Expanding & Factorising (AQA GCSE Further Maths): Revision Note

Basic expanding & factorising

How do I expand brackets?

• Expanding brackets means multiplying all the terms inside a bracket by the term outside of the bracket

• For GCSE Mathematics you will have learnt how to expand a single bracket

  • Multiply each term inside the bracket by the term outside the bracket

$\begin{array}{rcl}2x\left(3x-4y+5\right)& =& 2x\times 3x-2x\times 4y+2x\times 5\\ & =& 6x^2-8xy+10x\end{array}$

• You will also have learnt how to expand double brackets

  • You can turn this into two single brackets by multiplying each term inside one bracket by the other bracket

  • For Level 2 Further Mathematics you might get more than two terms in a bracket

    • Therefore using a grid or turning the expression into single brackets will be more helpful than using FOIL

$\begin{array}{rcl}\left(2x+5\right)\left(3x-4y+5\right)& =& 2x\left(3x-4y+5\right)+5\left(3x-4y+5\right)\\ & =& 6x^2-8yx+10x+15x-20y+25\\ & =& 6x^2+-8xy+25x-20y+25\end{array}$

• A bracket that is raised to the power of 2 can also be written as a double bracket

  • ${\left(x+y+z\right)}^2=\left(x+y+z\right)\left(x+y+z\right)$

  • Write as a double bracket and then expand

    • Do not just square each term inside the bracket

• Be very careful when working with negatives

• Remember to simplify expressions where possible

How do I factorise out a term from an expression?

• Factorising is the opposite of expanding

• Firstly find the highest common factor of each term in the expression and put this outside a bracket

  • The highest common factor could be a single number or a variable or both 

    • $12x+8=4\left(...+...\right)$

    • $2x^2-5x=x\left(...-...\right)$

    • $6x^2+8xy+4x=2x\left(...+...+...\right)$

  • If the terms involve the same variable(s) to different powers then the highest common factor will include the variable(s) with the smallest power

    • $4x^5+5x^4=x^4\left(...+...\right)$

    • $6x^4{y}^7-8x^9{y}^5=2x^4{y}^5\left(...-...\right)$

  • The highest common factor could also contain brackets

    • ${\left(x+1\right)}^2+5\left(x+1\right)=\left(x+1\right)\left(...+...\right)$

    • $\left(x+1\right)\left(x+2\right)\left(x+3\right)+\left(x+1\right)\left(x+2\right)\left(x+4\right)=\left(x+1\right)\left(x+2\right)\left(...+...\right)$

• Then find what you need to multiply the highest common factor by to get each term

  • You might need to use the index law: $x^a\times x^b=x^{a+b}$

Expanding triple brackets

How do I expand three brackets?

• Multiply out any two brackets using a standard method and simplify this answer (collect any like terms)

• Replace the two brackets above with one long bracket containing the expanded result

• Expand this long bracket with the third (unused) bracket

  • This step often looks like (x + a)(x² + bx + c)

  • Every term in the first bracket must be multiplied with every term in the second bracket

    • This leads to six terms 

  • A grid can often help to keep track of all six terms, for example (x + 2)(x² + 3x + 1)

    •  

      	x2	+3x	+1
      x	x3	3x2	x
      +2	2x2	6x	2

    • add all the terms inside the grid (diagonals show like terms) to get x³ + 2x² + 3x² + 6x + x + 2

    • collect like terms to get the final answer of x³ + 5x² + 7x + 2

• Simplify the final answer by collecting like terms (if there are any)

• It helps to put negative terms in brackets when multiplying

Factorising by grouping

How do I factorise expressions with common brackets?

• To factorise 3x(t + 4) + 2(t + 4), both terms have a common bracket, (t + 4)

  • the whole bracket, (t + 4), can be "taken out" like a common factor

    • (t + 4)(3x + 2)

  • this is like factorising 3xy + 2y to y(3x + 2)

    • y represents (t + 4) above

How do I factorise by grouping?

• Some questions may require you to form the common bracket yourself

  • for example, factorise xy + px + qy + pq

    • "group" the first pair of terms, xy + px, and factorise, x(y + p)

    • "group" the second pair of terms, qy + pq, and factorise, q(y + p),

  • now factorise x(y + p) + q(y + p) as above

    • (y + p)(x + q)

• This is called factorising by grouping

• The groupings are not always the first pair of terms and the second pair of terms, but two terms with common factors