# Use the binomial theorem to expand the following, and simplify. (3x + y)^5

## Question:

Use the binomial theorem to expand the following, and simplify.

{eq}\left(3x + y\right)^5 {/eq}

## Expansion with Two Variables:

The mathematical expression that we are going to use in this particular problem to expand the function {eq}(a+b)^m {/eq} that contains two different varaibles a and b is shown below:

{eq}(a+b)^m=^{m}\textrm{C}_{0}a^mb^0+^{m}\textrm{C}_{1}a^{m-1}b^1+^{m}\textrm{C}_{2}a^{m-2}b^2+^{m}\textrm{C}_{3}a^{m-3}b^3+\dots\\ {/eq}, where,

• m shows the positive exponent.

The given function with two variables and the positive exponent is:

{eq}f(x, y)=\left(3x + y\right)^5 {/eq}

Comparing the above function with the function {eq}(a+b)^m {/eq}, we have:

{eq}a=3x\\ b=y\\ m=5 {/eq}

Using the above values and the formula of expansion for the given function, we get:

{eq}\begin{align*} (3x+y)^5&=^{5}\textrm{C}_{0}(3x)^5y^0+^{5}\textrm{C}_{1}(3x)^{5-1}y^1+^{5}\textrm{C}_{2}(3x)^{5-2}y^2+^{5}\textrm{C}_{3}(3x)^{5-3}y^3+^{5}\textrm{C}_{4}(3x)^{5-4}y^4+^{5}\textrm{C}_{5}(3x)^{5-5}y^5\\ &=\frac{5!}{(5-0)!0!}(3^5x^5)(1)+\frac{5!}{(5-1)!1!}(3x)^{4}y+\frac{5!}{(5-2)!2!}(3x)^{3}y^2+\frac{5!}{(5-3)!3!}(3x)^{2}y^3+\frac{5!}{(5-4)!4!}(3x)^{1}y^4+\frac{5!}{(5-5)!5!}(3x)^{0}y^5&\because ^{n}\textrm{C}_{r}=\frac{n!}{(n-r)!r!}\\ &=\frac{5!}{5!}(243x^5)+\frac{5!}{4!}(81x^{4}y)+\frac{5!}{3!2!}(27x^{3}y^2)+\frac{5!}{2!3!}(9x^{2}y^3)+\frac{5!}{1!4!}(3xy^4)+\frac{5!}{0!5!}(1)y^5\\ &=243x^5+\frac{5(5-1)!}{4!}(81x^{4}y)+\frac{5(5-1)(5-2)!}{3!(2\times 1)}(27x^{3}y^2)+\frac{5(5-1)(5-2)!}{(2\times 1)3!}(9x^{2}y^3)+\frac{5(5-1)!}{4!}(3xy^4)+y^5&\because m!=m(m-1)(m-2)\dots\\ &=243x^5+5(81x^{4}y)+10(27x^{3}y^2)+10(9x^{2}y^3)+5(3xy^4)+y^5\\ &=\boxed{243x^5+405x^{4}y+270x^{3}y^2+90x^{2}y^3+15xy^4+y^5} \end{align*} {/eq}