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?

Let $p(x)q(x)\in P[x]+\langle x\rangle$

Now $p(x)q(x)\in P[x]+\langle x\rangle$

$\implies p(x)q(x)=f(x)+\langle x\rangle $ where $f(x)\in P[x]$

$\implies p(x)q(x)-f(x)\in \langle x\rangle$

$\implies p(0)q(0)=f(0)\in P $

Now since $P$ is Prime ideal $\implies $ either $p(0)\in P$ or $q(0)\in P$

Suppose $p(0)\in P$ then $p(x)=p(0)+xh(x) $ for some $h(x)\in R[x]$

Thus $p(x)\in P[x]+\langle x\rangle$