Answer:
the product of given expression [tex]\frac{y^2}{y-3}[/tex] and [tex]\frac{y^2-y-6}{y^2+y}[/tex] is [tex]\frac{y(y+2)}{(y+1)},y\neq -1[/tex]
Step-by-step explanation:
Consider the given expression [tex]\frac{y^2}{y-3}[/tex] and [tex]\frac{y^2-y-6}{y^2+y}[/tex]
We have to write the product of given expressions in simplest form,
We solve the two terms separately,
Consider the second expression
[tex]\frac{y^2-y-6}{y^2+y}[/tex]
Taking y common from denominator we have [tex]\frac{y^2-y-6}{y(y+1)}[/tex]
Numerator consist of a quadratic equation, [tex]{y^2-y-6}[/tex]
We can solve it using middle term splitting method,
-y can be written as 2y -3y
Thus, it becomes,
[tex]{y^2-y-6}=y^2+2y-3y-6[/tex]
Taking y common from first two terms and -3 common from last two terms we have,
[tex]y^2+2y-3y-6=y(y+2)-3(y+2)=(y-3)(y+2)[/tex]
thus, the second expression becomes ,
[tex]\frac{y^2-y-6}{y(y+1)}=\frac{(y-3)(y+2)}{y(y+1)}[/tex]
Now , we multiply the two expression ,we get,
[tex]\Rightarrow \frac{y^2}{y-3}\times \frac{y^2-y-6}{y^2+y}[/tex]
[tex]\Rightarrow \frac{y^2}{y-3}\times \frac{(y-3)(y+2)}{y(y+1)} [/tex]
Simplifying , we get,
[tex]\Rightarrow y \times \frac{(y+2)}{(y+1)}\\\\\\ \Rightarrow\frac{y(y+2)}{(y+1)},y\neq -1[/tex]
Thus, the product of given expression [tex]\frac{y^2}{y-3}[/tex] and [tex]\frac{y^2-y-6}{y^2+y}[/tex] is [tex]\frac{y(y+2)}{(y+1)},y\neq -1[/tex]
