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Question
Divide: (4m2n3 + 16m4n2 – mn) by 2mn
Sum
Solution
`(4"m"^2"n"^3 + 16"m"^4"n"^2 - "mn")/(2"mn") = (4"m"^2"n"^3)/(2"mn") + (16"m"^4"n"^2)/(2"mn") - "mn"/(2"mn")`
= `4/2"m"^(2-1)"n"^(3-1) + 16/2 "m"^(4-1)"n"^(2-1) - 1/2"m"^(1-1)"n"^(1-1)`
= `2"m"^1 "n"^2 + 8"m"^3 "n"^1 - 1/2 "m"^0"n"^0`
= `2"mn"^2 + 8"m"^3"n" - 1/2`
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Chapter 3: Algebra - Exercise 3.2 [Page 85]
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