Chapter 12.6, Problem 44E

To determine

To find: The factor using Binomial Theorem ${\left(x-1\right)}^5+5{\left(x-1\right)}^4+10{\left(x-1\right)}^3+10{\left(x-1\right)}^2+5\left(x-1\right)+1$

Answer to Problem 44E

  $x^5$

Explanation of Solution

Given information:

  ${\left(x-1\right)}^5+5{\left(x-1\right)}^4+10{\left(x-1\right)}^3+10{\left(x-1\right)}^2+5\left(x-1\right)+1$

Calculation:

According to the Binomial theorem, expansion of ${\left(x+y\right)}^n$ with non-negative integer $n$ is:

  ${\left(x+y\right)}^n=\left(\begin{array}{l}n\\ 0\end{array}\right)x^n{y}^0+\left(\begin{array}{l}n\\ 1\end{array}\right)x^{n-1}{y}^1+\mathrm{.............}+\left(\begin{array}{l}n\\ n-1\end{array}\right)x^1{y}^{n-1}+\left(\begin{array}{l}n\\ n\end{array}\right)x^0{y}^n$

Where $n\ge 0$ , and $\left(\begin{array}{l}n\\ k\end{array}\right)$ denotes the binomial coefficient for $k=0,1,\mathrm{2........}n-1,n$

The number of terms in expansion of ${\left(x+y\right)}^n$ is $n+1$

Given

  ${\left(x-1\right)}^5+5{\left(x-1\right)}^4+10{\left(x-1\right)}^3+10{\left(x-1\right)}^2+5\left(x-1\right)+1$

Let the factor of ${\left(x-1\right)}^5+5{\left(x-1\right)}^4+10{\left(x-1\right)}^3+10{\left(x-1\right)}^2+5\left(x-1\right)+1$ be ${\left(a+b\right)}^n$

Here, the number of terms is $6$ .

  $\begin{array}{l}\Rightarrow n+1=6\\ \Rightarrow n=5\end{array}$

Then the factor becomes ${\left(a+b\right)}^5$ .

We know that, ${\left(a+b\right)}^5=a^5+\mathrm{.....}+b^5$

Comparing the first term,

  $\begin{array}{l}{a}^5={\left(x-1\right)}^5\\ \Rightarrow a=\sqrt[5]{{\left(x-1\right)}^5}\\ \Rightarrow a=x-1\end{array}$

Comparing the last term,

  $\begin{array}{l}{b}^5=1^5\\ \Rightarrow b=\sqrt[5]{1^5}\\ \Rightarrow b=1\end{array}$

Therefore,

  ${\left(a+b\right)}^5={\left(x-1+1\right)}^5=x^5$

The factor of ${\left(x-1\right)}^5+5{\left(x-1\right)}^4+10{\left(x-1\right)}^3+10{\left(x-1\right)}^2+5\left(x-1\right)+1$ is $x^5$