Quadratics when the coefficient of x² ≠ 1 - Higher - Algebraic expressions - AQA - GCSE Maths Revision - AQA

Factorising quadratics when the coefficient of x squared ≠ 1 - Higher

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Quadratic expressions can be written in the form \(ax^2 + bx + c\), where \(a\), \(b\) and \(c\) are numbers.

\(a\) is called the coefficient of \(x^2\) and \(b\) is the coefficient of \(x\). \(c\) is a constant term – it is a number that is not multiplied by the variable \(x\).

For example, for the quadratic expression \(6x^2 + 13x + 6\), \(a\) = 6, \(b\) = 13 and \(c\) = 6.

To factorise this quadratic, first multiply the of \(x^2\) by the constant term (\(c\)).

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6 × 6 = 36. Find two numbers which have a product of 36 and a sum of 13. These are 4 and 9 as 4 × 9 = 36 and 4 + 9 = 13
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Rewrite 13x as 4x + 9x as these are the two numbers found using the product and sum rule
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Factorise the first two terms and the last two terms separately. The bracket created should always be the same. The two brackets have now been found. The first bracket is the common factor of (3x + 2). The second bracket is the factorised terms outside of each bracket (2x + 3)

Example

Factorise \(6x^2 - 7x - 3\).

First, multiply the coefficient of \(x^2\) by the constant term (\(c\)).

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6 × -3 = -18. Find two numbers which have a product of -18 and a sum of -7. 2 × -9 = -18 and 2 + -9 = -7 so the numbers are 2 and -9
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Rewrite the centre term as these two numbers, so -7x becomes 2x - 9x
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Factorise each pair of terms. Check that the bracket created is the same. The brackets have now been found. The first bracket is the common factor of (3x + 1) and the second bracket is the factorised terms outside of each bracket (2x - 3)

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Algebraic expressions - AQAQuadratics when the coefficient of x² ≠ 1 - Higher

Factorising quadratics when the coefficient of x squared ≠ 1 - Higher

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