The Logic: Division Always Has a Pattern#
When you divide $10$ by $3$, you get $3$ remainder $1$. We can write this as:
$$10 = 3 \times 3 + 1$$Polynomials follow the exact same logic. When you divide a polynomial $f(x)$ by $(x - c)$, there’s always a quotient $Q(x)$ and a remainder $R$:
$$f(x) = (x - c) \cdot Q(x) + R$$This equation is always true — for every value of $x$.
1. The Remainder Theorem#
The Theorem#
If a polynomial $f(x)$ is divided by $(x - c)$, the remainder is $f(c)$.
Why It Works#
Start from the division identity:
$$f(x) = (x - c) \cdot Q(x) + R$$Now substitute $x = c$:
$$f(c) = (c - c) \cdot Q(c) + R = 0 \cdot Q(c) + R = R$$The entire first term vanishes, leaving just the remainder. That’s the whole proof — one substitution.
Key insight: You don’t need to do any division at all to find the remainder. Just plug in the value.
Worked Example 1 — Finding a Remainder#
Find the remainder when $f(x) = 2x^3 - 5x^2 + 3x - 7$ is divided by $(x - 2)$.
Simply calculate $f(2)$:
$$f(2) = 2(8) - 5(4) + 3(2) - 7 = 16 - 20 + 6 - 7 = -5$$$$\boxed{\text{Remainder} = -5}$$No long division needed.
Worked Example 2 — Division by $(x + 3)$#
Find the remainder when $f(x) = x^3 + 2x^2 - x + 4$ is divided by $(x + 3)$.
Be careful with the sign: $(x + 3) = (x - (-3))$, so $c = -3$.
$$f(-3) = (-3)^3 + 2(-3)^2 - (-3) + 4 = -27 + 18 + 3 + 4 = -2$$$$\boxed{\text{Remainder} = -2}$$2. The Factor Theorem#
The Theorem#
$(x - c)$ is a factor of $f(x)$ if and only if $f(c) = 0$.
This is just a special case of the Remainder Theorem: if the remainder is zero, the division is exact — meaning $(x - c)$ divides evenly into $f(x)$.
The Chain of Logic#
$$f(c) = 0 \quad \Leftrightarrow \quad \text{Remainder} = 0 \quad \Leftrightarrow \quad (x - c) \text{ is a factor} \quad \Leftrightarrow \quad x = c \text{ is a root/x-intercept}$$These four statements are all saying the same thing.
Worked Example 3 — Testing for Factors#
$$f(1) = 1 - 6 + 11 - 6 = 0\;\checkmark$$Is $(x - 1)$ a factor of $f(x) = x^3 - 6x^2 + 11x - 6$?
Yes, $(x - 1)$ is a factor.
$$f(-2) = (-8) - 6(4) + 11(-2) - 6 = -8 - 24 - 22 - 6 = -60 \neq 0$$Is $(x + 2)$ a factor?
No, $(x + 2)$ is not a factor.
Worked Example 4 — Testing Multiple Values#
Find a factor of $f(x) = 2x^3 + x^2 - 13x + 6$.
Test systematically:
| $c$ | $f(c)$ | Factor? |
|---|---|---|
| $1$ | $2 + 1 - 13 + 6 = -4$ | No |
| $-1$ | $-2 + 1 + 13 + 6 = 18$ | No |
| $2$ | $16 + 4 - 26 + 6 = 0$ | Yes! |
$(x - 2)$ is a factor of $f(x)$.
Which values to try? The Rational Root Theorem says: try $\pm$ (factors of the constant term) $\div$ (factors of the leading coefficient). For $2x^3 + \dots + 6$: try $\pm 1, \pm 2, \pm 3, \pm 6, \pm \frac{1}{2}, \pm \frac{3}{2}$.
3. Finding Unknown Coefficients#
A classic exam question: the polynomial has an unknown coefficient, and you’re told that a specific binomial is a factor (or gives a specific remainder).
Worked Example 5 — Finding $k$ When Given a Factor#
$(x - 3)$ is a factor of $f(x) = x^3 + kx^2 - 2x - 12$. Find $k$.
Since $(x - 3)$ is a factor, $f(3) = 0$:
$$27 + 9k - 6 - 12 = 0$$$$9k + 9 = 0$$$$\boxed{k = -1}$$Worked Example 6 — Finding $k$ When Given a Remainder#
When $f(x) = 2x^3 - x^2 + kx + 5$ is divided by $(x + 1)$, the remainder is $3$. Find $k$.
