Remainder Theorem
The remainder theorem says that when a polynomial \(P(x)\) is divided by \(x-a\), the remainder is \(P(a)\). So dividing \(x^3+2x^2-7\) by \(x-2\) leaves a remainder of \(P(2)=9\), with no long division needed. In NSW Year 10 Advanced maths the theorem is used to find remainders and unknown coefficients (NESA outcome MA5‑POL‑P‑01).
You will find remainders for divisors such as \(x+3\) and \(3x-1\), including for very high powers of \(x\). You will find one or two unknown coefficients from given remainders, solve problems where two remainders are equal or one is twice the other, and find a remainder when a term is added to the polynomial. Harder questions find three coefficients from three remainders. It is part of Year 10 Advanced Logarithms and Polynomials.
Theory
The remainder theorem: when a polynomial \(P(x)\) is divided by \((x - a)\), the remainder is \(P(a)\). This Year 10 Advanced maths page shows how to find remainders without long division, handle divisors like \(2x - 1\), find unknown coefficients and find the remainder on division by a quadratic.
Dividing \(P(x)\) by \((x - a)\) gives \(P(x) = (x - a)\,Q(x) + R\), where \(R\) is a number because the divisor is linear.
Substitute \(x = a\): the first term becomes \(0 \times Q(a) = 0\), so \(P(a) = R\). This is the remainder theorem.
So the remainder is found by substituting the value of \(x\) that makes the divisor zero. For \((x + 4)\) substitute \(-4\); for \((2x - 3)\) substitute \(\dfrac{3}{2}\); dividing by \(x\) itself leaves \(P(0)\), the constant term.
Long division gives the quotient as well; the remainder theorem gives only the remainder, but much faster.
This comes up if you are heading for Advanced or Extension maths in Years 11 and 12.
How to use the remainder theorem
- Find the value: set the divisor equal to zero and solve for \(x\).
- Substitute that value into \(P(x)\), with brackets around negatives and fractions.
- Simplify: the result is the remainder.
Problems with unknowns
- One unknown: set \(P(a)\) equal to the given remainder and solve.
- Two unknowns: use two remainders to get two simultaneous equations. Dividing by \(x\) gives \(P(0)\), which is often one unknown on its own.
- Three unknowns: three remainders give three equations; start with \(P(0)\), then add and subtract the others.
- Comparing remainders: write each remainder in terms of the unknown and form the equation the question describes (equal, twice, and so on).
- Quadratic divisor: the remainder is \(Ax + B\). Substitute the two zeros of the divisor to find \(A\) and \(B\) (Example 4).
(i) \(x^4 - 2x^3 + 5x - 1\) is divided by \((x + 2)\)
(ii) \(9x^3 - 3x^2 + 2x + 4\) is divided by \((3x - 1)\)
(iii) \((x - 3)^4 + x\) is divided by \((x - 2)\)
(i) \(x + 2 = x - (-2)\), so substitute \(x = -2\).
| \(P(-2)\) | \(=\) | \((-2)^4 - 2(-2)^3 + 5(-2) - 1\) |
| \(\) | \(=\) | \(16 + 16 - 10 - 1\) |
| \(\) | \(=\) | \(\mathbf{21}\) |
(ii) \(3x - 1 = 0\) when \(x = \dfrac{1}{3}\).
| \(P\left(\genfrac{}{}{}{1}{1}{3}\right)\) | \(=\) | \(9\left(\genfrac{}{}{}{1}{1}{27}\right) - 3\left(\genfrac{}{}{}{1}{1}{9}\right) + 2\left(\genfrac{}{}{}{1}{1}{3}\right) + 4\) |
| \(\) | \(=\) | \(\genfrac{}{}{}{1}{1}{3} - \genfrac{}{}{}{1}{1}{3} + \genfrac{}{}{}{1}{2}{3} + 4\) |
| \(\) | \(=\) | \(\mathbf{4\genfrac{}{}{}{1}{2}{3}}\) |
(iii) There is no need to expand. Substitute \(x = 2\) straight in.
| \(P(2)\) | \(=\) | \((2 - 3)^4 + 2\) |
| \(\) | \(=\) | \(1 + 2 = \mathbf{3}\) |
(i) When \(x^3 + kx^2 - 2x + 7\) is divided by \((x - 2)\) the remainder is \(19\). Find \(k\).
