The Remainder Theorem

Long division of polynomials is sometimes unavoidable, but for linear divisors there is a stunning shortcut. The Remainder Theorem says: when you divide $p(x)$ by $(x - a)$, the remainder is just $p(a)$. One substitution; no division. This single theorem powers half the chapter.

Definition

If $p(x)$ is a polynomial of degree $\ge 1$ and $a$ is any real number, then dividing $p(x)$ by the linear polynomial $(x - a)$ gives a unique quotient $q(x)$ and a unique constant remainder $r$ such that $p(x) = (x - a) \cdot q(x) + r.$ The number $r$ is the remainder.

Statement, proof, and uses

Remainder Theorem. Let $p(x)$ be a polynomial of degree $\ge 1$ and let $a$ be a real number. The remainder when $p(x)$ is divided by $(x - a)$ is $p(a)$.

Why it works. Start with the division statement $p(x) = (x - a) \, q(x) + r.$ Substitute $x = a$: $p(a) = (a - a) \, q(a) + r = 0 + r = r.$ So $r = p(a)$. The whole proof is one substitution. Notice we did not need to know $q(x)$ at all , we just needed the form of the division.

Use 1: compute a remainder without doing the division. What is the remainder when $p(x) = x^3 - 4x + 1$ is divided by $x - 2$? Answer: $p(2) = 8 - 8 + 1 = 1$. No long division required.

Use 2: check divisibility. $(x - a)$ divides $p(x)$ exactly when the remainder is zero, i.e. when $p(a) = 0$. This is the Factor Theorem , the next lesson. Already we can spot divisibility instantly.

Use 3: find unknown coefficients. Suppose the polynomial $p(x) = x^3 + kx - 6$ leaves remainder $2$ when divided by $(x - 1)$. Then $p(1) = 1 + k - 6 = k - 5 = 2$, so $k = 7$.

Important: write the divisor as $(x - a)$. If the divisor is $(x + 3)$, write it as $(x - (-3))$, so $a = -3$. The remainder is then $p(-3)$. Similarly for $(x - 5)$, the value $a = 5$.

Divisor of the form $(ax + b)$. Here the linear factor is not exactly $(x - a)$, but we can rewrite: $ax + b = a\!\left(x + \dfrac{b}{a}\right) = a\!\left(x - \left(-\dfrac{b}{a}\right)\right)$. So the "zero" of the divisor is $x = -\dfrac{b}{a}$, and the remainder when $p(x)$ is divided by $(ax + b)$ is $p\!\left(-\dfrac{b}{a}\right)$.

Example with $(2x - 3)$. Find the remainder when $p(x) = 2x^2 - 5x + 1$ is divided by $(2x - 3)$. The zero of the divisor is $x = \dfrac{3}{2}$. Hence the remainder is $p(\tfrac{3}{2}) = 2 \cdot \tfrac{9}{4} - 5 \cdot \tfrac{3}{2} + 1 = \tfrac{9}{2} - \tfrac{15}{2} + 1 = -3 + 1 = -2$.

Worked examples

Example 1. Find the remainder when $p(x) = x^3 - 6x^2 + 11x - 6$ is divided by $(x - 1)$.

$p(1) = 1 - 6 + 11 - 6 = 0$. Remainder is $0$ , and as a bonus, $(x-1)$ is a factor.

Example 2. Find the remainder when $p(x) = x^4 - 3x^2 + 4$ is divided by $(x + 2)$.

$a = -2$. $p(-2) = 16 - 12 + 4 = 8$. Remainder is $8$.

Example 3. If $p(x) = ax^3 + 3x^2 - 13$ leaves remainder $-3$ on division by $(x - 2)$, find $a$.

$p(2) = 8a + 12 - 13 = 8a - 1 = -3$. So $8a = -2$, $a = -\tfrac{1}{4}$.

Example 4. Find the remainder of $p(x) = 4x^3 - 12x^2 + 14x - 3$ on division by $(2x - 1)$.

Zero of divisor: $x = \tfrac{1}{2}$. $p(\tfrac{1}{2}) = 4 \cdot \tfrac{1}{8} - 12 \cdot \tfrac{1}{4} + 14 \cdot \tfrac{1}{2} - 3 = \tfrac{1}{2} - 3 + 7 - 3 = \tfrac{3}{2}$.

Example 5. Without long division, show that $(x - 3)$ is a factor of $p(x) = x^3 - 3x^2 - x + 3$.

$p(3) = 27 - 27 - 3 + 3 = 0$. By the Remainder Theorem the remainder is $0$, so $(x-3)$ divides $p(x)$ exactly.

Try it yourself

1. Find the remainder when $x^3 - 2x^2 + 3x - 1$ is divided by $(x - 1)$.
2. Find the remainder when $2x^3 + x^2 - 5x + 6$ is divided by $(x + 2)$.
3. Find the remainder when $x^4 - x^3 + 4x - 7$ is divided by $(x - 2)$.
4. If $(x - 1)$ leaves remainder $5$ on dividing $p(x) = x^3 + kx + 1$, find $k$.
5. Without dividing, decide whether $(x + 1)$ divides $x^3 + x^2 + x + 1$.
6. Find the remainder when $4x^3 - 8x^2 + x + 1$ is divided by $(2x - 1)$.
7. If $(x - 3)$ leaves remainder $0$ on dividing $p(x) = x^3 - 4x^2 + 3x + k$, find $k$.
8. Is $(x - 2)$ a factor of $x^4 - 3x^3 + 2x + 4$? Justify using the Remainder Theorem.
9. Find the remainder when $x^5 + 5$ is divided by $(x + 1)$.
10. State the Remainder Theorem in your own words.

Pitfalls / Insight

• Get the sign right. Divisor $(x + 3)$ means $a = -3$, not $a = 3$. Be careful.
• The theorem assumes the divisor has degree $1$. It does not work for $(x^2 - 1)$ in this form.
• A remainder is always a constant (degree $0$) here. The quotient may be high degree, but the remainder cannot be $x + 2$ when the divisor is linear.

Insight. The Remainder Theorem turns "polynomial division by a linear factor" into "evaluate the polynomial at one point". This trick is the secret ingredient in every factorisation that follows. Master substitution; you will not need long division for most problems in this chapter.