$f(-1) = 3$:
$$2(-1)^3 - (-1)^2 + k(-1) + 5 = 3$$$$-2 - 1 - k + 5 = 3$$$$2 - k = 3$$$$\boxed{k = -1}$$Worked Example 7 — Two Unknowns#
$(x - 1)$ and $(x + 2)$ are both factors of $f(x) = x^3 + ax^2 + bx - 6$. Find $a$ and $b$.
$f(1) = 0$: $1 + a + b - 6 = 0 \Rightarrow a + b = 5\;\dots(1)$
$f(-2) = 0$: $-8 + 4a - 2b - 6 = 0 \Rightarrow 4a - 2b = 14 \Rightarrow 2a - b = 7\;\dots(2)$
Add (1) and (2): $3a = 12 \Rightarrow a = 4$
From (1): $b = 5 - 4 = 1$
$$\boxed{a = 4, \quad b = 1}$$Check: $f(x) = x^3 + 4x^2 + x - 6$. $f(1) = 1 + 4 + 1 - 6 = 0\;\checkmark$. $f(-2) = -8 + 16 - 2 - 6 = 0\;\checkmark$.
4. Connection Between Theorems and Graphs#
| Algebraic statement | Graphical meaning |
|---|---|
| $f(c) = 0$ | The graph of $f$ crosses or touches the $x$-axis at $x = c$ |
| $(x - c)$ is a factor | $x = c$ is an $x$-intercept |
| $(x - c)^2$ is a factor | $x = c$ is a turning point on the $x$-axis (graph touches, doesn’t cross) |
| $f(c) = 5$ | The point $(c; 5)$ is on the graph |
This connection is critical for Graphing Cubic Functions — you use the Factor Theorem to find the $x$-intercepts, then calculus to find the turning points.
5. Summary: When to Use Which Theorem#
| Question asks… | Use… | Method |
|---|---|---|
| “Find the remainder when…” | Remainder Theorem | Calculate $f(c)$ |
| “Is $(x - c)$ a factor?” | Factor Theorem | Check if $f(c) = 0$ |
| “Find a factor of $f(x)$” | Factor Theorem | Test $c = \pm 1, \pm 2, \dots$ until $f(c) = 0$ |
| “Find $k$ if $(x - c)$ is a factor” | Factor Theorem | Set $f(c) = 0$ and solve for $k$ |
| “Find $k$ if the remainder is $R$” | Remainder Theorem | Set $f(c) = R$ and solve for $k$ |
| “Factorise $f(x)$ completely” | Factor Theorem + Division | Find one factor, divide, then factorise the quotient |
For the full factorisation method (finding all three roots of a cubic), see Solving Cubics.
🚨 Common Mistakes#
| Mistake | Why it’s wrong | Fix |
|---|---|---|
| Sign error with $(x + c)$ | $(x + 3)$ means $c = -3$, not $c = 3$ | Always rewrite as $(x - (-3))$ first |
| Substitution arithmetic | One wrong sign ruins the whole answer | Write out every term separately before adding |
| Confusing remainder with factor | Remainder $= 5$ does NOT mean it’s a factor | Factor means remainder $= 0$ specifically |
| Not testing enough values | Giving up after $\pm 1$ don’t work | Try $\pm 2, \pm 3$, and fractions if the leading coefficient $\neq 1$ |
| Forgetting to state the reason | “$(x - 2)$ is a factor” needs justification | Write: “$f(2) = 0$, $\therefore (x - 2)$ is a factor (Factor Theorem)” |
💡 Pro Tips for Exams#
1. Start with $\pm 1$#
These are the easiest values to substitute — you can often do them mentally. If neither works, try $\pm 2$, then $\pm 3$.
2. The “Sum of Coefficients” Shortcut#
$f(1) =$ the sum of all coefficients. For $x^3 - 6x^2 + 11x - 6$: $1 - 6 + 11 - 6 = 0$. If the coefficients sum to zero, $(x - 1)$ is always a factor.
Similarly, $f(-1)$ alternates signs: change the signs of the odd-power coefficients and add. If this gives zero, $(x + 1)$ is a factor.
3. Two Unknowns = Two Equations#
If you need to find two unknown coefficients, you need two pieces of information (two factors, or one factor and one remainder). Set up simultaneous equations from $f(c_1) = 0$ and $f(c_2) = 0$.