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# A simple method to factorise an expression into a double bracket. Expressions of the form x² + bx + c

Expressions of the form x² + bx + c can be factorised into a double bracket. Basically you are looking for two numbers that multiply to give c and add to give b.

**Question 1 on double
bracket factorisation**

Factorise:

x² -5x + 6

To begin with write down all the factor pairs of +6 (the number at the end). These are:

1 × 6

2 × 3

-1 × -6

-2 × -3

Notice that all the pairs have the same signs when the number at the end is positive.

Now one of these factor pairs must add to give you -5 (the number before x)

1 + 6 = 7

2+ 3 = 5

-1 + -6 = -7

-2 + -3 = -5

So the last pair gives -5.

Therefore the expression can be written down as (x-2)(x-3) or (x-3)(x-2) as it doesn’t matter which bracket you write down first.

**Question 2 on double
bracket factorisation**

Factorise:

x² + 6x -7

To start off with write down all the factor pairs of -7 (the number at the end). These are:

1 × -7

-1 × 7

Notice that the pairs have different signs when the number at the end is negative.

Now one of these factor pairs must add to give you 6 (the number before x)

1 + -7 = -6

-1 + 7 = 6

So the last pair gives +6.

Therefore the expression can be written down as (x-1)(x+7).

**Question 3 on double
bracket factorisation**

Factorise:

x² -9x -10

To start off with write down all the factor pairs of -10 (the number at the end). These are:

1 × -10

-1 × 10

2 × -5

-2 × 5

Again since the number at the end is negative all the pairs have different signs.

Yet again, select the pair of factors that gives you the number before x which is -9.

1 + -10 = -9

-1 +10 = 9

2 + -5 = -3

-2 + 5 = 3

So the first pair gives you -9.

Therefore the expression can be written down as (x+1)(x-10).

If you are finding these examples difficult then take a look at this other article on double bracket factorisation.

For harder examples on double bracket factorising then click here.

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