Why these are isomorphic each other for given these rings? For the ring $\mathbb{Z}_{15}[x]$ and its ideal $<3x^2 + 5x>$
Find the order of the $\mathbb{Z}_{15}[x]/<3x^2 + 5x>$

In my answer sheet it said 
$\mathbb{Z}_{15}[x]/<3x^2 + 5x> \simeq (\mathbb{Z}_{3}[x]/<5x>) \times (\mathbb{Z}_{5}[x]/<3x^2>)$
Hence the order is 75. 
I don't understand Why does $\mathbb{Z}_{15}[x]/<3x^2 + 5x> \simeq (\mathbb{Z}_{3}[x]/<5x>) \times (\mathbb{Z}_{5}[x]/<3x^2>)$. 
Why those are isomorphic each other?
 A: As Chris hinted, it is easy to lift up  $\,\Bbb Z_{15} \cong \Bbb Z_3\times \Bbb Z_5\,$ by CRT. Let's examine the idea more closely.
Notice  in $\,R = \Bbb Z/15\!:\   (3)+(5)=(1)\,\Rightarrow\, (3)\cap (5) = (3)(5) = (0)$ $\smash{\overset{\small\rm CRT}\Rightarrow}\, R^{\phantom{|^|}}\!\!\! \cong R/3\times R/5$
The above ideal equalities extend to $\,E = R[x]/(3x^3+5x),\,$ thus also $\smash{\overset{\small\rm CRT}\Rightarrow}\, E^{\phantom{|^|}}\!\!\! \cong E/3\times E/5$
Ring isomorphism theorems $\Rightarrow E/3  \cong \Bbb Z_3[x]/5x,\,$ $\,E/5  \cong \Bbb Z_5[x]/3x^2$

For variety here's another way: we apply CRT in $R=\Bbb Z_{\color{#c00}{15}}[x]\,$  with $\,I+J=(3,5x) + (5,3x^2) \supseteq (5,3)= (1).\,$ By ring isomorphism theorems $\,R/I = R/(3,5x)^{\phantom{|^|}}\!\!\! \cong \Bbb Z_3[x]/(5x),\,$ $\,R/J = R/(5,3x^2) \cong \Bbb Z_5[x]/(3x^2)$ 
$I\!+\!J=(1)\,\Rightarrow\,I\cap J = IJ = (3,5x)(5,3x^2) =(9x^2,-5x)= (3x^2,5x)\,$ by $\,2(9x^2)=3x^2$
hence $\,IJ =(3x^2,5x)=(3x^2\!+\!5x)\ $ by $\ (6,-5)(3x^2\!+\!5x)=(3x^2,5x)\,$ by $\,\color{#c00}{15=0}\,$ in $\,R$.
Conclude $\ R/(3x^2\!+\!5x) = R/(I\cap J)\overset{\rm\small CRT_{\phantom |}\!} = R/I\times R/J = \Bbb Z_3[x]/(5x)\times \Bbb Z_5[x]/(3x^2)$
