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:

$$\begin{gathered} \left(x^{2}y-1\right)\left(3-2x^{2}y\right) \\ =x^{2}y\left(3-2x^{2}y\right)-1\times \left(3-2x^{2}y\right) \\ =3x^{2}y-2x^4{y}^2-3+2x^{2}y \\ =5x^{2}y-2x^4{y}^2-3 \end{gathered}$$

$\therefore$ $\left(x^{2}y-1\right)\left(3-2x^{2}y\right)=5x^{2}y-2x^4{y}^2-3$

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

$$\begin{gathered} \text{LHS =}\left(x^{2}y-1\right)\left(3-2x^{2}y\right) \\ =\left[{\left(-1\right)}^2\left(-2\right)-1\right]\left[3-2{\left(-1\right)}^2\left(-2\right)\right] \\ =\left[1\times \left(-2\right)-1\right]\left[3-2\times 1\times \left(-2\right)\right] \\ =\left(-2-1\right)\left(3+4\right) \\ =-3\times 7 \\ =-21 \end{gathered}$$

$$\begin{gathered} \text{RHS}=5x^{2}y-2x^4{y}^2-3 \\ =5{\left(-1\right)}^2\left(-2\right)-2{\left(-1\right)}^4{\left(-2\right)}^2-3 \\ =\left[5\times 1\times \left(-2\right)\right]-\left[2\times 1\times 4\right]-3 \\ =-10-8-3 \\ =-21 \end{gathered}$$

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

Thus, the answer is $5x^{2}y-2x^4{y}^2-3$.