--- title: "Polynomial Division and the Factor Theorem" description: "Divide a polynomial by a linear expression and read the remainder, then use the fact that a root gives a factor to break a cubic apart." canonical: https://lightmysky.com/learn/mathematics/polynomial-division-and-the-factor-theorem-mt_haL2UkGa5C source: https://lightmysky.com/learn/mathematics/polynomial-division-and-the-factor-theorem-mt_haL2UkGa5C.md retrieved: 2026-09-02 --- > **Agent view.** This is the Markdown twin of the page, for tools and assistants. > When to use this site, and the call that answers each job: https://lightmysky.com/agent-instructions.md > API description (OpenAPI 3.1): https://lightmysky.com/openapi.json · Authentication: https://lightmysky.com/auth.md > Pricing: https://lightmysky.com/pricing.md · Catalog: https://lightmysky.com/llms.txt · Full catalog: https://lightmysky.com/llms-full.txt > Every machine-readable file on this domain: https://lightmysky.com/.well-known/ai-catalog.json > Ask for Markdown with `Accept: text/markdown`, a `.md` address, or `?mode=agent`. # Polynomial Division and the Factor Theorem Divide a polynomial by a linear expression and read the remainder, then use the fact that a root gives a factor to break a cubic apart. Subject: Mathematics · Area: Algebra · Ages 16 to 17 Page: https://lightmysky.com/learn/mathematics/polynomial-division-and-the-factor-theorem-mt_haL2UkGa5C ## Ready when they can - Divide x³ - 2x² - 5x + 6 by (x - 1) and state the quotient - Use f(2) = 0 to write down a factor of a cubic - Factorise a cubic fully after finding its first root by substitution ## Lesson: Polynomial Division and the Factor Theorem Last stop you asked which inputs a rule accepts. Now the outputs do some work for you. The Ridgeway crew has a cubic for the concrete in the bowl, and it will not open up the way a quadratic does. There is no pair of numbers to spot. The way in has two moves. Find one input that makes the cubic zero, turn it into a factor, then divide that factor out. What is left is a quadratic. Which inputs are worth trying? When the coefficients are whole numbers, start with the ones that divide the constant term. Divide any polynomial f(x) by (x - k) and the answer can be written f(x) = (x - k)Q(x) + R, with Q the quotient and R the remainder. Put x = k into that line. The bracket becomes zero, so the first term disappears and f(k) = R. That is the remainder theorem: the remainder on dividing by (x - k) is f(k), with no division done. The factor theorem is the case f(k) = 0. Then (x - k) divides exactly and is a factor. Watch the sign. A root of 3 gives the factor (x - 3). *(drawing: One test to find a factor, one division to use it, then familiar ground.)* **Example.** Divide x³ - 2x² - 5x + 6 by (x - 1). Work term by term. x³ divided by x is x², and x² times (x - 1) is x³ - x². Subtracting leaves -x², and bringing down -5x gives -x² - 5x. Now -x² divided by x is -x, and -x times (x - 1) is -x² + x. Subtracting leaves -6x, and bringing down 6 gives -6x + 6. Last, -6x divided by x is -6, and -6 times (x - 1) is -6x + 6. Subtracting leaves nothing, so the quotient is x² - x - 6. A leading 2x³ would give 2x² instead. *(drawing: Each stage kills the leading term and hands the rest on.)* **Example.** The division has done the hard part. So far x³ - 2x² - 5x + 6 = (x - 1)(x² - x - 6), and that quotient is an ordinary quadratic. Find the pair that multiplies to -6 and adds to -1, which is -3 and 2. So x² - x - 6 = (x - 3)(x + 2), and the cubic is (x - 1)(x - 3)(x + 2). The roots are 1, 3 and -2. Check one of them in the original: at x = 3, 27 - 18 - 15 + 6 = 0. **Example.** Not every division is exact. Divide x³ + 2x² - 3x + 4 by (x - 2). The test gives f(2) = 8 + 8 - 6 + 4 = 14, so the remainder is 14 and (x - 2) is not a factor. The long division agrees: the quotient is x² + 4x + 5 with 14 left over, which we write as x³ + 2x² - 3x + 4 = (x - 2)(x² + 4x + 5) + 14. Use the theorem when only the remainder is wanted, and the division when you need the quotient as well. **Recap.** A cubic gives up its factors in two moves. Test values of k until f(k) = 0, which makes (x - k) a factor, then divide by (x - k) to leave a quadratic and factorise that the usual way. When the division is not exact, the remainder is f(k), so one substitution tells you what is left over. ## Practice 24 questions on this page, each with its working shown. ## Needs first - [Domain and Range of a Function](https://lightmysky.com/learn/mathematics/domain-and-range-of-a-function-mt_b_4zoHz8Jc) - [Solving Quadratic Equations by Factorising](https://lightmysky.com/learn/mathematics/solving-quadratic-equations-by-factorising-mt_tg_sqRzDjU) ## Opens up - [Sketching a Curve from Its Factorised Form](https://lightmysky.com/learn/mathematics/sketching-a-curve-from-its-factorised-form-mt_x4tZLdH8fh)