Question

What is the product in simplest form? State any restrictions on the variable.

Answer 1

first take common terms from numerator and denominator and cancel them out

Answer 2

how? i been out of school for a long time so can u show me

Answer 3

ok do you know how to factorise a quadratic ?

Answer 4

yea

Answer 5

so try factorising quadratic in the numerator

Answer 6

@dietrich_harmon please reply ?

Answer 7

\[{{y^2}\over{y-3}}\times{{y^2-y-6}\over{y^2+y}}\]\[{\color{blue}{y(y)}\over{y-3}}\times{\color{blue}{(y-3)(y+2)}\over\color{blue}{y(y+1)}}\]\[{\cancel{y}^1 {(y)}\over\cancel{y-3}^1}\times{\cancel{(y-3)}^1{(y+2)}\over\cancel{y}^1{(y+1)}}\]\[{y\over1}\times{{y+2}\over{y+1}}\]\[={{y^2+2y}\over{y+1}}\]

Answer 8

or in simplest form it would be:\[{y(y+2)}\over{y+1}\]

Answer 9

were there any restrictions on this ?

Answer 10

yes, \(y\neq1\) because that would make the denominator \(0\) and so there is an asymptote at \(y=0\)