What is #-3x^4y^2 * -x^0y^-4 * 2x#?

1 Answer
Azimet
Jan 25, 2017

#6x^5y^-2#

Explanation:

Using the property of exponents, that #a^b * a^c = a^(b+c)#:

#-3x^2y^2 * -x^0y^-4 * 2x = -(-3) * 2 * x^(4 + 1 - 0) * y^(2-4)#

#= 6x^5y^-2# or #(6x^5)/y^2#. Note that the minus sign in front of #x^0# does matter, even if it's attached to #x^0#, which is equal to #1# anyway.

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