Factoring all real #s

Math professor gave me this problem.
3x²-19y²
Attempted to solve it, thinking it was Prime.
(√3x)² - (√19y)²
(√3x - √19y)(√3x + √19y)

My check doesn't work out though. (Using FOIL)
(√3x - √19y)(√3x + √19y)
3x + √57xy - √57xy - 19y
3x - 19y

Now the professor marked it as correct (I didn't put the check on the paper I turned in) but I would like to know if it is solvable and if so, how to do it. You can copy my symbols if you don't know how to insert them.

You did your check wrong. √3x multiplied by itself is 3x², same with - √19y and √19y, it's -19y².
When you check, you do
(√3x-√19y)(√3x+√19y)
3x²+√57xy-√57xy-19y²
3x²-19y²,it checks.
When you square an exponent (i.e., 2x²), you get (√2x)², because √2x times √2x is 2x².
So, your answer is correct.

Factor: 3x^2 - 19y^2
= (√3x + √19y)(√3x - √19y)
= 3x + √57xy - √57xy + 19y
= 3x + 19y

Your check was wrong...you didn't multiply the x's or the y's to get the squares back.

(√3x - √19y)(√3x + √19y)
3x² + √57xy - √57xy - 19y²
3x² - 19y²

(√3x - √19y)(√3x + √19y)

√3x times √3x is not, 3x, it's 3x²
√3 * √3 * x * x = 3x²
same for the 19y²
It works out

You are factoring only.
Your steps are correct.