Answer
(a) The solutions are $x=-y$ and $x=\frac{y}{2}$.
(b) The solutions are $x=-y$, $x=y$ and $x=-4y$.
Work Step by Step
(a) Given: $(x+y)(2x-y)=0$
Using zero-product property, we have
$x+y=0$ or $2x-y=0$
Solving for $x$, we get
$x=-y$ or $x=\frac{y}{2}$
The solutions are $x=-y$ and $x=\frac{y}{2}$.
(b) Given: $(x^{2}-y^{2})(4x+16y)=0$
This can be written as
$(x+y)(x-y)(4x+16y)=0$
Using zero-product property, we have
$x+y=0$ or $x-y=0$ or $4x+16y=0$
Solving for $x$, we get
$x=-y$ or $x=y$ or $x=-4y$
The solutions are $x=-y$, $x=y$ and $x=-4y$.
