Chapter 9.5, Problem 20E

To determine

To expand: The expression ${\left(y-2\right)}^5$ by using the Binomial Theorem.

Answer to Problem 20E

The expansion of the expression ${\left(y-2\right)}^5$ is $y^5-10y^4+40y^3-80y^2+80y-32$ .

Explanation of Solution

Given information: The expression is ${\left(y-2\right)}^5$ .

Theorem used:Binomial Theorem: ${\left(x+y\right)}^n=x^n+n{x}^{n-1}y+\cdots +nx{y}^{n-1}+y^n$ .

Calculation:

Here, $n=3$ .

First write out the Pascal’s triangle such that the row that begins with 1, 5.

  $\begin{array}{ccccccccccc}& & & & & 1& & & & & \\ & & & & 1& & 1& & & & \\ & & & 1& & 2& & 1& & & \\ & & 1& & 3& & 3& & 1& & \\ & 1& & 4& & 6& & 4& & 1& \\ 1& & 5& & 10& & 10& & 5& & 1\end{array}$

From the Pascal’s triangle, it is observed that the binomial coefficients of the expression are 1, 5, 10, 10, 5 and 1.

Expand the expression ${\left(y-2\right)}^5$ by using formula in Binomial theorem as follows.

  $\begin{array}{c}{\left(y-2\right)}^5=\left(1\right)y^5+5y^{5-1}\left(-2\right)+10y^{5-2}{\left(-2\right)}^2+10y^{5-3}{\left(-2\right)}^3+5y^{5-4}{\left(-2\right)}^4+\left(1\right){\left(-2\right)}^5\\ =y^5-10y^4+10y^3\left(4\right)+10y^2\left(-8\right)+5y\left(16\right)-32\\ =y^5-10y^4+40y^3-80y^2+80y-32\end{array}$

Thus, the expansion of the expression ${\left(y-2\right)}^5$ is $y^5-10y^4+40y^3-80y^2+80y-32$ .