The Remainder and Factor Theorems
Theory
The remainder theorem: when a polynomial \(P(x)\) is divided by \((x - a)\), the remainder is \(P(a)\). The factor theorem: \((x - a)\) is a factor of \(P(x)\) exactly when \(P(a) = 0\). This Year 10A maths page shows how to find remainders without long division, handle divisors like \(2x - 1\), test for factors and find unknown coefficients.
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. The remainder is found by substituting the value of \(x\) that makes the divisor zero. For \((x + 4)\) substitute \(-4\); for \((2x - 3)\) substitute \(\dfrac{3}{2}\).
If \(P(a) = 0\) there is nothing left over, so \((x - a)\) divides exactly: it is a factor. This is the factor theorem, and it works both ways: \((x - a)\) is a factor exactly when \(P(a) = 0\).
A number \(a\) with \(P(a) = 0\) is a zero of the polynomial and a root of the equation \(P(x) = 0\). On a graph, zeros are the \(x\)-intercepts.
Long division gives the quotient as well; the remainder theorem gives only the remainder, but much faster.
This prepares you for Mathematical Methods and Specialist Mathematics in Years 11 and 12.
| Fact about \(P(x)\) | Equation it gives |
|---|---|
| remainder \(7\) on division by \((x - 1)\) | \(P(1) = 7\) |
| remainder \(-2\) on division by \((x + 5)\) | \(P(-5) = -2\) |
| \((x - 3)\) is a factor | \(P(3) = 0\) |
| \((2x + 1)\) is a factor | \(P\left(-\dfrac{1}{2}\right) = 0\) |
How to use the remainder and factor theorems
- Find the value: set the divisor (or possible factor) 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. A result of \(0\) means the divisor is a factor; anything else means it is not.
Problems with unknowns
- Which is a factor? Test each option; only a factor gives \(0\). For an option like \((2x - 5)\), substitute \(x = \dfrac{5}{2}\).
- One unknown: set \(P(a)\) equal to the given remainder (or to \(0\) for a factor) and solve.
- Two unknowns: two remainders, two factors, or one of each give two simultaneous equations. Solve them by elimination.
- Unknown leading coefficient: it is treated like any other unknown: substitute and collect the terms in \(a\) and \(b\).
- Factorising: once a factor \((x - a)\) is found, divide to get the other factor. This is used to solve polynomial equations.
(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) \((x + 2)\) is a factor of \(x^3 + 4x^2 + x - 6\)
(ii) \((3x - 2)\) is a factor of \(3x^3 - 2x^2 + 6x - 4\)
(iii) \((x - 1)\) is a factor of \(x^4 + x^3 - 3x + 2\)
A factor gives a zero remainder, so substitute the value that makes the factor zero.
(i) Substitute \(x = -2\).
| \(P(-2)\) | \(=\) | \(-8 + 16 - 2 - 6 = 0\) |
\(P(-2) = 0\), so \((x + 2)\) is a factor.
(ii) \(3x - 2 = 0\) at \(x = \dfrac{2}{3}\).
| \(P\left(\genfrac{}{}{}{1}{2}{3}\right)\) | \(=\) | \(3\left(\genfrac{}{}{}{1}{8}{27}\right) - 2\left(\genfrac{}{}{}{1}{4}{9}\right) + 6\left(\genfrac{}{}{}{1}{2}{3}\right) - 4\) |
| \(\) | \(=\) | \(\genfrac{}{}{}{1}{8}{9} - \genfrac{}{}{}{1}{8}{9} + 4 - 4 = 0\) |
So \((3x - 2)\) is a factor.
(iii) Substitute \(x = 1\).
| \(P(1)\) | \(=\) | \(1 + 1 - 3 + 2 = 1\) |
\(P(1) = 1 \ne 0\), so \((x - 1)\) is not a factor; it leaves remainder \(1\).
(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: \(P(-2) = -8 + 4 - 8 - 2 = -14\).
(i) Find \(a\) and \(b\).
(ii) Factorise \(P(x)\) fully.
(i) Each factor gives a zero.
| \(P(1)\) | \(=\) | \(1 + a + b - 9 = 0\) |
| \(\) | \(\Rightarrow\) | \(a + b = 8\) |
| \(P(-3)\) | \(=\) | \(-27 + 9a - 3b - 9 = 0\) |
| \(\) | \(\Rightarrow\) | \(3a - b = 12\) |
| \(4a\) | \(=\) | \(20\) |
| \(a\) | \(=\) | \(\boldsymbol{5},\quad b = \boldsymbol{3}\) |
(ii) \(P(x) = x^3 + 5x^2 + 3x - 9\). The product \((x - 1)(x + 3) = x^2 + 2x - 3\) is a factor, so the third factor is linear: \((x + c)\) with \(-3c = -9\), so \(c = 3\).
| \(P(x)\) | \(=\) | \(\boldsymbol{(x - 1)(x + 3)^2}\) |
The factor \((x + 3)\) appears twice. Check: expanding gives \(x^3 + 5x^2 + 3x - 9\).
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 \(P(a) = R\).
What is the factor theorem?
\((x - a)\) is a factor of \(P(x)\) if and only if \(P(a) = 0\). In other words, \(a\) is a zero exactly when \((x - a)\) divides \(P(x)\) with no remainder.
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 show that something is a factor?
Substitute the value of x that makes the possible factor zero and show the polynomial equals 0. State that by the factor theorem it is a factor.
How do you find unknown coefficients using remainders or factors?
Each remainder or factor gives an equation: substitute the value that makes the divisor zero and set the result equal to the remainder, or to 0 for a factor. Two unknowns need two conditions, solved simultaneously.