Question

x
$x ^{2} + 5xy + 6y ^{2}$
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Answer 1

Answer:

x² + 5xy + 6y²

= x² + (3 + 2)xy + 6y²

= x² + 3xy +2xy + 6y²

= x(x + 3y) + 2y(x + 3y)

= (x + 3y)(x + 2y)

______________________

For finding its zeros, then

(x + 3y)(x + 2y) = 0

Either | Or

x + 3y = 0 | x + 2y = 0

or, x = -3y | or, x = -2y