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Find the product of $(5x^2y-4xy^2)$ and $(7x^2+8y^2)$ and verify the result for $x=1, y=-2$.
Given:
Given terms are $(5x^2y-4xy^2)$ and $(7x^2+8y^2)$.
To do:
We have to find the product of the given terms.
Solution:
$(5x^2y-4xy^2) \times (7x^2+8y^2)=5x^2y(7x^2+8y^2)-4xy^2(7x^2+8y^2)$
$=5(7)x^{(2+2)}y+5(8)x^2y^{(1+2)}-4(7)x^{(1+2)}y^2-4(8)xy^{(2+2)}$
$=35x^4y+40x^2y^3-28x^3y^2-32xy^4$
Verification for $x=1, y=-2$:
LHS
$(5x^2y-4xy^2) \times (7x^2+8y^2)=[5(1)^2(-2)-4(1)(-2)^2]\times[7(1)^2+8(-2)^2]$
$=(-10-16)\times(7+32)$
$=(-26)\times(39)$
$=-1014$
RHS
$35x^4y+40x^2y^3-28x^3y^2-32xy^4=35(1)^4(-2)+40(1)^2(-2)^3-28(1)^3(-2)^2-32(1)(-2)^4$
$=-70-320-112-512$
$=-1014$
LHS$=$RHS
Hence verified.
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