Factoring x and y in quadradic form

[torquerotates]

Homework Statement

factor 4x^2-6xy+10y^2

Homework Equations

The Attempt at a Solution

completing the square i get,

4[(1/4)y^2-(3/4)xy+(9/16)y^2]-4(9/16)y^2+10y^2

but i don't know what to do with the bracketed term. It can't be factored easily

 

[HallsofIvy]

I presume that is supposed to be $x^2$, not $y^2$ at the first but then I have no idea why you have "1/4"

$4x^2- 6xy+ 10y^2= 4(x^2- (3/2)xy)+ 10y^2$
$= 4(x^2- (3/2)xy+ (9/16)y^2- (9/16)y^2)+ 10y^2$
$= 4(x^2- (3/2)xy+ (9/16)y^2)- (9/4)y^2+ 10y^2$
$= 4(x- (3/4)y)^2+ (31/4)y^2$
In order to factor that we need a difference of two squares so write it as
$= 4(x- (3/4)y)^2- (-31/4)y^2$

$= 4(x- (3/4)y)^2- ((\sqrt{31}i/2)y)^2$

$$= (2(x- (3/4)y)- (i\sqrt{31}/2)y)(2(x-(3/4)y)+ (i\sqrt{31}/2)y)$$

 

[epenguin]

Are you sure you transcribed it right?

If it were 4x^2-6xy-10y^2 it would all work out nice.

The people who invent these probs tend to avoid such things as this is giving - it is easier for them too.

But then maybe they also sometimes set traps.

Easy to check.