Answer
$(2x+3)^5=\binom{5}{0}(2x)^53^0+\binom{5}{1}(2x)^41^1+\binom{5}{2}(2x)^{3}3^2+\binom{5}{3}(2x)^{2}3^3+\binom{5}{4}(2x)^{1}3^4+\binom{5}{5}(2x)^{0}3^{5}=32x^5+240x^4+720x^3+1080x^2+810x+243$
Work Step by Step
The binomial theorem expands an algebraic expression in the form of:
$(x+y)^n=\binom{n}{0}x^ny^0+\binom{n}{1}x^{n-1}y^1+\binom{n}{2}x^{n-2}y^2+...+\binom{n}{i}x^{n-i}y^i+...+\binom{n}{n-1}x^{1}y^{n-1}+\binom{n}{n}x^0y^n=\sum_{i=0}^{n}\binom{n}{i}x^{n-i}y^i$
Here, x becomes 2x, $y=3; n=5$
$(2x+3)^5=\binom{5}{0}(2x)^53^0+\binom{5}{1}(2x)^41^1+\binom{5}{2}(2x)^{3}3^2+\binom{5}{3}(2x)^{2}3^3+\binom{5}{4}(2x)^{1}3^4+\binom{5}{5}(2x)^{0}3^{5}=32x^5+240x^4+720x^3+1080x^2+810x+243$
