Why is the radical of a Clifford algebra generated by the kernel of the associated symmetric form?

Question

I was recently reading through Jacobson's Basic Algebra. I got to the section on Clifford algebras, and have the following question.

Let $Cl_\omega$ be the Clifford algebra with bilinear symmetric form $\omega$ on a vector space $V$.

I hear that the Jacobson radical $\text{rad}(Cl_\omega)$ is generated by $\ker\omega$, but it's not immediately clear to my why it is. How can one see this? Thanks.