Polynomial Remainder Theorem

In algebra, the Polynomial Remainder Theorem states that the remainder upon dividing any polynomial $P(x)$ by a linear polynomial $x-a$, both with complex coefficients, is equal to $P(a)$.


We use Euclidean polynomial division with dividend $P(x)$ and divisor $x-a$. The result states that there exists a quotient $Q(x)$ and remainder $R(x)$ such that \[P(x) = (x-a) Q(x) + R(x),\] with $\deg R(x) < \deg (x-a)$. We wish to show that $R(x)$ is equal to the constant $f(a)$. Because $\deg (x-a) = 1$, $\deg R(x) < 1$. Hence, $R(x)$ is a constant, $r$. Plugging this into our original equation and rearranging a bit yields \[r = P(x) - (x-a) Q(x).\] After substituting $x=a$ into this equation, we deduce that $r = P(a)$; thus, the remainder upon diving $P(x)$ by $x-a$ is equal to $P(a)$, as desired. $\square$


The strategy used in the above proof can be generalized to divisors with degree greater than $1$. A more general method, with any dividend $P(x)$ and divisor $D(x)$, is to write $R(x) = D(x) Q(x) - P(x)$, and then substitute the zeroes of $D(x)$ to eliminate $Q(x)$ and find values of $R(x)$. Example 2 showcases this strategy.


Here are some problems with solutions that utilize the Polynomial Remainder Theorem and its generalization.

Example 1

What is the remainder when $x^2+2x+3$ is divided by $x+1$?

Solution: Although one could use long or synthetic division, the Polynomial Remainder Theorem provides a significantly shorter solution. Note that $P(x) = x^2 + 2x + 3$, and $x-a = x+1$. A common mistake is to forget to flip the negative sign and assume $a = 1$, but simplifying the linear equation yields $a = -1$. Thus, the answer is $P(-1)$, or $(-1)^2 + 2(-1) + 3$, which is equal to $2$. $\square$.

More examples

See also

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