The Remainder Theorem
The remainder theorem is the shortest useful theorem in the course: one sentence to state, one line to prove, and it replaces an entire page of long division with a single substitution.
Every division you have done can be written $P(x) = (x-a)Q(x) + R$. There is one value of $x$ that makes the first term on the right disappear completely, no matter what $Q$ is. Which value, and what does the equation then say?
The remainder when $P(x)$ is divided by $x - a$ is $P(a)$. Not approximately, not usually: exactly, always. The proof is to substitute $x = a$ into the division identity and notice that the quotient is multiplied by zero.
$$P(x) \div (x - a) \quad \text{leaves remainder} \quad P(a)$$
The divisor $ax - b$ works the same way, with the substitution $x = \dfrac{b}{a}$: solve divisor $= 0$ every time rather than trying to remember a formula. And read the theorem in reverse: if $P(a) = 0$ then the remainder is zero, which is the factor theorem of the next lesson.
Know
- The statement of the remainder theorem for a divisor $x - a$
- The extension to a divisor $ax - b$, with substitution $x = \tfrac{b}{a}$
- That the theorem gives the remainder only, never the quotient
Understand
- Why the proof is a single substitution into the division identity
- Why the theorem cannot be applied directly to a divisor of degree two or more
Can Do
- Verify the theorem by comparing a long division with a substitution
- Find a remainder by substitution, including for divisors of the form $ax - b$
- Use given remainders to set up and solve equations for unknown coefficients
Before proving anything, watch it happen. Take $P(x) = x^3 - 4x^2 + 5x - 1$ and divide by $x - 3$.
The long way. $x^3 \div x = x^2$; multiply and subtract to leave $-x^2$; bring down $5x$; $-x^2 \div x = -x$; multiply and subtract to leave $2x$; bring down $-1$; $2x \div x = 2$; multiply and subtract to leave $5$. So
$$x^3 - 4x^2 + 5x - 1 = (x-3)(x^2 - x + 2) + 5$$
The short way. Substitute $x = 3$:
$$P(3) = 27 - 36 + 15 - 1 = 5$$
The same $5$, from three lines of arithmetic instead of nine of long division. Try it with a different polynomial and a different divisor and it happens again. That is verification, not proof: it shows the claim survives a test, and no number of tests establishes that it always holds. The next card supplies the argument that does.
Theorem. When a polynomial $P(x)$ is divided by $x - a$, the remainder is $P(a)$.
Proof. By the division identity there are polynomials $Q$ and $R$ with
$$P(x) = (x - a)Q(x) + R(x), \qquad \deg R < \deg(x-a) = 1$$
A polynomial of degree less than $1$ is a constant, so write the remainder as $R$. The identity holds for every value of $x$, so it holds at $x = a$:
$$P(a) = (a - a)Q(a) + R = 0 \times Q(a) + R = R$$
Therefore $R = P(a)$, which is the theorem. $\blacksquare$
Every step is doing something. The degree condition is what lets the remainder be a single constant $R$ rather than an unknown polynomial. The identity being true for all $x$ is what licenses the substitution. And $(a-a) = 0$ is what removes the quotient, which is why the theorem can tell you the remainder while knowing nothing whatsoever about $Q$.
Nothing in the proof needed the divisor's leading coefficient to be $1$. For a divisor $ax - b$, the value that makes it zero is $x = \dfrac{b}{a}$, so
$$\text{remainder} = P\!\left(\frac{b}{a}\right)$$
For example, the remainder when $P(x) = 4x^3 - 2x + 7$ is divided by $2x - 1$ is found by solving $2x - 1 = 0$, giving $x = \tfrac{1}{2}$:
$$P\!\left(\tfrac{1}{2}\right) = 4\left(\tfrac{1}{8}\right) - 2\left(\tfrac{1}{2}\right) + 7 = \tfrac{1}{2} - 1 + 7 = \tfrac{13}{2}$$
The advice worth internalising is not the formula but the habit: write down the divisor, set it equal to zero, and solve. That single step handles $x - 3$, $x + 5$, $2x - 1$ and $3x + 4$ identically, and it is immune to the commonest error in the topic, which is substituting $+3$ when the divisor is $x + 3$.
The theorem is most often used in reverse. A statement about a remainder becomes an equation about a value of $P$, and equations can be solved.
Suppose $P(x) = 2x^3 + kx^2 - x + 6$ leaves a remainder of $4$ when divided by $x + 1$. Then:
$$P(-1) = 4$$
$$2(-1)^3 + k(-1)^2 - (-1) + 6 = 4$$
$$-2 + k + 1 + 6 = 4 \quad \Longrightarrow \quad k = -1$$
Two unknowns need two remainder statements, producing simultaneous equations, exactly as in Lesson 5. The only new thing here is the name: what was "substitute into the identity" is now "apply the remainder theorem", and in a proof question you should cite it by name.
A remainder of zero is the most valuable case of all. It says $P(a) = 0$, and the next lesson turns that into a statement about factors.
Watch Me Solve It · 3 examples
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1Solve divisor = 0$x + 2 = 0 \ \Rightarrow \ x = -2$Never read the value off the divisor's appearance. Solve for it.
