How do you express the quotient of #(3x^2+13x+4)/((3x+1)/5)# in simplest form?

1 Answer
José F.
Apr 15, 2016

#5x+20, x!=-1/3#

Explanation:

#(3x^2+13x+4)/((3x+1)/5)=(3x^2+13x+4)-:(3x+1)/5=(3x^2+13x+4)xx5/(3x+1)#

#=5((3x^2+12x)+(x+4))/(3x+1)=5(3x(x+4)+(x+4))/(3x+1)=5(cancel((3x+1))(x+4))/cancel(3x+1)=5x+20, x!=-1/3#

When #3x+1=0# you obtain #0/0# in the full formula, while in simplified it is #55/3#. The statement #x!=-1/3# must be included to guarantee both full and simplified expressions are equivalent in their domains.

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