(ii) \(P(x) = x^3 + ax^2 + bx - 2\) leaves remainder \(4\) when divided by \((x - 1)\) and remainder \(-14\) when divided by \((x + 2)\). Find \(a\) and \(b\).
(i) The remainder is \(P(2)\).
| \(8 + 4k - 4 + 7\) | \(=\) | \(19\) |
| \(4k\) | \(=\) | \(8\) |
| \(k\) | \(=\) | \(\mathbf{2}\) |
(ii) Two remainders give two equations.
| \(P(1) = 1 + a + b - 2\) | \(=\) | \(4 \;\Rightarrow\; a + b = 5\) |
| \(P(-2) = -8 + 4a - 2b - 2\) | \(=\) | \(-14 \;\Rightarrow\; 2a - b = -2\) |
Add the two equations to remove \(b\).
| \(3a\) | \(=\) | \(3 \;\Rightarrow\; a = \mathbf{1}\) |
| \(b\) | \(=\) | \(5 - 1 = \mathbf{4}\) |
Check: \(P(-2) = -8 + 4 - 8 - 2 = -14\).
Write each remainder in terms of \(a\), then set them equal.
| \(P(1)\) | \(=\) | \(2 + a - 1 + 3 = a + 4\) |
| \(P(-2)\) | \(=\) | \(-16 + 4a + 2 + 3 = 4a - 11\) |
| \(a + 4\) | \(=\) | \(4a - 11\) |
| \(15\) | \(=\) | \(3a\) |
| \(a\) | \(=\) | \(\mathbf{5}\) |
Check: both remainders are \(9\). The same idea works for any condition, such as one remainder being twice the other: write \(P(c) = 2P(d)\).
(i) Find the remainder when \(P(x)\) is divided by \((x - 2)(x + 3)\).
(ii) Find the remainder when \(P(x) - 2x^2\) is divided by \((x - 2)\).
(i) The divisor is a quadratic, so the remainder has degree at most \(1\): call it \(Ax + B\).
| \(P(x)\) | \(=\) | \((x - 2)(x + 3)\,Q(x) + Ax + B\) |
Substitute the values that make the quadratic zero, so \(Q(x)\) disappears.
| \(P(2) = 2A + B\) | \(=\) | \(7\) |
| \(P(-3) = -3A + B\) | \(=\) | \(-8\) |
| \(5A\) | \(=\) | \(15 \;\Rightarrow\; A = 3,\ B = 1\) |
The remainder is \(\boldsymbol{3x + 1}\).
(ii) Let \(S(x) = P(x) - 2x^2\). Its remainder is \(S(2) = P(2) - 2(2)^2 = 7 - 8 = \mathbf{-1}\).
Common pitfalls
Frequently asked questions
What is the remainder theorem?
When a polynomial \(P(x)\) is divided by \(x - a\), the remainder equals \(P(a)\). You find it by substituting \(x = a\) instead of doing long division.
Why does the remainder theorem work?
\(P(x) = (x - a)Q(x) + R\). Putting \(x = a\) makes the first part zero, so \(P(a) = R\).
How do you find the remainder when dividing by x + 3?
Write \(x + 3\) as \(x - (-3)\) and substitute \(x = -3\) into the polynomial.
How do you use the remainder theorem with 2x - 1?
Set \(2x - 1 = 0\), so \(x = \dfrac{1}{2}\), and substitute \(x = \dfrac{1}{2}\). The answer may be a fraction.
How do you find unknown coefficients using remainders?
Each remainder gives an equation: substitute the value that makes the divisor zero and set the result equal to the remainder. Two unknowns need two remainders, solved simultaneously.