Question

How do you factor `2x ^ { 2} + 3x y + 5y ^ { 2}`?

Answer 1

`2x^2+3xy+5y^2 = 1/8(4x+(3-sqrt(31)i)y)(4x+(3+sqrt(31)i)y)`

Explanation:

Given:

`2x^2+3xy+5y^2`

Note that this homogeneous quadratic in `x` and `y` is of the form:

`ax^2+bxy+cy^2`

with `a=2`, `b=3` and `c=5`.

This has discriminant `Delta` given by the formula:

`Delta = b^2-4ac`

`color(white)(Delta) = color(blue)(3)^2-4(color(blue)(2))(color(blue)(5))`

`color(white)(Delta) = 9-40`

`color(white)(Delta) = -31`

Since `Delta < 0` this quadratic has no linear factors with real coefficients.

It is still possible to factor it, but only with non-real complex coefficients.

We find:

`8(2x^2+3xy+5y^2) = 16x^2+24xy+40y^2`

`color(white)(8(2x^2+3xy+5y^2)) = (4x)^2+2(4x)(3y)+(3y)^2+31y^2`

`color(white)(8(2x^2+3xy+5y^2)) = (4x+3y)^2+(sqrt(31)y)^2`

`color(white)(8(2x^2+3xy+5y^2)) = (4x+3y)^2-(sqrt(31)iy)^2`

`color(white)(8(2x^2+3xy+5y^2)) = ((4x+3y)-sqrt(31)iy)((4x+3y)+sqrt(31)iy)`

`color(white)(8(2x^2+3xy+5y^2)) = (4x+(3-sqrt(31)i)y)(4x+(3+sqrt(31)i)y)`

So:

`2x^2+3xy+5y^2 = 1/8(4x+(3-sqrt(31)i)y)(4x+(3+sqrt(31)i)y)`