Tensor product of exact complexes is exact

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Let $M_\circ = \dots \to M_n \dots \to M_0 \to 0$ and $N_\circ = \dots \to N_n \dots \to N_0 \to 0$ be exact complexes of modules over a ring $A$ such that each module is flat.

Is it then true that $(M\otimes N)_\circ = \dots M_n\otimes_AN_n \dots \to M_0\otimes_A N_0 \to 0$ is exact?

I can get an exact double complex but I don't know how to use that to conclude what I want.

(The motivation is to show that if $P_\circ$ and $Q_\circ$ are polynomial simplicial resolutions of rings $B,C$ flat over $A$, then $P_\circ\otimes_AQ_\circ$ is a polynomial simplicial resolution of $B \otimes_A C$.

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No. For instance, let both complexes be $0\to A\to A\to 0$ but with one of them in degrees $0$ and $1$ and the other in degrees $1$ and $2$. Then the tensor product will be nonzero only in degree $1$ and so will not be exact.