Polynomials
Senior 2 · Mathematics
Polynomials
Lesson 5 of 5
25 min
Factorising is expanding in reverse: it turns a sum into a product. A product is zero only when one of its factors is zero, so factorising is how you solve polynomial equations. It also simplifies algebraic fractions and helps you see where a graph crosses the x-axis.
This lesson gathers every method you need at S2, from the simplest to the most powerful.
Take out the highest common factor of all the terms. 6x3 − 9x2 = 3x2(2x − 3). Always try this first.
With four terms, group them in pairs that share a factor.
Worked example 1
xy + 3x + 2y + 6 = x(y + 3) + 2(y + 3) = (y + 3)(x + 2).
For x2 + bx + c, find two numbers whose product is c and whose sum is b.
Worked example 2
x2 + 7x + 12: product 12, sum 7 gives 3 and 4. So (x + 3)(x + 4).
Worked example 3
x2 − x − 12: product −12, sum −1 gives −4 and 3. So (x − 4)(x + 3).
For ax2 + bx + c with a ≠ 1, find two numbers with product ac and sum b, split the middle term, then group.
Worked example 4
2x2 + 7x + 3: ac = 6, and 6 + 1 = 7. Split: 2x2 + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (x + 3)(2x + 1).
From the remainder theorem: if P(a) = 0, then x − a is a factor of P(x). For a polynomial with integer coefficients and leading coefficient 1, any integer root must divide the constant term. Find one root by trial, divide by synthetic division (this layout is also called Horner's or Ruffini's method), then factorise the quotient.
Worked example 5
Factorise P(x) = x3 − 2x2 − 5x + 6.
The constant is 6, so try ±1, ±2, ±3, ±6. P(1) = 1 − 2 − 5 + 6 = 0, so x − 1 is a factor. Divide with c = 1:
| c = 1 | 1 | −2 | −5 | 6 |
| 1 | −1 | −6 | ||
| 1 | −1 | −6 | 0 |
The quotient is x2 − x − 6 = (x − 3)(x + 2). So P(x) = (x − 1)(x − 3)(x + 2). The roots are 1, 3 and −2.
Tip
Exam technique. Use a fixed order: (1) common factor, (2) count the terms: two terms, look for a difference of squares or cubes; three terms, sum and product; four terms, grouping; (3) for a cubic, test divisors of the constant term. Then check that no bracket can be factorised again, and expand your answer to confirm it.
Factorise 12a2b − 18ab2.
Solution: HCF is 6ab: 6ab(2a − 3b).
Factorise 2x3 − 18x completely.
Solution: 2x(x2 − 9) = 2x(x − 3)(x + 3).
Factorise x2 − 9x + 20.
Solution: Product 20, sum −9: −4 and −5. (x − 4)(x − 5).
Factorise 3x2 + 10x + 8.
Solution: ac = 24; 6 + 4 = 10. 3x2 + 6x + 4x + 8 = 3x(x + 2) + 4(x + 2) = (x + 2)(3x + 4).
Show that x + 2 is a factor of x3 + x2 − 4x − 4, then factorise fully.
Solution: P(−2) = −8 + 4 + 8 − 4 = 0, so x + 2 is a factor. Synthetic division with c = −2 gives 1, −1, −2, remainder 0. So P(x) = (x + 2)(x2 − x − 2) = (x + 2)(x − 2)(x + 1).
Remember
Sources and credits
The Rwandan worked examples, the practice items and the exercises were written by Ganzaa.
