Question

Expand ${(3x^2+y)}^5$.

Short answer

The value of${(3x^2+y)}^5$is$y^5+15x^2{y}^4+90x^4{y}^3+270x^6{y}^2+405x^{8}y+243x^{10}$.

Step-by-step solution

Step 1.  Given information.

The given expression is ${(3x^2+y)}^5$.

Step 2. Expand the given expression.

Using the formula ${(a+b)}^n=\sum _{x=0}^n\left(\begin{array}{c}n\\ x\end{array}\right){\left(a\right)}^{n-x}{\left(b\right)}^x$

${(3x^2+y)}^5=\sum _{x=0}^5\left(\begin{array}{c}5\\ x\end{array}\right){\left(3x^2\right)}^{5-x}{\left(y\right)}^x$

$$\begin{gathered} =\left(\begin{array}{c}5\\ 0\end{array}\right){\left(3x^2\right)}^{5-0}{\left(y\right)}^0+\left(\begin{array}{c}5\\ 1\end{array}\right){\left(3x^2\right)}^{5-1}{\left(y\right)}^1+\left(\begin{array}{c}5\\ 2\end{array}\right){\left(3x^2\right)}^{5-2}{\left(y\right)}^2+\left(\begin{array}{c}5\\ 3\end{array}\right){\left(3x^2\right)}^{5-3}{\left(y\right)}^3+\left(\begin{array}{c}5\\ 4\end{array}\right){\left(3x^2\right)}^{5-4}{\left(y\right)}^4+\left(\begin{array}{c}5\\ 5\end{array}\right){\left(3x^2\right)}^{5-5}{\left(y\right)}^5 \\ =1{\left(3x^2\right)}^5+5{\left(3x^2\right)}^4{\left(y\right)}^1+10{\left(3x^2\right)}^3{\left(y\right)}^2+10{\left(3x^2\right)}^2{\left(y\right)}^3+5{\left(3x^2\right)}^1{\left(y\right)}^4+1{\left(3x^2\right)}^0{\left(y\right)}^5 \\ =243x^{10}+405x^{8}y+270x^6{y}^2+90x^4{y}^3+15x^2{y}^4+y^5 \end{gathered}$$