+9)+19 \]

\[ =2x^3+7x^2+9x-4x^2-14x-18+19 \]

\[ =2x^3+3x^2-5x+1 \]

The division is correct.

The Euclidean Division Theorem

Theorem. Let \(A(x)\) and \(B(x)\) be two polynomials, with \(B(x)\neq 0\). Then there exist unique polynomials \(Q(x)\) and \(R(x)\) such that \(A(x)=B(x)Q(x)+R(x)\), with:

\[ R(x)=0 \quad \text{or} \quad \deg R(x)<\deg B(x) \]

This result is known as the Euclidean Division Theorem for polynomials (also called the Division Algorithm). Existence is guaranteed by the long division algorithm; uniqueness means that, given a fixed dividend and divisor, there cannot be two different quotients or two different remainders satisfying the conditions of the theorem.

Sketch of the uniqueness proof

Suppose there are two representations:

\[ A(x)=B(x)Q_1(x)+R_1(x) \]

and:

\[ A(x)=B(x)Q_2(x)+R_2(x) \]

with \(\deg R_1(x)<\deg B(x)\) and \(\deg R_2(x)<\deg B(x)\). Subtracting one from the other gives:

\[ B(x)(Q_1(x)-Q_2(x))=R_2(x)-R_1(x) \]

If \(Q_1(x)\neq Q_2(x)\), the left-hand side would have degree at least \(\deg B(x)\), while the right-hand side would have degree less than \(\deg B(x)\) — a contradiction. Therefore \(Q_1(x)=Q_2(x)\), and consequently \(R_1(x)=R_2(x)\).

Divisibility of Polynomials

Division provides a rigorous foundation for the notion of polynomial divisibility. We say that \(B(x)\) divides \(A(x)\), written \(B(x)\mid A(x)\), if there exists a polynomial \(Q(x)\) such that:

\[ A(x)=B(x)Q(x) \]

In terms of division:

\[ B(x)\mid A(x) \quad \iff \quad R(x)=0 \]

When the remainder is zero, the division is called exact. In this case, \(B(x)\) is a factor of \(A(x)\), and \(A(x)\) is a multiple of \(B(x)\).

The Remainder Theorem

Remainder Theorem. When a polynomial \(P(x)\) is divided by \(x-a\), the remainder equals \(P(a)\).

Indeed, by the Euclidean Division Theorem, we can write:

\[ P(x)=(x-a)Q(x)+R(x) \]

Since the divisor \(x-a\) has degree \(1\), the remainder must be a constant, which we denote \(r\):

\[ P(x)=(x-a)Q(x)+r \]

Substituting \(x=a\) yields:

\[ P(a)=(a-a)Q(a)+r=r \]

Therefore the remainder when dividing by \(x-a\) is \(P(a)\). An immediate consequence is:

\[ x-a\mid P(x) \quad \iff \quad P(a)=0 \]

In other words, \(a\) is a root of \(P(x)\) if and only if \(x-a\) is a factor of \(P(x)\).

Synthetic Division

Synthetic division (also known as Ruffini's rule) is a streamlined method for dividing a polynomial by a linear binomial of the form:

\[ x-a \]

It is not a technique separate from long division: synthetic division is simply a compact notation for the same algorithm, using only the coefficients of the polynomial rather than working with full monomials at every step.

Before applying synthetic division, the polynomial must be written with all powers of \(x\) present in descending order. Any missing term must be represented by a coefficient of \(0\).

For example:

\[ x^3-4x+1=x^3+0x^2-4x+1 \]

The coefficients to use are:

\[ 1,\ 0,\ -4,\ 1 \]

If the divisor is \(x-a\), the number placed on the left of the synthetic division scheme is \(a\). For instance, \(x-3\Rightarrow 3\), while \(x+3=x-(-3)\Rightarrow -3\).

Synthetic Division Step by Step

Divide \( P(x)=x^3-6x^2+11x-6 \) by \( x-1 \).

