Chapter 6.2, Problem 43WE

To determine

To calculate: The simplified form of the expression $\frac{n^2-3n+2}{n^2+3n+2}\cdot \frac{8n+8}{4n+8}$.

Answer to Problem 43WE

The simplified form of the expression $\frac{n^2-3n+2}{n^2+3n+2}\cdot \frac{8n+8}{4n+8}$ is $\frac{2\left(n-1\right)\left(n-2\right)}{{\left(n+2\right)}^2}$.

Explanation of Solution

Given information:

The expression $\frac{n^2-3n+2}{n^2+3n+2}\cdot \frac{8n+8}{4n+8}$.

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 $\frac{n^2-3n+2}{n^2+3n+2}\cdot \frac{8n+8}{4n+8}$.

Take the trinomial $n^2-3n+2$.

Recall that 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.

For the above polynomial $-2$ and $-1$ are two numbers such that their sum is $-3$ and product is $2$.

Therefore, the trinomial is rewritten as,

  $n^2-3n+2=n^2-2n-n+2$

Take out the common terms,

  $\begin{array}{c}{n}^2-3n+2=n^2-2n-n+2\\ =n\left(n-2\right)-1\left(n-2\right)\\ =\left(n-2\right)\left(n-1\right)\end{array}$

Take the trinomial $n^2+3n+2$.

Recall that 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.

For the above polynomial $2$ and $1$ are two numbers such that their sum is $3$ and product is $2$.

Therefore, the trinomial is rewritten as,

  $n^2+3n+2=n^2+2n+n+2$

Take out the common terms,

  $\begin{array}{c}{n}^2+3n+2=n^2+2n+n+2\\ =n\left(n+2\right)+1\left(n+2\right)\\ =\left(n+2\right)\left(n+1\right)\end{array}$

Take out the common factor that is $8$ from the numerator of second term and 4 from denominator of second term,

  $\frac{n^2-3n+2}{n^2+3n+2}\cdot \frac{8n+8}{4n+8}=\frac{\left(n-2\right)\left(n-1\right)}{\left(n+2\right)\left(n+1\right)}\cdot \frac{8\left(n+1\right)}{4\left(n+2\right)}$

Strike off the similar terms,

  $\begin{array}{c}\frac{n^2-3n+2}{n^2+3n+2}\cdot \frac{8n+8}{4n+8}=\frac{\left(n-2\right)\left(n-1\right)}{\left(n+2\right)\left(n+1\right)}\cdot \frac{8\left(n+1\right)}{4\left(n+2\right)}\\ =\frac{\left(n-2\right)\left(n-1\right)}{\left(n+2\right)}\cdot \frac{2}{\left(n+2\right)}\end{array}$

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

Apply it,

  $\begin{array}{c}\frac{n^2-3n+2}{n^2+3n+2}\cdot \frac{8n+8}{4n+8}=\frac{\left(n-2\right)\left(n-1\right)}{\left(n+2\right)\left(n+1\right)}\cdot \frac{8\left(n+1\right)}{4\left(n+2\right)}\\ =\frac{\left(n-2\right)\left(n-1\right)}{\left(n+2\right)}\cdot \frac{2}{\left(n+2\right)}\\ =\frac{2\left(n-1\right)\left(n-2\right)}{{\left(n+2\right)}^2}\end{array}$

Thus, the simplified form of the expression $\frac{n^2-3n+2}{n^2+3n+2}\cdot \frac{8n+8}{4n+8}$ is $\frac{2\left(n-1\right)\left(n-2\right)}{{\left(n+2\right)}^2}$.