Factorising into Double Brackets

When we talk about factorising with double brackets, we're working with quadratic expressions. Factorising here means taking an expression like $x^2 + 5x + 6$ and rewriting it as two sets of brackets, like $(x + 2)(x + 3)$. 

It’s like finding what two numbers multiply together to make the expression.

 

What Does Factorising with Double Brackets Mean?

Factorising an expression into double brackets usually applies to quadratic expressions. The general form of a quadratic expression is: $$ax^2 + bx + c$$

where:

• $a$ is the coefficient of $x^2$,
• $b$ is the coefficient of $x$,
• $c$ is the constant term.

When factorising, we’re looking for two numbers that add to make $b$ and multiply to make $c$

 

Steps for Factorising Quadratic Expressions

1. Set up two brackets: Start with two sets of brackets, each containing an $x$ (since $x \times x = x^2$ ).
2. Find two numbers:
   • These two numbers need to add up to the middle term (the coefficient $b$ of $x$).
   • They also need to multiply to give the constant term $c$
3. Write the factors in the brackets: Place each number in one of the brackets with $x$

 

Example 

Factorise $x^2 - 7x + 12$

Step-by-Step Solution:

1. Set up two brackets: $(x + ?)(x + ?)$.

2. Find two numbers that:
   • Add up to -7,
   • Multiply to 12.

   • These numbers are -3 and -4, since: $$-3 + (-4) = -7 \text{ and }  -3 \times -4 = 12$$

3. Write the expression as: $$x^2 - 7x + 12 = (x - 3)(x - 4)$$

 

 

 

 

Example (When $a \not = 1$)

Factorise $2x^2 + 7x + 3$

Step-by-Step Solution:

1. Since $a = 2$, we’ll use the factor-pair method.

2. We need to find two numbers that:

   • Multiply to $2 \times 3 = 6$ (product of $a$ and $c$),
   • The numbers are 6 and 1, because: $$6 + 1 = 7  \text{and}  6 \times 1 = 6$$
3. Rewrite $7x$ as $6x + x$: $$2x^2 + 6x + x + 3$$
4. Factor by grouping: $$= 2x(x + 3) + 1 (x + 3)$$
5. Factor out $(x + 3)$: $$= (2x + 1)(x + 3) \\ \text{So,} \\ 2x^2 + 7x + 3 = (2x + 1)(x + 3)$$