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Find the product of the following.
$(3-8y+6y^2)$ and $(4-7y-2y^2)$
Given:
$(3-8y+6y^2)$ and $(4-7y-2y^2)$.
To do:
We have to find the product of the given terms.
Solution:
$(3-8y+6y^2)\times(4-7y-2y^2)=3(4-7y-2y^2)-8y(4-7y-2y^2)+6y^2(4-7y-2y^2)$
$=12-21y-6y^2-32y+56y^2+16y^3+24y^2-42y^3-12y^4$
$=-12y^4-26y^3+74y^2-53y+12$
Therefore, $(3-8y+6y^2)\times(4-7y-2y^2)=-12y^4-26y^3+74y^2-53y+12$.
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