Since the divisor is \(x-1\), the number on the left of the scheme is \(1\). The polynomial is complete and its coefficients are:

\[ 1,\ -6,\ 11,\ -6 \]

Initial layout

\[ \begin{array}{r|rrr|r} 1 & 1 & -6 & 11 & -6 \\ & & & & \\ \hline & & & & \end{array} \]

First step

Bring down the first coefficient unchanged:

\[ \begin{array}{r|rrr|r} 1 & 1 & -6 & 11 & -6 \\ & & & & \\ \hline & 1 & & & \end{array} \]

Second step

Multiply \(1\cdot 1=1\), write the result below the next coefficient, and add:

\[ -6+1=-5 \]

\[ \begin{array}{r|rrr|r} 1 & 1 & -6 & 11 & -6 \\ & & 1 & & \\ \hline & 1 & -5 & & \end{array} \]

Third step

Multiply \(1\cdot(-5)=-5\), write the result below the next coefficient, and add:

\[ 11+(-5)=6 \]

\[ \begin{array}{r|rrr|r} 1 & 1 & -6 & 11 & -6 \\ & & 1 & -5 & \\ \hline & 1 & -5 & 6 & \end{array} \]

Fourth step

Multiply \(1\cdot 6=6\), write the result below the last coefficient, and add:

\[ -6+6=0 \]

The completed scheme is:

\[ \begin{array}{r|rrr|r} 1 & 1 & -6 & 11 & -6 \\ & & 1 & -5 & 6 \\ \hline & 1 & -5 & 6 & 0 \end{array} \]

Reading the result

The entries in the bottom row, except for the last one, are the coefficients of the quotient polynomial:

\[ Q(x)=x^2-5x+6 \]

The last entry is the remainder:

\[ R=0 \]

Since the remainder is zero, the division is exact:

\[ x^3-6x^2+11x-6=(x-1)(x^2-5x+6) \]

Common Mistakes

1. Omitting missing terms

When using long division or synthetic division, the polynomial must include all powers of \(x\). Any missing power must be represented with a coefficient of \(0\).

2. Getting the sign wrong in the divisor

If the divisor is \(x-a\), the number used in synthetic division is \(a\). Therefore \(x-3\Rightarrow 3\), while \(x+3=x-(-3)\Rightarrow -3\).

3. Stopping too early

In long division, the process must continue until the remainder has degree strictly less than that of the divisor. If the remainder still has degree greater than or equal to the divisor, the division is not yet finished.

4. Confusing the remainder with a coefficient of the quotient

In the synthetic division scheme, the last entry in the bottom row is the remainder, not a coefficient of the quotient. This is why it is helpful to mark it off with a vertical line.

5. Applying synthetic division to unsuitable divisors

In its standard form, synthetic division applies directly only to divisors of the form \(x-a\). For divisors such as \(x^2+1\), long division must be used instead.

Exercises with Solutions

Exercise 1. Divide \(x^2+5x+6\) by \(x+2\).

Solution. Since \(x+2=x-(-2)\), we apply synthetic division with \(-2\):

\[ \begin{array}{r|rr|r} -2 & 1 & 5 & 6 \\ & & -2 & -6 \\ \hline & 1 & 3 & 0 \end{array} \]

Therefore \(Q(x)=x+3\) and \(R=0\). Hence \(x^2+5x+6=(x+2)(x+3)\).

Exercise 2. Divide \(x^2+1\) by \(x-1\).

Solution. Write the polynomial in complete form: \(x^2+1=x^2+0x+1\). Apply synthetic division with \(1\):

\[ \begin{array}{r|rr|r} 1 & 1 & 0 & 1 \\ & & 1 & 1 \\ \hline & 1 & 1 & 2 \end{array} \]

Therefore \(Q(x)=x+1\) and \(R=2\). Indeed, \(x^2+1=(x-1)(x+1)+2\).

