How do you multiply #\frac { 9y ^ { 2} + 6y + 1} { y ^ { 2} } \cdot \frac { 3y } { 3y + 1}#?

1 Answer
bp
Feb 1, 2017

#(9y+3)/y# or, 9 + #3/y#

Explanation:

Before multiplying, it would be wiser to simplify the expression. The quadratic expression in the numerator is #(3y+1)^2#. Now cancel out the common factors to get:
#(3y+1) /y * 3/1#

#(9y+3)/y#

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