Question

Use the Binomial Theorem to expand${\left(3x-2y\right)}^5$

Short answer

The expansion of the given binomial is${\left(3x-2y\right)}^5=243x^5-810x^{4}y+1080x^3{y}^2-720x^2{y}^3+240x{y}^4-32y^5$

Step-by-step solution

Step 1. Given information.

The given expression is ${\left(3x-2y\right)}^5$.

Step 2. Substitute the values and simplify.  

Evaluate the coefficients.

Remember,

$$\begin{gathered} a=3x \\ b=-2y \\ c=5 \end{gathered}$$

Now put the values,

${\left(3x-2y\right)}^5=\left(\begin{array}{c}5\\ 0\end{array}\right){\left(3x\right)}^5+\left(\begin{array}{c}5\\ 1\end{array}\right){\left(3x\right)}^{5-1}\left(-2y\right)+\left(\begin{array}{c}5\\ 2\end{array}\right){\left(3x\right)}^{5-2}{\left(-2y\right)}^2+\left(\begin{array}{c}5\\ 3\end{array}\right){\left(3x\right)}^{5-3}{\left(-2y\right)}^3+\left(\begin{array}{c}5\\ 4\end{array}\right){\left(3x\right)}^{5-4}{\left(-2y\right)}^4+\left(\begin{array}{c}5\\ 5\end{array}\right){\left(-2y\right)}^5$

Simplify the exponents

${\left(3x-2y\right)}^5=\left(\begin{array}{c}5\\ 0\end{array}\right){\left(3x\right)}^5+\left(\begin{array}{c}5\\ 1\end{array}\right){\left(3x\right)}^4\left(-2y\right)+\left(\begin{array}{c}5\\ 2\end{array}\right){\left(3x\right)}^3{\left(-2y\right)}^2+\left(\begin{array}{c}5\\ 3\end{array}\right){\left(3x\right)}^2{\left(-2y\right)}^3+\left(\begin{array}{c}5\\ 4\end{array}\right)\left(3x\right){\left(-2y\right)}^4+\left(\begin{array}{c}5\\ 5\end{array}\right){\left(-2y\right)}^5$

Remember that,

$\left(\begin{array}{c}n\\ n\end{array}\right)=1,\left(\begin{array}{c}n\\ 1\end{array}\right)=n,\left(\begin{array}{c}n\\ 0\end{array}\right)=1$

Simplify the coefficients.

${\left(3x-2y\right)}^5=1{\left(3x\right)}^5+5{\left(3x\right)}^4\left(-2y\right)+\frac{5!}{2!3!}{\left(3x\right)}^3{\left(-2y\right)}^2+\frac{5!}{3!2!}{\left(3x\right)}^2{\left(-2y\right)}^3+\frac{5!}{4!1!}\left(3x\right){\left(-2y\right)}^4+{\left(-2y\right)}^5$

Simplify,$243x^5+5.81x^4\left(-2y\right)+10.27x^3\left(4y^2\right)+10.9x^2\left(-8y^3\right)+5.3x\left(16y^4\right)+\left(-32y^5\right)$

Step 3. Conclusion.

Therefore, the expansion of the given expression is${\left(3x-2y\right)}^5=243x^5-810x^{4}y+1080x^3{y}^2-720x^2{y}^3+240x{y}^4-32y^5$