Show that $P[x]+\langle x\rangle $ is a prime ideal of $R[x]$. 
Show that if $P$ is a prime ideal of a  commutative ring $R$ with unity then $P[x]+\langle x\rangle $ is a prime ideal of $R[x]$.
Here $P[x]$ consists of all polynomials whose coefficients are in $P$.

I tried to show it using the fact that if $f(x)g(x)\in P[x]+\langle x\rangle $ then
either $f(x)\in P[x]+\langle x\rangle $ or $g(x)\in P[x]+\langle x\rangle $.
But I don't know how to show it.
Please help me out.

Consider the map
$$
\varphi\colon R[x]\to R/P, \qquad \varphi(f)=f(0)+P
$$
and prove it is a surjective ring homomorphism.
What's its kernel?
