Chapter 8.4, Problem 28E

To determine

To expand: and simplify the given expression using binomial theorem.

Answer to Problem 28E

The simplified form is $625-500y+150y^2-20y^3+y^4$ .

Explanation of Solution

Given information:

Given expression

  ${\left(y-5\right)}^4$

Formula used:

From binomial theorem expansion,

  ${\left(x+y\right)}^n=x^n+n{x}^{n-1}y+\mathrm{...}{+}_n{C}_r{x}^{n-r}{y}^r+\mathrm{...}+nx{y}^{n-1}+y^n$

And the coefficient of

  $x^{n-r}{y}^r$ is ${}_n{C}_r=\frac{n!}{\left(n-r\right)!r!}$

Calculation:

In the expansion of ${\left(x+y\right)}^n$

  ${\left(x+y\right)}^n=x^n+n{x}^{n-1}y+\mathrm{...}{+}_n{C}_r{x}^{n-r}{y}^r+\mathrm{...}+nx{y}^{n-1}+y^n$

Substitute $-5$ for $x$ and 4for $n$ , we get ${\left(-5+y\right)}^4$

From the above expansion

  $\begin{array}{c}{\left(-5+y\right)}^4{=}_4{C}_0{\left(-5\right)}^4{\left(y\right)}^0{+}_4{C}_1{\left(-5\right)}^3{\left(y\right)}^1{+}_4{C}_2{\left(-5\right)}^2{\left(y\right)}^2{+}_4{C}_3{\left(-5\right)}^1{\left(y\right)}^3{+}_4{C}_4{\left(-5\right)}^0{\left(y\right)}^4\\ =1{\left(-5\right)}^4{\left(y\right)}^0+4{\left(-5\right)}^3\left(y\right)+6{\left(-5\right)}^2{\left(y\right)}^2+4{\left(-5\right)}^1{\left(y\right)}^3+y^4\\ =625-500y+150y^2-20y^3+y^4\end{array}$