Consider the polynomial $ P(\xi)=\prod_{1\le j\le n}(\xi-\lambda_j). $ The inverse Fourier transform of $(\xi-\lambda_j)$ is $$ \int(\xi-\lambda_j) e^{2iπ x\xi} d\xi=(D_x-\lambda_j)(\delta_0)=\frac{\delta'_0}{2iπ}-\lambda_j\delta_0=T_j, \quad\text{support } T_j=\{0\}, $$ Let $F$ be the inverse Fourier transform of $P$: we have $$ F=T_1\ast\dots\ast T_n,\quad\text{support } F=\{0\},\quad \hat F= P. $$ If $\rho$ is smooth compactly supported, the function $F\ast \rho$ is smooth compactly supported and $$ \widehat{F\ast \rho}= P\hat \rho $$ and thus vanishes at the $\lambda_j$. There are generalizations using the Wierstrass factorization theorem to construct entire functions with prescribed zeroes.

Consider the polynomial $ P(\xi)=\prod_{1\le j\le n}(\xi-\lambda_j). $ The inverse Fourier transform of $(\xi-\lambda_j)$ is $$ \int(\xi-\lambda_j) e^{2iπ x\xi} d\xi=(D_x-\lambda_j)(\delta_0)=\frac{\delta'_0}{2iπ}-\lambda_j\delta_0=T_j, \quad\text{support } T_j=\{0\}, $$ Let $F$ be the inverse Fourier transform of $P$: we have $$ F=T_1\ast\dots\ast T_n,\quad\text{support } F=\{0\},\quad \hat F= P. $$ If $\rho$ is smooth compactly supported, the function $F\ast \rho$ is smooth compactly supported and $$ \widehat{F\ast \rho}= P\hat \rho $$ and thus vanishes at the $\lambda_j$. There are generalizations using the Weierstrass factorization theorem to construct entire functions with prescribed zeroes.