Question

What is the factored form of the expression over the complex numbers?

Answer 1

`121x^2+36y^2 = (11x+6iy)(11x-6iy)`

Explanation:

The difference of squares identity tells us that:

`A^2-B^2=(A-B)(A+B)`

We find:

`121x^2+36y^2 = (11x)^2+(6y)^2`

`color(white)(121x^2+36y^2) = (11x)^2-(-1)(6y)^2`

`color(white)(121x^2+36y^2) = (11x)^2-i^2(6y)^2`

`color(white)(121x^2+36y^2) = (11x)^2-(6iy)^2`

`color(white)(121x^2+36y^2) = (11x-6iy)(11x+6iy)`

`color(white)(121x^2+36y^2) = (11x+6iy)(11x-6iy)`

Answer 2

`(11x+6iy)(11x-6iy)`

Explanation:

Note that `(+i)^2=-1`

We need to 'get rid' of (whithout the signs) `2(11x xx6iy)`
Thus one has to be negative and the other positive so they cancel each other out. Thus let us investigate:

`color(white)("ddddd")color(green)(color(blue)((11x+6iy))(11x-6iy) )`

`color(green)(color(blue)(11x)(11x-6iy) color(white)("dddd")color(blue)(+6iy)(11x-6iy) )`

`121x^2cancel(-66ixy)color(white)("ddddd")cancel(+66ixy)-36i^2y^2`

but `i^2=-1` giving:

`121x+36y^2` as required