1. The problem gives two expressions: $(x+y)^2$ and $x^2 - xy - 2y^2$. 2. First, let's expand the square of the binomial $(x+y)^2$ using the formula $(a+b)^2 = a^2 + 2ab + b^2$. $$ (x+y)^2 = x^2 + 2xy + y^2 $$ 3. Next, consider the expression $x^2 - xy - 2y^2$. We can try to factor it. 4. To factor $x^2 - xy - 2y^2$, look for two numbers that multiply to $-2$ (the product of the coefficient of $x^2$ and $-2y^2$) and add up to $-1$ (the coefficient of $xy$). These numbers are $-2$ and $1$ because $-2 imes 1 = -2$ and $-2 + 1 = -1$. 5. Rewrite the middle term using these numbers: $$ x^2 - 2xy + xy - 2y^2 $$ 6. Group terms: $$ (x^2 - 2xy) + (xy - 2y^2) $$ 7. Factor each group: $$ x(x - 2y) + y(x - 2y) $$ 8. Factor out the common binomial factor: $$ (x + y)(x - 2y) $$ Final answers: $$ (x + y)^2 = x^2 + 2xy + y^2 $$ $$ x^2 - xy - 2y^2 = (x + y)(x - 2y) $$