Chapter 11.4, Problem 9MCQ

To determine

Multiply or divide $\frac{x^2-4}{x+3}\cdot \frac{x^2+7x+12}{x-2}$

Answer to Problem 9MCQ

The solution of the given equation is $\frac{x^4+7x^3+8x^2-28x-48}{x^2+x-6}$

Explanation of Solution

Given:

Given equation

  $\frac{x^2-4}{x+3}\cdot \frac{x^2+7x+12}{x-2}$

Concept Used:

• Least Common Denominator (LCD) is the smallest number that is divisible by the denominator of all the fractions that are being added or subtracted.
• Solving an equation containing rational expressions
• Multiply both sides of equations by LCD(least common denominator) of all denominators.
• Remove any grouping symbols and solve the resulting equation for the variable asked in equation

Calculation:

In order to solve the equation $\frac{x^2-4}{x+3}\cdot \frac{x^2+7x+12}{x-2}$ , observe that it is function of multiply, so simplify further as shown below

  $\begin{array}{c}\frac{x^2-4}{x+3}\cdot \frac{x^2+7x+12}{x-2}=\frac{\left(x^2-4\right)\cdot \left(x^2+7x+12\right)}{\left(x+3\right)\cdot \left(x-2\right)}\\ =\frac{x^2\left(x^2+7x+12\right)\left(-4\right)\left(x^2+7x+12\right)}{x\cdot x+x\cdot \left(-2\right)+3\cdot x+3\cdot \left(-2\right)}\\ =\frac{x^4+7x^3+12x^2-4x^2-28x-48}{x^2-2x+3x-6}\\ =\frac{x^4+7x^3+8x^2-28x-48}{x^2+x-6}\end{array}$

Thus, the solution of the given equation is $\frac{x^4+7x^3+8x^2-28x-48}{x^2+x-6}$