# Find the Following Product and Verify the Result for X = − 1, Y = − 2: (X2y − 1) (3 − 2x2y) - Mathematics

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

#### Solution

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 $5 x^2 y - 2 x^4 y^2 - 3$.

Concept: Multiplication of Algebraic Expressions
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#### APPEARS IN

RD Sharma Class 8 Maths
Chapter 6 Algebraic Expressions and Identities
Exercise 6.5 | Q 18 | Page 31