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Search for a charged partner of the<mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" display="inline"><mml:mi>X</mml:mi><mml:mo stretchy="false">(</mml:mo><mml:mn>3872</mml:mn><mml:mo stretchy="false">)</mml:mo></mml:math>in the<mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" display="inline"><mml:mi>B</mml:mi></mml:math>meson decay<mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" display="inline"><mml:mi>B</mml:mi><mml:mo>→</mml:mo><mml:msup><mml:mi>X</mml:mi><mml:mo>−</…

Search for a charged partner of the<mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" display="inline"><mml:mi>X</mml:mi><mml:mo stretchy="false">(</mml:mo><mml:mn>3872</mml:mn><mml:mo stretchy="false">)</mml:mo></mml:math>in the<mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" display="inline"><mml:mi>B</mml:mi></mml:math>meson decay<mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" display="inline"><mml:mi>B</mml:mi><mml:mo>→</mml:mo><mml:msup><mml:mi>X</mml:mi><mml:mo>−</…

We search for a charged partner of the $X(3872)$ in the decay $B\ensuremath{\rightarrow}{X}^{\ensuremath{-}}K$, ${X}^{\ensuremath{-}}\ensuremath{\rightarrow}J/\ensuremath{\psi}{\ensuremath{\pi}}^{\ensuremath{-}}{\ensuremath{\pi}}^{0}$, using $234\ifmmode\times\else\texttimes\fi{}{10}^{6}$ $B\overline{B}$ events collected at the $\ensuremath{\Upsilon}(4S)$ resonance with the BABAR detector at the PEP-II ${e}^{+}{e}^{\ensuremath{-}}$ asymmetric-energy storage ring. The resulting product branching fraction upper limits are $\mathcal{B}({B}^{0}\ensuremath{\rightarrow}{X}^{\ensuremath{-}}{K}^{+},{X}^{\ensuremath{-}}\ensuremath{\rightarrow}J/\ensuremath{\psi}{\ensuremath{\pi}}^{\ensuremath{-}}{\ensuremath{\pi}}^{0})&lt;5.4\ifmmode\times\else\texttimes\fi{}{10}^{\ensuremath{-}6}$ and $\mathcal{B}({B}^{\ensuremath{-}}\ensuremath{\rightarrow}{X}^{\ensuremath{-}}{\overline{K}}^{0},{X}^{\ensuremath{-}}\ensuremath{\rightarrow}J/\ensuremath{\psi}{\ensuremath{\pi}}^{\ensuremath{-}}{\ensuremath{\pi}}^{0})&lt;22\ifmmode\times\else\texttimes\fi{}{10}^{\ensuremath{-}6}$ at the $90%$ confidence level.