How do you find the quotient #(6x^2+10x)div2x#?

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Answer 1 · 2017-02-25T23:16:44

See the entire solution process below:

First, rewrite this expression as:

#(6x^2 + 10x)/(2x)#

Next, factor #color(red)(2x)# from the term in the numerator:

#((color(red)(2x) xx 3x) + (color(red)(2x) xx 5))/(2x)#

#(color(red)(2x)(3x + 5))/(2x)#

Then, cancel the common terms in the numerator and denominator:

#(cancel(color(red)(2x))(3x + 5))/color(red)(cancel(color(black)(2x)))#

#3x + 5#, when #2x != 0# or #x != 0#