# How do you expand this binomial using binomial theorem? (2x+3)^10

${\left(2 x + 3\right)}^{10} = {\left(2 x\right)}^{10} + 10 C 1 {\left(2 x\right)}^{9} \cdot 3 + 10 C 2 {\left(2 x\right)}^{8} \cdot {3}^{2} + 10 C 3 {\left(2 x\right)}^{7} \cdot {3}^{3} + 10 C 4 {\left(2 x\right)}^{6} \cdot {3}^{4} + 10 C 5 {\left(2 x\right)}^{5} \cdot {3}^{5} + 10 C 6 {\left(2 x\right)}^{6} \cdot {3}^{6} + 10 C 7 {\left(2 x\right)}^{7} \cdot {3}^{7} + 10 C 8 {\left(2 x\right)}^{8} \cdot {3}^{8} + 10 C 9 {\left(2 x\right)}^{9} \cdot {3}^{9} + {3}^{10}$
${\left(2 x + 3\right)}^{10} = {\left(2 x\right)}^{10} + 10 C 1 {\left(2 x\right)}^{9} \cdot 3 + 10 C 2 {\left(2 x\right)}^{8} \cdot {3}^{2} + 10 C 3 {\left(2 x\right)}^{7} \cdot {3}^{3} + 10 C 4 {\left(2 x\right)}^{6} \cdot {3}^{4} + 10 C 5 {\left(2 x\right)}^{5} \cdot {3}^{5} + 10 C 6 {\left(2 x\right)}^{6} \cdot {3}^{6} + 10 C 7 {\left(2 x\right)}^{7} \cdot {3}^{7} + 10 C 8 {\left(2 x\right)}^{8} \cdot {3}^{8} + 10 C 9 {\left(2 x\right)}^{9} \cdot {3}^{9} + {3}^{10}$