Chapter 9.5, Problem 29E

To determine

To expand: The expression ${\left(3-2z\right)}^4$ by using Pascal’s triangle to determine the coefficients.

Answer to Problem 29E

The expansion of the expression ${\left(3-2z\right)}^4$ is $16z^4-96z^3+216z^2-216z+81$ .

Explanation of Solution

Given information: The expression is ${\left(3-2z\right)}^4$ .

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

Calculation:

Here, $n=4$ .

First write out the Pascal’s triangle such that the row that begins with 1, 4.

  $\begin{array}{ccccccccc}& & & & 1& & & & \\ & & & 1& & 1& & & \\ & & 1& & 2& & 1& & \\ & 1& & 3& & 3& & 1& \\ 1& & 4& & 6& & 4& & 1\end{array}$

From the Pascal’s triangle, it is observed that the coefficients of the expression are 1, 4, 6, 4 and 1.

Expand the expression ${\left(3-2z\right)}^4$ by using formula in Binomial theorem as follows.

  $\begin{array}{c}{\left(3-2z\right)}^4=\left(1\right){\left(3\right)}^4+4{\left(3\right)}^3\left(-2z\right)+6{\left(3\right)}^2{\left(-2z\right)}^2+4{\left(3\right)}^1{\left(-2z\right)}^3+1{\left(-2z\right)}^4\\ =81-8\left(27\right)z+24\left(9\right)z^2-12z^3\left(8\right)+16z^4\\ =16z^4-96z^3+216z^2-216z+81\end{array}$

Thus, the expansion of the expression ${\left(3-2z\right)}^4$ is $16z^4-96z^3+216z^2-216z+81$ .