# Find the following products and verify the result for $x = -1, y = -2$:$(x^2y-1) (3-2x^2y)$

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Given:

$(x^2y-1) (3-2x^2y)$

To do:

We have to find the given product and verify the result for $x = -1, y = -2$.

Solution:

We know that,

$(a+b)\times(c+d)=a(c+d)+b(c+d)$

Therefore,

$(x^2y-1) (3-2x^2y)=x^2y(3 - 2x^2y) -1(3-2x^2y)$

$= x^2y(3) - x^2y(2x^2y)- 1(3)+ 1(2x^2y)$

$= 3x^2y-2x^{2+2}y^{1+1}-3+ 2x^2y$

$= 3x^2y-2x^4y^2-3+2x^2y$

$= 3x^2y + 2x^2y - 2x^4y^2 - 3$

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

LHS $= (x^2y - 1) (3 - 2x^2y)$

$= [(-1)^2(-2) -1] [3 - 2 (-1)^2 (-2)]$

$= [1 (-2) -1) [3 - 2(1) (-2)]$

$= (-2 - 1) (3 + 4)$

$= -3(7)$

$= -21$

RHS $= 5x^2y - 2x^4y^2 - 3$

$= 5(-1)^2 (-2) -2 (-1)^4 (-2)^2 -3$

$=5 (1) (-2) - 2 (1) (4) -3$

$= -10-8-3$

$= -21$

Therefore,

LHS $=$ RHS

Updated on 10-Oct-2022 13:19:30