Question

How to factorise: `3n^2-11n+6` ?

Answer 1

`(3n-2)(n-3)`

Explanation:

Let `Q=3n^2-11n+6`

`Q` is a quadratic function which implies that it will have factors of the form: `(pn+a)(qn+b)`

Hence, we must find the values of `p, q, a, b`

Since `3` is prime `-> p=3 and q=1` or vice versa which makes no difference to our factorisation.

Thus, `Q=(3n+a)(n+b)`

Now we are looking for two numbers `a and b` such that:

`axxb = +6 and a+3b = -11`

Since `ab>0 and a+3b<0 -> a<0 and b<0`

Let's assume `{a,b} in ZZ`

The negative factors of `+6` are: `-6 xx -1 and -3 xx -2`

Testing the above factors in turn against `a+3b` we find:

`-2 + 3(-3) = -11`

Hence: `a=-2 and b=-3`

Thus, `Q=(3n-2)(n-3)` is our required factorisation.

[NB: With some practice, relatively simple factorisation such as in this question, can usually be performed by inspection. However, the process is as above whether or not it need be expressed step by step.]

Answer 2

`3n^2-11n+6 = (3n-2)(n-3)`

Explanation:

Given:

`3n^2-11n+6`

Here are a couple of methods that we can use to factor it:

`color(white)()`
AC Method

Since the constant term is positive like the coefficient of the leading term, find a pair of factors of `AC = 3*6 = 18` with sum `B=11`.

The pair `9, 2` works in that `9*2 = 18` and `9+2 = 11`

Use this pair to split the middle term and factor by grouping:

`3n^2-11n+6 = (3n^2-9n)-(2n-6)`

`color(white)(3n^2-11n+6) = 3n(n-3)-2(n-3)`

`color(white)(3n^2-11n+6) = (3n-2)(n-3)`

When it works, the AC Method can be quite quick and easy, but if the factors have irrational or complex coefficients then you will need more powerful methods such as...

`color(white)()`
Completing the square

We can factor any quadratic in a single variable by completing the square then using the difference of squares identity:

`a^2-b^2 = (a-b)(a+b)`

In the given example, the leading term is not a perfect square and the middle term has an odd coefficient, so it makes the arithmetic simpler if we multiply by `3*2^2 = 12` first, then divide by it at the end.

We use the difference of squares identity with `a=(6n-11)` and `b=7`...

`12(3n^2-11n+6) = 36n^2-132n+72`

`color(white)(12(3n^2-11n+6)) = (6n)^2-2(6n)(11)+121-49`

`color(white)(12(3n^2-11n+6)) = (6n-11)^2-7^2`

`color(white)(12(3n^2-11n+6)) = ((6n-11)-7)((6n-11)+7)`

`color(white)(12(3n^2-11n+6)) = (6n-18)(6n-4)`

`color(white)(12(3n^2-11n+6)) = 6(n-3)(2)(3n-2)`

`color(white)(12(3n^2-11n+6)) = 12(n-3)(3n-2)`

So dividing both ends by `12` we find:

`3n^2-11n+6 = (n-3)(3n-2)`