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

Question

Savannah Anderson · Accepted Answer

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

Nathan Axel · Answer

undefined To factor completely $ x^2 - 2xy - 15y^2 $, you can use the following steps:

1. Factor out common factors, if any.
2. Rewrite the expression in the form $ ax^2 + bx + c $ if possible.
3. Factor the quadratic expression.

Let's start with step 1:

No common factors to factor out.

Moving on to step 2:

Rewrite the expression as $ x^2 - 5xy + 3xy - 15y^2 $.

Now, for step 3:

Factor by grouping: $ x(x - 5y) + 3y(x - 5y) $.
Common factor $ x - 5y $: $ (x + 3y)(x - 5y) $.

Therefore, the completely factored form of $ x^2 - 2xy - 15y^2 $ is $ (x + 3y)(x - 5y) $.