Remainder Theorem

Polynomial evaluation can replace a full long-division calculation when only the remainder is needed.

The theorem

If a polynomial $P(x)$ is divided by $x-k$, then the remainder is

Because the divisor has degree $1$, the remainder must be a constant. One substitution determines that constant exactly.

Why substitution reveals the remainder

Polynomial division gives

Setting $x=k$ makes the divisor term vanish:

This short proof also explains the factor theorem: the divisor is a factor precisely when $R=P(k)=0$.

Worked example: find a remainder

The division statement would have form $P(x)=(x-2)Q(x)+13$.

Handling a plus sign

If the divisor is $x+4$, rewrite it as $x-(-4)$. The required input is $k=-4$.

For $P(x)=x^2+3x-2$,

so division by $x+4$ leaves remainder $2$.

Divisors of the form ax−b

For a divisor $ax-b$, its zero is $x=b/a$. Substitute that value to obtain the remainder:

For example, division by $2x-3$ uses $x=3/2$. The value of the divisor becomes zero there, which is the key idea—not merely reading the constant with a sign change.

Finding an unknown coefficient

Suppose $P(x)=x^3+mx+5$ leaves remainder $11$ when divided by $x-2$. Then

So

Remainder conditions can therefore determine unknown parameters without full division.

Several remainder conditions

If a polynomial contains multiple unknown coefficients, each remainder statement creates an equation. For instance, remainders at $x=1$ and $x=-2$ produce $P(1)=r_1$ and $P(-2)=r_2$. Solve the resulting system for the unknowns.

Keep the input value separate from the stated remainder; they play different roles.

Checking synthetic division

Synthetic division by $k$ should end with $P(k)$. This makes the remainder theorem a fast error check for the division algorithm.

If the final synthetic value disagrees with direct substitution, recheck signs, missing coefficients, and arithmetic.

Graphical interpretation

The value $P(k)$ is the graph's height at $x=k$. If it is zero, the point lies on the $x$-axis and $x-k$ is a factor. If it is positive or negative, that height becomes the remainder.

This is an algebraic property, even though the graph gives it a useful visual meaning.

Limits of the shortcut

The theorem directly supplies the remainder for a linear divisor. If the divisor has degree $2$, the remainder may have degree at most $1$, such as $ax+b$, so one evaluation is not enough to determine it.

More advanced remainder problems use several substitutions or modular polynomial reasoning.

Common mistakes

Substituting the constant from the divisor. Solve the divisor equal to zero to find $k$.

Using $4$ for divisor $x+4$. The correct input is $-4$.

Reporting $P(k)$ as the quotient. It is only the remainder.

Assuming any nonzero remainder still gives a factor. A factor requires remainder zero.

Applying the single-value shortcut unchanged to a quadratic divisor. Its remainder need not be constant.

Quick self-check

• What value makes the divisor zero?
• Did I substitute that value into the complete polynomial?
• Were negative inputs enclosed in parentheses?
• Is the requested quantity the remainder or the quotient?
• Does synthetic division, if used, end with the same value?

Related topics

• Factor Theorem
• Polynomial Functions
• Synthetic Division
• Polynomial Long Division