Question

How do you factor the expression `9a^2 + 27aw +20w^2`?

Answer 1

Use the quadratic formula to find the zeros of the corresponding quadratic in one variable and hence find the factorisation:

`9a^2+27aw+20w^2 = (3a+4w)(3a+5w)`

Explanation:

Since this quadratic is homogeneous (all of the terms are of degree `2`), factoring it is similar to factoring `9x^2+27x+20`, which is of the form `ax^2+bx+c` with `a=9`, `b=27` and `c=20`.

This quadratic has zeros given by the quadratic formula:

`x = (-b+-sqrt(b^2-4ac))/(2a)`

`=(-27+-sqrt(27^2-(4*9*20)))/(2*9)`

`=(-27+-sqrt(729-720))/18`

`=(-27+-3)/18`

That is `x = -24/18 = -4/3` or `x = -30/18 = -5/3`

Hence:

`9x^2+27x+20 = (3x+4)(3x+5)`

and:

`9a^2+27aw+20w^2 = (3a+4w)(3a+5w)`

Answer 2

`(3a+5w)(3a+4w)`

Explanation:

Using factoring by grouping:

`9a^2+27aw+20w^2=9a^2+12aw+15aw+20w^2`

Find a common factor from the two pairs:

`=3a(3a+4w)+5w(3a+4w)`

Factor `(3a+4w)` from both the `3a` and `5w` terms.

`=(3a+5w)(3a+4w)`

Looking back, there is a way to determine that you should split `27aw` into `12aw` and `15aw`:

`27`, the middle coefficient, should be the sum of two integers whose product is also the product of the first and last coefficients. The only two integers that meet these criteria are `12` and `15`.