The remainder and factor theorems
Divide a cubic polynomial p(x) by (x − a) without long division: the remainder is just p(a). Change a and the polynomial, and find the values of a that make (x − a) a factor.
Core concept
When p(x) is divided by (x − a), the remainder is p(a); if p(a) = 0, then (x − a) is a factor of p(x) and a is a zero, where the graph meets the x-axis
Topic: Polynomials
Dividing a polynomial p(x) by (x − a) by long division takes many steps. The remainder theorem gives the remainder at once: it is p(a), the value of the polynomial when x = a. For p(x) = x³ − 6x² + 11x − 6 and a = 2, p(2) = 8 − 24 + 22 − 6 = 0, so the remainder is 0. A remainder of 0 means (x − a) divides p(x) exactly: this is the factor: A number that divides another exactly. 3 is a factor of 12. All key words theorem. Then a is a zero of p(x), and the graph of y = p(x) crosses the x-axis at x = a. Testing small whole numbers a = ±1, ±2, ±3 is a quick way to find the factors of a cubic and factorise it.
What you will learn
- Find the remainder of p(x) ÷ (x − a) by working out p(a).
- Use the factor theorem to test whether (x − a) is a factor.
- Factorise a cubic polynomial by finding its zeroes.
- Link zeroes to the points where the graph meets the x-axis.
Before you start
- Know how to substitute a number into an expression.
Materials
- A notebook for the substitutions
Equipment
- None
Procedure
- For p(x) = x³ − 6x² + 11x − 6, record a = 2.
- Set a = −1 and record.
- Change to p(x) = x³ − 2x² − x + 2 and record a = −1.
- Change to p(x) = 2x³ + x² − 5x + 2 and find a factor with a negative a.
How this simulation is simplified
- The divisor is always (x − a) with a whole number a.