There are different ways of factoring an equation depending on the type. A few are highlighted here:
1) Taking out a common factor
General formula: ab + ac = a ( b+ c)
Step 1: realise that all terms of the equation are divisible by the same number
Step 2: divide them by such number and take it out of the brackets
Example: factorise the equation → 7x +35
We notice that both terms are divisible by 7 since (7x/7) = x and (35/7) = 5
Therefore: 7 (x+5)
You can double check this by multiplying 7 by what’s in the brackets and you will obtain our original equation
2) Difference of two squares
General formula: a^2 - b^2
Step 1: realise that both terms of the equation are squares
Step 2: rewrite the equation as two brackets containing the variable x and the square root of the given number (one negative and one positive)
Example: factorise the equation → x^2 -16
We notice that both terms are squares → x^ 2 -16 = x ^2 - 4^2 Rewrite as two brackets → (x - 4) * (x +4)
You can double check this by multiplying the two expressions and you will obtain the original equation
3) Perfect squares General formula: a^2 + 2ab + b^2 = ( a + b ) ^2
Step 1: identity the a and b term
Step 2: check that 2 * a * b holds
Step 3: Rewrite as a perfect square
Example: factorise the equation → x^2 + 12x +36
We notice that 36 = 6 ^2 so 6 is our b term
We notice that x ^2 = x ^ 2 so x is our a term
We notice that 2 * a * b = 2 * 6 * x = 12x which holds
Therefore: (x + 6) ^2 If you expand this expression, our original equation is obtained
4) Trinomials
General formula: x^2 + ax + b = ( x + c ) * ( x * d)
Where cd = b and c + d = a
Example: factorise the equation → x^2 + 5x + 6
Find two numbers that when multiplied give you 6 and when added give you 5
The numbers in this case are 2 and 3 since 2+3 = 5 and 23 = 6 Rewrite as two brackets Therefore → (x + 2) * (x + 3)
If you multiply out these brackets, our original equation is obtained.

Expand and simplify (2x+5)(x^2 + 3x -1)

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