Exercise 3. Use the Remainder Theorem to find the remainder when \(P(x)=x^3-4x+7\) is divided by \(x-2\).

Solution. The remainder equals \(P(2)\). Writing \(P(x)=x^3+0x^2-4x+7\):

\[ P(2)=2^3+0\cdot 2^2-4\cdot 2+7=8-8+7=7 \]

The remainder is \(7\).

Exercise 4. Determine whether \(x-3\) divides \(P(x)=x^3-6x^2+11x-6\).

Solution. By the Remainder Theorem, \(x-3\) divides \(P(x)\) if and only if \(P(3)=0\). Computing:

\[ P(3)=27-54+33-6=0 \]

Therefore \(x-3\mid P(x)\).

Exercise 5. Divide \(2x^3-x^2+4x-3\) by \(x+1\).

Solution. Since \(x+1=x-(-1)\), we apply synthetic division with \(-1\):

\[ \begin{array}{r|rrr|r} -1 & 2 & -1 & 4 & -3 \\ & & -2 & 3 & -7 \\ \hline & 2 & -3 & 7 & -10 \end{array} \]

Therefore \(Q(x)=2x^2-3x+7\) and \(R=-10\). Hence:

\[ 2x^3-x^2+4x-3=(x+1)(2x^2-3x+7)-10 \]

Exercise 6. Find the value of \(k\) such that \(x-2\) divides \(P(x)=x^3+kx^2-4x+4\).

Solution. We need \(P(2)=0\). Computing:

\[ P(2)=8+4k-8+4=4k+4 \]

Setting \(4k+4=0\) gives \(k=-1\).

Exercise 7. Divide \(x^4-1\) by \(x^2+1\).

Solution. Since the divisor is not of the form \(x-a\), synthetic division does not apply directly. We use long division instead. Writing the dividend in complete form:

\[ x^4-1=x^4+0x^3+0x^2+0x-1 \]

The computation yields:

\[ Q(x)=x^2-1,\qquad R(x)=0 \]

Therefore:

\[ x^4-1=(x^2+1)(x^2-1) \]

Exercise 8. Divide \(x^3-4x+1\) by \(x+2\) using synthetic division.

Solution. Write the polynomial in complete form:

\[ x^3-4x+1=x^3+0x^2-4x+1 \]

Since \(x+2=x-(-2)\), we apply synthetic division with \(-2\):

\[ \begin{array}{r|rrr|r} -2 & 1 & 0 & -4 & 1 \\ & & -2 & 4 & 0 \\ \hline & 1 & -2 & 0 & 1 \end{array} \]

Therefore \(Q(x)=x^2-2x\) and \(R=1\). Hence:

\[ x^3-4x+1=(x+2)(x^2-2x)+1 \]

Exercise 9. Let \(P(x)\) be a polynomial. Prove that \(x-a\) divides \(P(x)-P(a)\).

Proof. Let \(H(x)=P(x)-P(a)\). By the Remainder Theorem, \(x-a\) divides \(H(x)\) if and only if \(H(a)=0\). But:

\[ H(a)=P(a)-P(a)=0 \]

Therefore:

\[ x-a\mid P(x)-P(a) \]

Conclusion

Polynomial division is far more than a computational routine. Through long division, one sees how a polynomial can be progressively reduced by eliminating its leading terms; through synthetic division, one sees how the same procedure can be carried out more efficiently when the divisor is of the form \(x-a\).

The Euclidean Division Theorem guarantees that the quotient and remainder always exist and are unique. The Remainder Theorem then shows that, when dividing by \(x-a\), the remainder is simply \(P(a)\) — and from this springs the fundamental connection between roots, factors, and divisibility.

For this reason, a solid understanding of polynomial division is an understanding of one of the central mechanisms of algebra: the ability to decompose, analyze, and reconstruct polynomials from their internal structure.