Remainder Theorem

When P(x) is divided by x−k, the remainder is the single value P(k).

Zeros connect polynomial factors to x-intercepts. A smooth curve crosses the horizontal axis at three highlighted and labelled zeros, showing how algebraic roots become geometric x-intercepts.

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

$$ P(k). $$

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

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

$$ P(-4)=16-12-2=2, $$

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:

R=P\left(\frac ba\right).

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

$$ P(2)=11. $$

$$ 8+2m+5=11,qquad2m=-2,qquad m=-1. $$

Remainder conditions can therefore determine unknown parameters without full division.

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.