Factorising where $a=1$

Factorising where a=1

It is possible to factorise a quadratic expression, such as x^2+3x+2, into a factorised form*, such as (x+2)(x+1).

For a simple quadratic where a (the coefficient of x^2) is 1, we find two numbers which multiply to give 2, and add to give 3. In this case: 2 and 1.

Example: factorise x^2+5x+6

• We need two numbers which multiply to give 6, and add to give 5.
• The numbers 2 and 3 work, since 2\times3=6 and 2+3=5.
• Therefore, we can factorise x^2+5x+6 as (x+2)(x+3).

Factorising where a\ne1

When a is not equal to 1, we can use a method called splitting the middle term (or simply called ‘the split method’).

First, we multiply a and c together. Then, we find two numbers which multiply to give ac, and add to give b. We then split the middle term using these two numbers, and factor by grouping.

Example: factorise 2x^2+7x+3

• a=2, b=7 and c=3.
• ac=2\times3=6.
• We need two numbers which multiply to give 6, and add to give 7
• The numbers 6 and 1 work, since 6\times1=6 and 6+1=7.
• Split the middle term into 6x and 1x:
  • 2x^2+6x+1x+3
• Use the distributive law to factor by grouping:
  • 2x(x+3)+1(x+3)
• Factor out the common factor (x+3):
  • (x+3)(2x+1)
• Answer: (x+3)(2x+1).

Example: factorise 3x^2-11x+10

• a=3, b=-11 and c=10.
• ac=3\times10=30.
• We need two numbers which multiply to give 30, and add to give $-11
• The numbers -5 and -6 work, since -5\times-6=30 and -5+(-6)=-11.
• Split the middle term into -5x and -6x:
  • 3x^2-5x-6x+10
• Use the distributive law to factor by grouping:
  • x(3x-5)-2(3x-5)
• Factor out the common factor (3x-5):
  • (3x-5)(x-2)
• Answer: (3x-5)(x-2).

Solving quadratic equations by factorising

Please see solving-quadratics-by-factorising.