Question

Find the following product and verify the result for x = − 1, y = − 2:
(x²y − 1) (3 − 2x²y)

Answer 1

To multiply, we will  use distributive law as follows:

\[\left( x^2 y - 1 \right)\left( 3 - 2 x^2 y \right)\]

\[ = x^2 y\left( 3 - 2 x^2 y \right) - 1 \times \left( 3 - 2 x^2 y \right)\]

\[ = 3 x^2 y - 2 x^4 y^2 - 3 + 2 x^2 y\]

\[ = 5 x^2 y - 2 x^4 y^2 - 3\]

\[\therefore\] \[\left( x^2 y - 1 \right)\left( 3 - 2 x^2 y \right) = 5 x^2 y - 2 x^4 y^2 - 3\]

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

\[\text { LHS } =\left( x^2 y - 1 \right)\left( 3 - 2 x^2 y \right)\]

\[ = \left[ \left( - 1 \right)^2 \left( - 2 \right) - 1 \right]\left[ 3 - 2 \left( - 1 \right)^2 \left( - 2 \right) \right]\]

\[ = \left[ 1 \times \left( - 2 \right) - 1 \right]\left[ 3 - 2 \times 1 \times \left( - 2 \right) \right]\]

\[ = \left( - 2 - 1 \right)\left( 3 + 4 \right)\]

\[ = - 3 \times 7\]

\[ = - 21\]

\[\text { RHS } = 5 x^2 y - 2 x^4 y^2 - 3\]

\[ = 5 \left( - 1 \right)^2 \left( - 2 \right) - 2 \left( - 1 \right)^4 \left( - 2 \right)^2 - 3\]

\[ = \left[ 5 \times 1 \times \left( - 2 \right) \right] - \left[ 2 \times 1 \times 4 \right] - 3\]

\[ = - 10 - 8 - 3\]

\[ = - 21\]

Because LHS is equal to RHS, the result is verified.
Thus, the answer is \[5 x^2 y - 2 x^4 y^2 - 3\].