what is the method on how to factorise quadratics?

Answer 1

The first step is to write the quadratic in the standard form: ax2 + bx + c = 0 where a > 0, and b and c are constants. Also, that a, b and c are coprime.

[If a = 0 then the equation is linear, not quadratic and if a < 0 change the sign of each term so that it is then in the above form.]

Preliminary tidying up:

If a, b and c have a common factor, factorise out the this number. For example, 6x2 + 15x + 6 = 3*(2x2 + 5x + 2) and it is the expression inside the bracket that this method addresses.

You need to find two [positive] factors of abs(a*c), call them p and q, where p ≥ q. You also need the following:

• if c > 0 their sum must be equal to the absolute value of b and
• if c < 0 their difference must be equal to the absolute value of b.

That is,

• if c > 0, p + q = abs(b)
• if c < 0, p - q = abs(b)

This is a trial-and-error procedure, although there are ways to make the search more efficient. Not enough space here (or my patience!) for details.

Write the factorised equation as (ax ± p)*(ax ± q) = 0.

Give p (the bigger factor) the same sign as b. And

• if c > 0, then q has the same sign as p, while
• if c < 0, then q has the opposite sign to p.

Finally, in each of the factors (brackets) remove (ie delete) any common factor.

Example:

6x2 = 9x + 6

In standard form, this is 6x2 - 9x - 6 = 0

This equation has 3 as a common factor so factorise it out to give

3*(2x2 - 3x - 2) = 0 .. .. .. .. .. .. .. .. (*)

The equation that needs to be factorised is 2x2 - 3x - 2 = 0

a = 2, b = -3, c = -2

ac = 2*(-2) = -4 so abs(ac) = 4

also abs(b) = 3

Therefore we need factors, p and q, of abs(ac) = 4 and, since c is negative p - q = abs(b) = 3.

That is factors of 4 whose difference is 3. Answer: 4 and 1.

p = 4 is the bigger factor so give it the sign of b: that is p = -4

c is negative so give q the opposite sign: that is, q = +1

The equation to be factorised was 2x2 - 3x - 2

Re-write as (2x ± p)(2x ± q) = 0

Substitute for p and q to give: (2x - 4)*(2x + 1)

The first bracket has 2 as a common factor so get rid of it:

(x - 2)*(2x + 1)

Finally, bring back the factor 3, which was removed at (*)

Answer:

3*(x - 2)*(2x + 1)

Note: the method, described in words, is far more complicated that it is in reality. It is easy after you have done some examples. Even simpler if a = 1 (in terms of UK GCSE that will cover most A grade questions, but not A*)