Hence, the product is:
[tex]\dfrac{z(z+2)}{z+3}[/tex] such that: z≠ -1,0 and -3.
We are asked to represent the product in the simplest form along with the restrictions applied to z.
We have to evaluate the expression:
[tex]\dfrac{z^2}{z+1}\times \dfrac{z^2+3z+2}{z^2+3z}\\\\=\dfrac{z^2}{z+1}\times \dfrac{z^2+3z+2}{z(z+3)}[/tex]
Hence,
z≠ -1,0 and -3.
Since, otherwise the denominator will be equal to zero and hence the product will not be defined.
Now, we know that:
[tex]z^2+3z+2=z^2+2z+z+2\\\\z^2+3z+2=z(z+2)+1(z+2)\\\\z^2+3z+2=(z+1)(z+2)[/tex]
Hence,
[tex]\dfrac{z^2}{z+1}\times \dfrac{z^2+3z+2}{z^2+3z}=\dfrac{z^2}{z+1}\times \dfrac{(z+1)(z+2)}{z(z+3)}\\\\=\dfrac{z(z+2)}{z+3}[/tex]
( since z and (z+1) term is cancelled as it was same in numerator and denominator)
Hence, the product is:
[tex]\dfrac{z(z+2)}{z+3}[/tex] such that: z≠ -1,0 and -3.
