Complete the working below to show that [tex]\frac{12 x^4}{9 x^2-36} \div \frac{4 x^5}{3(x-2)}[/tex] simplifies to [tex] \frac{1}{x(x+p)}[/tex] where [tex]p[/tex] is an integer.
[tex]\begin{array}{l}
\frac{3}{\square} \\
\frac{1}{\square}
\end{array}[/tex]
Answer (1)
Rewrite the division as multiplication by the reciprocal.
Factor the denominator 9 x 2 − 36 as 9 ( x − 2 ) ( x + 2 ) .
Cancel common factors in the expression.
Simplify the expression to obtain x ( x + 2 ) 1 , thus p = 2 , and the final answer is 2 .
Explanation
Understanding the Problem We are given the expression 9 x 2 − 36 12 x 4 ÷ 3 ( x − 2 ) 4 x 5 and we want to simplify it to the form x ( x + p ) 1 where p is an integer. Our goal is to find the value of p .
Rewriting the Division First, we rewrite the division as multiplication by the reciprocal: 9 x 2 − 36 12 x 4 ÷ 3 ( x − 2 ) 4 x 5 = 9 x 2 − 36 12 x 4 × 4 x 5 3 ( x − 2 ) .
Factoring the Denominator Next, we factor the denominator 9 x 2 − 36 . We can factor out a 9: 9 x 2 − 36 = 9 ( x 2 − 4 ) . Then, we recognize x 2 − 4 as a difference of squares, so we can factor it as x 2 − 4 = ( x − 2 ) ( x + 2 ) . Therefore, 9 x 2 − 36 = 9 ( x − 2 ) ( x + 2 ) .
Substituting the Factored Form Now, we substitute the factored form into the expression: 9 ( x − 2 ) ( x + 2 ) 12 x 4 × 4 x 5 3 ( x − 2 ) .
Simplifying the Expression We simplify the expression by canceling common factors. First, we simplify the constants: 9 × 4 12 × 3 = 36 36 = 1. Next, we simplify the x terms: x 5 x 4 = x 1 . Then, we cancel the ( x − 2 ) terms: x − 2 x − 2 = 1. So, the expression becomes: 1 1 × x 1 × x + 2 1 = x ( x + 2 ) 1 .
Determining the Value of p Finally, we compare the simplified expression x ( x + 2 ) 1 with the target form x ( x + p ) 1 to determine the value of p . We see that p = 2 .
Conclusion Therefore, the simplified expression is x ( x + 2 ) 1 , and p = 2 .
Examples
Understanding how to simplify rational expressions is crucial in many areas of mathematics and physics. For example, when analyzing electrical circuits, you might encounter complex fractions involving impedances. Simplifying these fractions helps in determining the overall behavior of the circuit and designing it effectively. Similarly, in fluid dynamics, simplifying expressions involving flow rates and pressures can aid in understanding the behavior of fluids in various systems.
