An algebraic expression combines numbers, variables, and operations. A polynomial in one variable \(x\) is a sum of terms of the form \(a x^n\) where \(a\) is a constant and \(n\) is a non-negative integer. Examples: \(3x^2 - 5x + 1\), \(x^3 + 2x\), and \(7\).
In \(5x^2 - 3x + 4\), the terms are \(5x^2\), \(-3x\), and \(4\). The coefficient of \(x^2\) is \(5\); the coefficient of \(x\) is \(-3\). The degree of the polynomial is the highest power of \(x\) — here, \(2\). Like terms have the same variable part (same power of \(x\)); only their coefficients may differ.
Line up like terms and add or subtract their coefficients: \[ (3x^2 + 2x - 1) + (x^2 - 5x + 4) = 4x^2 - 3x + 3. \] When subtracting, distribute the minus sign to every term in the bracket: \[ (2x^2 + x) - (x^2 - 3x + 2) = 2x^2 + x - x^2 + 3x - 2 = x^2 + 4x - 2. \]
Use the distributive law: each term in the first bracket multiplies each term in the second. For two binomials, \[ (a+b)(c+d) = ac + ad + bc + bd. \] Example: \[ (x+3)(x-2) = x^2 - 2x + 3x - 6 = x^2 + x - 6. \]
A binomial times a trinomial works the same way — multiply each of the two terms by each of the three: \[ (x+1)(x^2 - x + 2) = x^3 - x^2 + 2x + x^2 - x + 2 = x^3 + x + 2. \]
After expanding, combine like terms so the answer is simplified. From above, \(x^2 - 2x + 3x - 6\) becomes \(x^2 + x - 6\). Constants, \(x\)-terms, and \(x^2\)-terms are collected separately.
Factorisation is the reverse of expansion: write an expression as a product of simpler factors. Choose a method by inspecting the shape of the expression:
| Situation | Method |
|---|---|
| Every term shares a common factor | Highest common factor (HCF) |
| Four terms, pairs share factors | Grouping |
| Matches \(a^2-b^2\) or \((a\pm b)^2\) | Standard identities |
| Quadratic in \(x\): \(ax^2+bx+c\) | Two-number search or ac-method |
Factor out the largest factor common to every term: \[ 6x^2 + 9x = 3x(2x + 3), \qquad 4x^3 - 8x^2 = 4x^2(x - 2). \] Always check whether the bracket can be factorised further.
With four terms, split into two pairs and factor each pair: \[ ax + ay + bx + by = a(x+y) + b(x+y) = (a+b)(x+y). \] Example: \[ 2x^2 + 6x - x - 3 = 2x(x+3) - 1(x+3) = (2x-1)(x+3). \]
Identities
\[ a^2 - b^2 = (a+b)(a-b) \] \[ (a+b)^2 = a^2 + 2ab + b^2 \] \[ (a-b)^2 = a^2 - 2ab + b^2 \]
Recognising these forms speeds up factorisation. For example, \(x^2 - 9 = x^2 - 3^2 = (x+3)(x-3)\) and \(x^2 + 6x + 9 = (x+3)^2\).
To factorise a monic quadratic \(x^2 + bx + c\), find two numbers that multiply to \(c\) and add to \(b\): \[ x^2 + 5x + 6 = (x+2)(x+3) \quad\text{since } 2\times 3 = 6,\; 2+3 = 5. \]
For a non-monic quadratic \(ax^2 + bx + c\) with \(a \neq 1\), use the ac-method: find two numbers that multiply to \(ac\) and add to \(b\), split the middle term, then factor by grouping: \[ 2x^2 + 7x + 3 = 2x^2 + 6x + x + 3 = 2x(x+3) + 1(x+3) = (2x+1)(x+3). \]
This chapter stops at factorising expressions. Solving equations of the form \(ax^2+bx+c=0\) is covered in the chapter on quadratic equations.
Use the area model to see how \((a+b)(c+d)\) splits into four partial products. Then practise choosing a factorisation method and checking your factors in the explorer below.