Find the following product and verify the result for x = − 1,

To multiply, we will use distributive law as follows:

(3x − 5y) (x + y)

$=3 x(x+y)-5 y(x+y)$

$=3 x^{2}+3 x y-5 x y-5 y^{2}$

$=3 x^{2}-2 x y-5 y^{2}$

$\therefore(3 x-5 y)(x+y)=3 x^{2}-2 x y-5 y^{2}$

Now, we put $x=-1$ and $y=-2$ on both sides to verify the result.

LHS $=(3 x-5 y)(x+y)$

$=\{3(-1)-5(-2)\}\{-1+(-2)\}$

$=(-3+10)(-3)$

$=(7)(-3)$

$=-21$

$\mathrm{RHS}=3 x^{2}-2 x y-5 y^{2}$

$=3(-1)^{2}-2(-1)(-2)-5(-2)^{2}$

$=3 \times 1-4-5 \times 4$

$=3-4-20$

$=-21$

Because LHS is equal to RHS, the result is verified.

Thus, the answer is $3 x^{2}-2 x y-5 y^{2}$.