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2Substitute, with brackets on every power$P(-2) = 3(-2)^4 - 2(-2)^3 + (-2) - 10$
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3Evaluate the powers first$(-2)^4 = 16, \qquad (-2)^3 = -8$Even power positive, odd power negative.
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4Finish the arithmetic$P(-2) = 48 + 16 - 2 - 10 = 52$The remainder is $52$. Note that $-2(-8) = +16$, which is where signs are usually lost.
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1Solve divisor = 0$2x + 1 = 0 \ \Rightarrow \ x = -\tfrac{1}{2}$The leading coefficient of the divisor changes the value substituted, but nothing else.
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2Substitute$P\!\left(-\tfrac{1}{2}\right) = 8\left(-\tfrac{1}{8}\right) + 4\left(\tfrac{1}{4}\right) - 6\left(-\tfrac{1}{2}\right) + 1$$\left(-\tfrac{1}{2}\right)^3 = -\tfrac{1}{8}$ and $\left(-\tfrac{1}{2}\right)^2 = \tfrac{1}{4}$.
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3Simplify each term$= -1 + 1 + 3 + 1 = 4$The coefficients were chosen to clear the fractions, which is common in exam questions with non-monic divisors.
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4State the answer with its meaningThe remainder is $4$. Since it is not zero, $2x + 1$ is not a factor of $P(x)$.
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1Translate each statement using the theorem$P(1) = -4, \qquad P(2) = 8$Each divisor is linear, so each remainder is a value of $P$.
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2Form the first equation$1 + a + b - 10 = -4$$a + b = 5$
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3Form the second equation$8 + 4a + 2b - 10 = 8$$4a + 2b = 10 \ \Rightarrow \ 2a + b = 5$Divide through by $2$ to keep the numbers small.
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4Solve and check$a = 0, \qquad b = 5$Subtracting the equations gives $a = 0$, then $b = 5$. Check: $P(1) = 1 + 0 + 5 - 10 = -4$ and $P(2) = 8 + 0 + 10 - 10 = 8$. Both hold.
Brain Trainer · 5 problems
Five items using the remainder theorem. Work each one, then reveal the answer.
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1 Find the remainder when $x^3 + 2x - 4$ is divided by $x - 1$.
Substitute $x = 1$: $1 + 2 - 4$.$-1$ -
2 Find the remainder when $x^4 - 3x^2 + 1$ is divided by $x + 1$.
Substitute $x = -1$: $1 - 3 + 1$.$-1$ -
3 What value would you substitute for the divisor $3x - 2$?
Solve $3x - 2 = 0$.$x = \tfrac{2}{3}$ -
4 Find the remainder when $x^{20} + x^{10}$ is divided by $x + 1$.
Both powers are even, so $(-1)^{20} + (-1)^{10} = 1 + 1$.$2$ -
5 $P(x) = x^2 + kx + 3$ leaves remainder $9$ on division by $x - 2$. Find $k$.
$4 + 2k + 3 = 9$.$k = 1$
Multiple Choice · 5 questions
The remainder theorem states that the remainder when $P(x)$ is divided by $x - a$ is:
The remainder when $x^3 - 2x^2 + 4$ is divided by $x + 1$ is:
To find the remainder when $P(x)$ is divided by $4x - 3$, substitute:
Applying the remainder theorem with $x = 5$ to a quartic $P(x)$ tells you:
To find the remainder when $P(x)$ is divided by $x^2 - 9$, the remainder theorem:
Short Answer · 3 questions
(a) Divide $P(x)$ by $x - 2$ using long division, and state the quotient and remainder.
(b) Find $P(2)$ by substitution.
(c) State what (a) and (b) together demonstrate, and explain carefully why this is a verification and not a proof.
(d) Prove the remainder theorem for a general polynomial $P$ and divisor $x - a$.
(a) Given that the remainder on division by $x + 1$ is $6$, form an equation in $p$ and $q$.
(b) Given also that the remainder on division by $x - 2$ is $30$, form a second equation.
(c) Find $p$ and $q$.
(d) With those values, find the remainder on division by $2x - 1$.
(b) $P(x)$ leaves remainder $5$ on division by $x - 3$. Find the remainder when $P(x) + 2x$ is divided by $x - 3$, with reasons.
(c) $P(x)$ and $S(x)$ leave remainders $4$ and $-1$ respectively on division by $x - 1$. Find the remainder when $P(x)S(x)$ is divided by $x - 1$.
(d) A student claims that if $P(x)$ leaves remainder $0$ on division by $x - 2$ and also on division by $x - 3$, then it leaves remainder $0$ on division by $x^2 - 5x + 6$. Decide whether the claim is true and justify your decision.
(b) A polynomial $P$ of degree $3$ satisfies $P(1) = P(2) = P(3) = 0$ and $P(0) = -12$. Find $P(x)$ completely.
(c) Explain why a polynomial of degree $n$ that leaves remainder zero on division by $n+1$ different linear polynomials $x - a_1, \ldots, x - a_{n+1}$ must be the zero polynomial.
The theorem
Remainder on division by $x - a$ is $P(a)$
The proof
Substitute $x = a$ into the identity
Non-monic divisors
Solve divisor $= 0$, then substitute
Its limit
Remainder only, and linear divisors only
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