Chapter 1.4, Problem 30E

To determine

To calculate: The product of the rational expression,

  $\frac{x^2+2xy+y^2}{x^2-y^2}\cdot \frac{2x^2-xy-y^2}{x^2-xy-2y^2}$

Answer to Problem 30E

The product of rational expression is $\frac{\left(2x+y\right)}{\left(x-2y\right)}$

Explanation of Solution

Given information:

The expression is given as:

  $\frac{x^2+2xy+y^2}{x^2-y^2}\cdot \frac{2x^2-xy-y^2}{x^2-xy-2y^2}$

Formula used:

For the rationalexpression:

Fractions property: $\frac{A}{B}\cdot \frac{C}{D}=\frac{AC}{BD}$

Product formula: $\left(A+B\right)\left(A-B\right)=A^2-B^2$

Factoring trinomials: The factor of rationalexpression which contain three terms is of the from $x^2+bx+c$ ,

  $\left(x+r\right)\left(x+s\right)=x^2+\left(r+s\right)x+rs$

Choose the values of $r$ and $s$ which satisfied these equations $r+s=b$ and $rs=c$

Calculation:

Consider therationalexpression,

  $\frac{\left(x^2+2xy+y^2\right)\left(2x^2-xy-y^2\right)}{\left(x^2-y^2\right)\left(x^2-xy-2y^2\right)}$

Use the fraction property,

  $\frac{A}{B}\cdot \frac{C}{D}=\frac{AC}{BD}$

Factor the terms in numerator by Factoring trinomials rule,

  $\begin{array}{l}\frac{\left(x^2+xy+xy+y^2\right)\left(2x^2-2xy+xy-y^2\right)}{\left(x^2-y^2\right)\left(x^2-xy-2y^2\right)}\\ \frac{\left(x+y\right)\left(x+y\right)\left(2x+y\right)\left(x-y\right)}{\left(x^2-y^2\right)\left(x^2-xy-2y^2\right)}\end{array}$

Factor the terms in denominator by Factoring trinomials rule and product property,

  $\begin{array}{l}\frac{\left(x+y\right)\left(x+y\right)\left(2x+y\right)\left(x-y\right)}{\left(x+y\right)\left(x-y\right)\left(x^2+xy-2xy-2y^2\right)}\\ \frac{\left(x+y\right)\left(x+y\right)\left(2x+y\right)\left(x-y\right)}{\left(x+y\right)\left(x-y\right)\left(x-2y\right)\left(x+y\right)}\end{array}$

Cancel common factors from the numerators and denominator,

  $\frac{\left(2x+y\right)}{\left(x-2y\right)}$

Thus, the product of rationalexpression is $\frac{\left(2x+y\right)}{\left(x-2y\right)}$