Question

2. Multiply and verify the result when x = 3 and y = 1. .
i. (x²y - 1)(3 – 2x²y)
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Answer 1

$\bold {Question:}$

2.) Multiply and verify the result when x = 3 and y = 1.

(i)$( {x}^{2} y - 1)(3 - 2 {x}^{2}y)$

$\bold {Solution:}$

$( x^2 y - 1)(3 - 2 x^2y)\\ = >x^2y (3-2x^2y)-1 (3-2x^2y)\\= >3x^2y-2x^4y^2-3+2x^2y\\ = >5x^2y-2x^4y^2-3$

$\bold {Verification:}$

Here,x = 3 and y = 1.

$= >( x^2y - 1)(3 - 2 x^2y)= 5x^2y-2x^4y^2-3 \\= >(3^2\times1-1)(3-2\times 3^2\times 1)= ( 5\times3^2\times 1)-(2\times 3^4\times1^2)-3 \\= >(9\times1-1)(3-2\times9\times 1)=(5\times 9\times1)-(2\times 81\times 1)-3\\ = > (9-1)(3-18)=45-162-3\\ = >8×(-15) = 45-165\\ = >120=-120$

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Answer 2

Hi this is your answer

this answer you are finding na

tell me it is wrong or correct