Chapter 6.2, Problem 56WE

To determine

To calculate: The simplified form of the expression $\frac{4n^2-4}{1+n^2}\cdot \frac{1-n}{2n}\cdot \frac{1-2n^2+n^4}{2+2n}$.

Answer to Problem 56WE

The simplified form of the expression $\frac{4n^2-4}{1+n^2}\cdot \frac{1-n}{2n}\cdot \frac{1-2n^2+n^4}{2+2n}$ is $-\frac{{\left(n-1\right)}^4{\left(n+1\right)}^2}{n\left(1+n^2\right)}$.

Explanation of Solution

Given information:

The expression $\frac{4n^2-4}{1+n^2}\cdot \frac{1-n}{2n}\cdot \frac{1-2n^2+n^4}{2+2n}$.

Formula used:

Fractions are multiplied as $\frac{a}{b}\cdot \frac{c}{d}=\frac{ab}{cd}$.

To factor a polynomial split the middle term in such a way that sum of two numbers is the middle term and product of two numbers is same is product of first and last term.

Calculation:

Consider the expression $\frac{4n^2-4}{1+n^2}\cdot \frac{1-n}{2n}\cdot \frac{1-2n^2+n^4}{2+2n}$.

Take the trinomial $1-2n^2+n^4$.

Recall that square of difference of two numbers a and bis ${\left(a-b\right)}^2=a^2+b^2-2ab$.

Therefore,

  $\begin{array}{c}1-2n^2+n^4=1-2\left(1\right)\left(n^2\right)+{\left(n^2\right)}^2\\ ={\left(n^2-1\right)}^2\end{array}$

Take the binomial $4n^2-4$.

Recall that difference of square of two numbers a and bis $a^2-b^2=\left(a-b\right)\left(a+b\right)$.

Apply it,

  $\begin{array}{c}\frac{4n^2-4}{1+n^2}\cdot \frac{1-n}{2n}\cdot \frac{1-2n^2+n^4}{2+2n}=\frac{{\left(2n\right)}^2-2^2}{1+n^2}\cdot \frac{1-n}{2n}\cdot \frac{{\left(n^2-1\right)}^2}{2+2n}\\ =\frac{\left(2n+2\right)\left(2n-2\right)}{1+n^2}\cdot \frac{1-n}{2n}\cdot \frac{{\left(n^2-1\right)}^2}{2+2n}\end{array}$

Strike off the similar terms,

  $\begin{array}{c}\frac{4n^2-4}{1+n^2}\cdot \frac{1-n}{2n}\cdot \frac{1-2n^2+n^4}{2+2n}=\frac{{\left(2n\right)}^2-2^2}{1+n^2}\cdot \frac{1-n}{2n}\cdot \frac{{\left(n^2-1\right)}^2}{2+2n}\\ =\frac{\left(2n+2\right)\left(2n-2\right)}{1+n^2}\cdot \frac{1-n}{2n}\cdot \frac{{\left[\left(n+1\right)\left(n-1\right)\right]}^2}{2+2n}\\ =\frac{\left(n-1\right)}{1+n^2}\cdot \frac{1-n}{n}\cdot \frac{{\left(n-1\right)}^2{\left(n+1\right)}^2}{1}\end{array}$

Recall that fractions are multiplied as $\frac{a}{b}\cdot \frac{c}{d}=\frac{ab}{cd}$.

Apply it,

  $\begin{array}{c}\frac{4n^2-4}{1+n^2}\cdot \frac{1-n}{2n}\cdot \frac{1-2n^2+n^4}{2+2n}=\frac{\left(n-1\right)}{1+n^2}\cdot \frac{1-n}{n}\cdot \frac{{\left(n-1\right)}^2{\left(n+1\right)}^2}{1}\\ =-\frac{\left(n-1\right)\left(n-1\right)}{n\left(1+n^2\right)}\cdot \frac{{\left(n-1\right)}^2{\left(n+1\right)}^2}{1}\\ =-\frac{{\left(n-1\right)}^4{\left(n+1\right)}^2}{n\left(1+n^2\right)}\end{array}$

Thus, the simplified form of the expression $\frac{4n^2-4}{1+n^2}\cdot \frac{1-n}{2n}\cdot \frac{1-2n^2+n^4}{2+2n}$ is $-\frac{{\left(n-1\right)}^4{\left(n+1\right)}^2}{n\left(1+n^2\right)}$.