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.
What a learner can do afterwards
- 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
1 · Read
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).
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.
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.
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.
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.
2 · Watch
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Where it sits
Where this leads
Jobs that lean on this skill. Follow one to see everything it is built on.
24 questions wait behind this lesson, each with its answer explained. Every answer feeds the sky: stars light as they are learned, and dim when it is time to come back.