Factor Theorem
The factor theorem says that \(x-a\) is a factor of a polynomial \(P(x)\) exactly when \(P(a)=0\). Testing values finds a first factor, and division then finds the rest, as in \(x^3-7x-6=(x+1)(x+2)(x-3)\). This is NSW Year 10 Advanced maths (NESA outcome MA5‑POL‑P‑01).
You will test which linear expressions are factors, find unknown coefficients that make a given expression a factor, and factorise cubics and quartics fully. You will solve cubic equations, find two unknowns from two known factors, and use a factor and a remainder together. A box's dimensions are found from its volume. Harder questions find a remaining quadratic factor and prove that \(x-k\) is always a factor. It is part of Year 10 Advanced Logarithms and Polynomials.
Every question with a fully worked solution.
- Factor Theorem - Video - The Factor Theorem Watch
Theory
The factor theorem: \((x - a)\) is a factor of \(P(x)\) exactly when \(P(a) = 0\). This Year 10 Advanced maths page shows how to test for factors, choose values to try, factorise cubics fully, solve cubic equations and find unknown coefficients.
By the remainder theorem, dividing \(P(x)\) by \((x - a)\) leaves remainder \(P(a)\). 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.
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.
For a factor such as \((2x - 1)\), the zero is \(x = \dfrac{1}{2}\): test \(P\left(\dfrac{1}{2}\right) = 0\).
To factorise a cubic, find one factor by testing values, divide (or match terms) to get a quadratic, then factorise the quadratic.
This comes up if you are heading for Advanced or Extension maths in Years 11 and 12.
| Fact about \(P(x)\) | Equation it gives |
|---|---|
| \((x - 3)\) is a factor | \(P(3) = 0\) |
| \((x + 4)\) is a factor | \(P(-4) = 0\) |
| \((2x + 1)\) is a factor | \(P\left(-\dfrac{1}{2}\right) = 0\) |
| remainder \(7\) on division by \((x - 1)\) | \(P(1) = 7\) |
How to factorise a polynomial with the factor theorem
- List candidates: the factors of the constant term (positive and negative). If the leading coefficient is not \(1\), also try those factors divided by factors of the leading coefficient.
- Test \(P(a)\) for each candidate until one gives \(0\). Then \((x - a)\) is a factor.
- Divide \(P(x)\) by \((x - a)\), by long division or by matching the first and last terms of the quadratic factor.
- Factorise the quotient. For a quartic, repeat the process on the cubic quotient.
- Check by expanding, or by multiplying the constants.
Related problems
- Which is a factor? Test each option; only a factor gives \(0\).
- Solving \(P(x) = 0\): factorise fully, then set each factor equal to zero. A cubic has at most \(3\) roots.
- One unknown: substitute the zero and solve \(P(a) = 0\) for the unknown.
- Two conditions: two factors, or a factor and a remainder, give two simultaneous equations.
- Missing quadratic factor: multiply the two known linear factors, then match the leading and constant terms (Example 4).
- Applications: if a volume is a cubic whose factors are the dimensions, factorise it, then substitute to find each length.
(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 \(1\).
Any whole-number zero divides the constant \(6\): try \(\pm1, \pm2, \pm3, \pm6\).
| \(P(1)\) | \(=\) | \(1 - 2 - 5 + 6 = 0\) |
So \((x - 1)\) is a factor. Divide to find the other factor.
| \(x^2\) | \(-x\) | \(-6\) | ||
| \(x - 1\) | \(x^3\) | \(-2x^2\) | \(-5x\) | \(+6\) |
| \(-\) | \(x^3\) | \(-x^2\) | ||
| \(-x^2\) | \(-5x\) | |||
| \(-\) | \(-x^2\) | \(+x\) | ||
| \(-6x\) | \(+6\) | |||
| \(-\) | \(-6x\) | \(+6\) | ||
| \(0\) |
| \(P(x)\) | \(=\) | \((x - 1)(x^2 - x - 6)\) |
| \(\) | \(=\) | \(\boldsymbol{(x - 1)(x - 3)(x + 2)}\) |
Check: the constants multiply to \((-1)(-3)(2) = 6\).
Try \(x = 1\): \(3 + 4 - 13 + 6 = 0\). So \((x - 1)\) is a factor.
Find the quadratic factor by matching terms: it must start with \(3x^2\) (to give \(3x^3\)) and end with \(-6\) (to give \(+6\)).
| \(3x^3 + 4x^2 - 13x + 6\) | \(=\) | \((x - 1)(3x^2 + bx - 6)\) |
| \(x^2 \text{ terms: } b - 3\) | \(=\) | \(4 \;\Rightarrow\; b = 7\) |
| \(\) | \(=\) | \((x - 1)(3x^2 + 7x - 6)\) |
| \(\) | \(=\) | \((x - 1)(3x - 2)(x + 3)\) |
Set each factor equal to zero: \(x = \boldsymbol{1}\), \(x = \boldsymbol{\dfrac{2}{3}}\) or \(x = \boldsymbol{-3}\).
Check the \(x\) terms of \((x - 1)(3x^2 + 7x - 6)\): \(x(-6) + (-1)(7x) = -13x\). Correct.
(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 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 factorise a cubic polynomial?
Find one zero by testing factors of the constant term, which gives a linear factor. Divide the cubic by that factor to get a quadratic, then factorise the quadratic.
Which numbers should you test with the factor theorem?
Factors of the constant term, both positive and negative. If the leading coefficient is not 1, also try fractions: a factor of the constant divided by a factor of the leading coefficient.
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.
What is the difference between a zero and a root?
A zero belongs to a polynomial: a value of \(x\) where \(P(x) = 0\). A root belongs to an equation: a solution of \(P(x) = 0\). They are the same numbers.