After a few synthetic divisions you may notice something: the remainder from dividing by $x - r$ is always $p(r)$. That is not a coincidence, and it gives two powerful shortcuts.
Key idea
Remainder theorem: when $p(x)$ is divided by $x - r$, the remainder is $p(r)$. Factor theorem: $x - r$ is a factor of $p(x)$ exactly when $p(r) = 0$.
Why it works
Division always gives $p(x) = (x - r)\,q(x) + R$, where the remainder $R$ is a number because it has lower degree than $x - r$. This is true for every $x$, so put in $x = r$: $p(r) = (r - r)\,q(r) + R = 0 + R = R$. If that remainder is 0, then $p(x) = (x - r)\,q(x)$, which says exactly that $x - r$ is a factor. Graphically, $p(r) = 0$ means the graph crosses or touches the $x$-axis at $r$.
Worked examples
Worked example: A remainder without dividing
Find the remainder when $p(x) = x^4 - 3x^3 + 2x - 5$ is divided by $x - 2$.
- By the remainder theorem, it is $p(2)$.
- $p(2) = 16 - 24 + 4 - 5 = -9$.
Answer: $-9$
Worked example: Testing a factor
Is $x + 1$ a factor of $x^3 + 2x^2 - x - 2$?
- $x + 1 = x - (-1)$, so compute $p(-1) = -1 + 2 + 1 - 2 = 0$.
- It is 0, so yes.
Answer: Yes: $x + 1$ is a factor.
Worked example: Finding a missing coefficient
Find $k$ so that $x - 2$ is a factor of $x^3 + kx^2 - 4x + 4$.
- Set $p(2) = 0$: $8 + 4k - 8 + 4 = 0$.
- $4k + 4 = 0$, so $k = -1$.
Answer: $k = -1$
Common mistake
Evaluating at the wrong sign: for the divisor $x + 1$ you need $p(-1)$, not $p(1)$. The number to use is the one that makes the divisor 0.