Question

Factorise the following:

x + 3y – x² + 9y²

Answer 1

Given expression: x + 3y – x² + 9y²

Step-wise calculation:

1. Rearrange terms grouping squares and linear terms:

–x² + 9y² + x + 3y

2. Rewrite as:

–(x² – x) + (9y² + 3y)

3. Complete the square for each group:

For x² – x:

`x^2 - x = x^2 - x + (1/2)^2 - (1/2)^2`

`x^2 - x = (x - 1/2)^2 - 1/4`

For 9y² + 3y:

Factor out 9:

`9y^2 + 3y = 9(y^2 + 1/3 y)`

Complete the square inside the bracket:

`y^2 + 1/3 y = y^2 + 1/3 y + (1/6)^2 - (1/6)^2`

`y^2 + 1/3 y = (y + 1/6)^2 - 1/36`

So:

`9(y + 1/6)^2 - 9 xx 1/36 = 9(y + 1/6)^2 - 1/4`

4. Substitute back:

`-((x - 1/2)^2 - 1/4) + (9(y + 1/6)^2 - 1/4)`

= `-(x - 1/2)^2 + 1/4 + 9(y + 1/6)^2 - 1/4`

= `-(x - 1/2)^2 + 9(y + 1/6)^2`

The expression factorises into:

`9(y + 1/6)^2 - (x - 1/2)^2`

This is a difference of squares and can be factorised as:

`(3(y + 1/6) - (x - 1/2))(3(y + 1/6) + (x - 1/2))`

Or more simply:

`(3y + 1/2 - x + 1/2)(3y + 1/2 + x - 1/2)`

Which simplifies to:

(3y – x + 1)(3y + x)