IGCSE Maths / Quadratic equations, curves and inequalities
Solving quadratics by factorising revision notes
Edexcel International GCSE Maths (4MA1), chapter 22: Quadratic equations, curves and inequalities, specification point 2.7a. Free lesson.
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Factorising solves a quadratic by writing it as two brackets multiplied together. First make one side zero, then set each bracket to zero to find both possible values of x.
Why one side must be zero
To factorise is to rewrite an expression as brackets multiplied together. This is useful for solving a quadratic when the equation has the form of two brackets multiplied together equalling zero. If a product is zero, at least one of its factors must be zero. Get all the terms on one side so the other side is zero. Then factorise and set each bracket to zero. Each bracket gives an answer.
Move every term to one side
For x² = 5x + 24, the x² and 5x need to be on the same side before you factorise. Subtract 5x from both sides, then subtract 24 from both sides. The result has the required form.
Factorise and find both answers
For x² − 5x − 24 = 0, find two numbers that multiply to −24 and add to −5. These are −8 and 3, so the expression factorises into (x − 8)(x + 3). Set each bracket to zero. You must give both answers.
Keep the zero answer when x is a factor
In x² = 5x, dividing both sides by x seems to give x = 5, but it loses the possible answer x = 0. Instead, move the terms to one side and factorise. This keeps both possible answers. x² = 5x becomes x² − 5x = 0. Factorise: x(x − 5) = 0. Set each factor to zero: x = 0 or x = 5.
When the coefficient of x² is not 1
For 2x² + 7x + 3 = 0, multiply the coefficient of x² by the constant term: 2 × 3 = 6. Find a pair that multiplies to 6 and adds to 7, which is 6 and 1. Split the middle term, then factorise by taking out a common factor from each pair.
Use the four steps in order
Factorising questions follow the same method. If there is a number in front of x², use it with the number at the end when choosing the pair for splitting the middle term. For x² − 7x + 12 = 0, the pair is −3 and −4 because they multiply to 12 and add to −7. So (x − 3)(x − 4) = 0, giving x = 3 or x = 4.