Question

For the following exercises, find $R\left(x\right)=\frac{f\left(x\right)}{g\left(x\right)}$where $f\left(x\right)$and $g\left(x\right)$are given.

role="math" localid="1646212760399" $$\begin{gathered} f\left(x\right)=\frac{24x^2}{2x+8} \\ g\left(x\right)=\frac{4x^3+28x^2}{x^2+11x+28} \end{gathered}$$

Short answer

The required rational expression is$\frac{3(x+4)}{(x-4)}$.

Step-by-step solution

Step 1. Given information

The given expressions are$$\begin{gathered} f\left(x\right)=\frac{24x^2}{2x+8} \\ g\left(x\right)=\frac{4x^3+28x^2}{x^2+11x+28} \end{gathered}$$

Step 2. Rewrite the Expression as a Product 

• Substitute the given values into $R\left(x\right)=\frac{f\left(x\right)}{g\left(x\right)}$.

$R\left(x\right)=\frac{{\displaystyle \frac{24x^2}{2x-8}}}{\frac{4x^3+28x^2}{x^2+11x+28}}$

• Write the expression of $R\left(x\right)$as the product of numerator and the reciprocal of denominator.

$R\left(x\right)=\frac{24x^2}{2x-8}\times \frac{x^2+11x+28}{4x^3+28x^2}$

Step 3. Factor the Numerator and Denomiantor 

• Express the numerator and denominator of the obtained expression as factors.

$R\left(x\right)=\frac{6\times 4\times x^2}{2(x-4)}\times \frac{(x+7)(x+4)}{4\times x^2\times (x+7)}$

Step 4. Cancel out the Common Factors 

• Simplify the obtained expression by canceling out the common factors.

$\begin{array}{rcl}R\left(x\right)& =& \frac{\cancel{2}\times 3\times \cancel{4}\times \cancel{x^2}}{\cancel{2}(x-4)}\times \frac{\cancel{(x+7)}(x+4)}{\cancel{4}\times \cancel{x^2}\times \cancel{(x+7)}}\\ & =& \frac{3}{x-4}\times x+4\\ & =& \frac{3(x+4)}{(x-4)}\end{array}$

• So, the simplified expression is$R\left(x\right)=\frac{3(x+4)}{(x-4)}$.