How do you multiply #x^ { 0} y ^ { 2} \cdot 3y x ^ { 5}#?

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Answer 1 · 2018-05-07T11:42:22

#3*x^5*y^3#

#x^0y^2*3yx^5=#

#3*x^(0+5)*y^(2+1)=#

#3*x^5*y^3#

Answer 2 · 2018-05-07T11:43:49

#=3y^3x^5#

Using the power rule:

#a^n*a^m=a^(n+m)#

#x^0*x^5=x^5#

(where #x^0# is also equivalent to #1#)

#y^2*3y=y^2*3y^1=3y^3#

Therefore, the solution to #x^0y^2*3yx^5# is:

#=3y^3x^5#