How factorising works
You are looking for two brackets that multiply back to the original expression. For 2x² + 7x + 3 they are (x + 3) and (2x + 1).
The rule of thumb is to find two numbers that multiply to a × c and add to b, then split the middle term and group.
Worked example
For 2x² + 7x + 3, a × c is 6 and the two numbers that multiply to 6 and add to 7 are 6 and 1.
Split the middle term into 6x + x, group as 2x(x + 3) + 1(x + 3), and the answer is (x + 3)(2x + 1).
Its discriminant is 25, a perfect square, which is why whole-number brackets exist.
Taking out a common factor first
Always check for a number that divides all three coefficients. 6x² + 9x + 3 has a common factor of 3, so it becomes 3(2x² + 3x + 1).
Leaving the common factor out is the most common way to lose marks, because the brackets alone do not multiply back to the original.
When it will not factorise
If the discriminant is not a perfect square, the roots are irrational and no whole-number brackets exist.
The page says so instead of inventing brackets, and the quadratic page will give the roots in surd form.
Inputs for the factorise calculator
Coefficient of x²
The number multiplying x². It cannot be zero.
Coefficient of x
The number multiplying x, including its sign.
Constant
The number on its own. Whole numbers only, because the brackets are searched for exactly.
CALCULATOR QUESTIONS
Factorise calculator questions
How do I factorise 2x² + 7x + 3?
Find two numbers multiplying to 6 and adding to 7, which are 6 and 1. Splitting and grouping gives (x + 3)(2x + 1).
What is the difference between factorising and solving?
Factorising rewrites the expression. Solving finds the values of x that make it zero, which are the numbers each bracket is set to zero for.
Why take out a common factor first?
The brackets are only complete once the common factor is outside them. 6x² + 9x + 3 is 3(2x + 1)(x + 1), not (2x + 1)(x + 1).
What if it does not factorise?
Then the roots are irrational, so no whole-number brackets exist. Use the quadratic formula instead.