Deriving the Remainder Theorem

Why a Remainder Can Be Found by Evaluation

A polynomial division can take several lines of algebra. But when the divisor is linear, finding the remainder does not require finding the quotient first. The key is that the remainder has a very simple form: it is a constant. Evaluating the division identity at one carefully chosen input then reveals that constant immediately.

This result is useful when a problem asks only for a remainder, or when the quotient would be lengthy. To use it correctly, identify the number that makes the divisor zero. For a divisor \(x-a\), that number is \(a\), so the relevant value is \(p(a)\).

Polynomial Division by a Linear Polynomial

Let \(p(x)\) be a polynomial with real coefficients, and let \(a\) be a real number. Polynomial division gives polynomials \(q(x)\) and \(r(x)\) such that

where \(r(x)\) is either zero or has degree less than the degree of the divisor. Since \(x-a\) has degree \(1\), the remainder must have degree less than \(1\). A nonzero polynomial of degree less than \(1\) is a constant. Thus \(r(x)=r\) for some real number \(r\).

Polynomial division guarantees that such a quotient and remainder exist and are unique. For a linear divisor, existence can be seen by repeatedly cancelling the leading term: if the current polynomial has degree \(m\geq 1\) and leading coefficient \(c\), subtract \(cx^{m-1}(x-a)\). This removes its degree-\(m\) term. Repeating lowers the degree until only a constant, or zero, remains. The accumulated terms form the quotient.

For uniqueness, suppose there are two division identities with quotients \(q_1,q_2\) and constant remainders \(r_1,r_2\). Subtracting them gives

If \(q_1-q_2\) were nonzero, the left side would have degree at least \(1\), while the right side is constant or zero. That is impossible. Therefore \(q_1=q_2\), and the equation then gives \(r_1=r_2\).

Deriving the Remainder Theorem

Once the division identity is written down, the central step is to evaluate it at \(x=a\). This is allowed because polynomials can be evaluated at every real input. Since \(a-a=0\), the quotient term disappears:

The theorem depends on using the divisor \(x-a\), or a nonzero constant multiple of it. Its essential feature is that the divisor becomes zero at \(x=a\). Substituting that input into the division identity removes the quotient term and leaves just the remainder.

Worked Examples

A Common Sign Error

The most frequent mistake is to use the wrong input when the divisor is written with a plus sign. For division by \(x+4\), the divisor is \(x-(-4)\), so the relevant input is \(-4\). It is not \(4\).

For example, let \(p(x)=x^2+3x+1\). A wrong calculation would substitute \(4\) and report \(p(4)=16+12+1=29\). The correction is to solve \(x+4=0\), which gives \(x=-4\):

Indeed, the division identity is

so the remainder is \(5\). A reliable habit is to set the divisor equal to zero before substituting into the polynomial.

Using the Theorem Efficiently

When asked for a remainder on division by a linear polynomial, use this short procedure:

The theorem does not say that the quotient equals \(p(a)\), nor that the remainder is the polynomial’s value at an arbitrary input. It specifically identifies the remainder with the value at the zero of the linear divisor. If the divisor is \(2x-6\), for instance, the input is \(3\), not \(6\).

The idea behind the result is also a useful way to check the answer: in the division identity, substituting the divisor’s zero makes the entire quotient term vanish. What remains must be the remainder. This reasoning applies even when the quotient is complicated, because its value is multiplied by zero.

Summary

Dividing a polynomial by \(x-a\) leaves a constant remainder because the divisor has degree \(1\). Writing \(p(x)=(x-a)q(x)+r\) and evaluating at \(x=a\) gives \(p(a)=r\). Therefore, evaluate the polynomial at the zero of the divisor to find the remainder. A nonzero constant multiple of \(x-a\) has the same zero and the same remainder, although its quotient is scaled.