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Factorising techniques for OCR GCSE Maths

Factorising techniques

This page covers the following topics:

1. Factorising linear equations
2. Factorising quadratic equations
3. The quadratic equation

When solving linear equations we generally want to rearrange it, and make some variable the 'subject' of the formula. Something is the subject of the formula if it appears by itsel on one side of the equation. For example, if y = 2x + 3, then y is the subject. To solve an equation for y, we want to isolate it like that on one side, and then evaluate the terms on the other side with numbers we know.

Factorising linear equations

When given a quadratic such as (x + 1)(x + 2), we can multiply out the brackets using FOIL (meaning first, outside, inside, last), giving xยฒ + 2x + x + 2, which is just xยฒ + 3x + 2. However, for solving a quadratic it's more useful to do this process in reverse. Given some quadratic like xยฒ + 3x + 2, we want to find 2 values to fit in (x + _)(x + _). We know these values need to equal 2 when multiplied together, and when summed they need to equal 3. Once we spot the values 1 and 2 for this, we have (x + 1)(x + 2), and it's very easy to solve an equation like (x + 1)(x + 2) = 0, because this just gives our solutions x = โ€“1 and โ€“2.

Factorising quadratic equations

We have a quadratic equation of the form axยฒ + bx + c = 0, where a, b, and c are just coefficients. The solutions to the quadratic are given by the quadratic formula: x = (โ€“b +โ€“ โˆš(bยฒ โ€“ 4ac))/2a.

The quadratic equation

1

Factorise xยฒ + 5x + 6.

We want something of the form (x + _)(x + _) and we know when multiplied together, the values in the gaps should equal 6. They could either be 1 and 6, or 2 and 3 (or also โ€“1 and โ€“6, or โ€“2 and โ€“3). However the only pair of values here that add up to 5 is 2 and 3, so we have (x + 2)(x + 3).

Factorise xยฒ + 5x + 6.

2

Solve 22x + xยฒ = โ€“21.

One needs to rearrange and be aware of the ordering of terms. We have a = 1, b = 22 and c = 21, which in the quadratic formula gives our solutions x = โ€“1 and x = โ€“21.

Solve 22x + xยฒ = โ€“21.

3

Solve s โ€“ 4 = b + 3 for b when s = 4.

Can first rearrange to b = s โ€“ 7. Then substitute 4 in for s, which gives the solution b = โ€“3.

Solve s โ€“ 4 = b + 3 for b when s = 4.

4

Suppose we have x โ€“ 6 = 7 โ€“ y. Rearrange the equation, making y the subject.

One needs to add y to both sides to make the negative y positive on the other side of the equation. The rest of the rearranging then gives y = 13 โ€“ x.

Suppose we have x โ€“ 6 = 7 โ€“ y. Rearrange the equation, making y the subject.

5

Factorise xยฒ + x โ€“ 2.

One needs to notice that โ€“1 and 2 multiplied together give โ€“2, but when added together give 1 (the coefficient of x in our equation). The factorised equation is therefore (x โ€“ 1)(x + 2).

Factorise xยฒ + x โ€“ 2.

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