How do you factor completely #x^2-2xy-15y^2#?

1 Answer
Jul 3, 2016

Answer:

#(x-5y)(x+3y)#

Explanation:

#x^2-2xy-15y^2#
Looking at the given algebraic expression we recognize from the first two terms that to factor the expression we have to apply the property:
#color(blue)((x-y)^2=x^2- 2xy+y^2)#
But in the given expression we need the term #y^2# so we can add it and subtract so that as if #0# is added to the expression.
Let's add #y^2# then subtract it
#=x^2-2xy-15y^2+y^2-y^2#
#=x^2-2xy+y^2-15y^2-y^2#
#=(x-y)^2-16y^2#
#=(x-y)^2-(4y)^2#
Checking the last step reached it is the difference of two squares that says:
#color(blue)(a^2-b^2=(a-b)(a+b))#
where in our case:#a=(x-y)# and #b=4y#
Then,
#(x-y)^2-(4y)^2#
#=(x-y-4y)(x-y+4y)#
#=(x-5y)(x+3y)#