To show that the expression [tex]\(\frac{6 x^3}{9 x^2 - 144} \div \frac{2 x^4}{3(x - 4)}\)[/tex] can be written in the form [tex]\(\frac{1}{x(x + r)}\)[/tex] where [tex]\(r\)[/tex] is an integer, let's go through the steps together.
1. Rewrite the division as multiplication by the reciprocal:
[tex]\[
\frac{6 x^3}{9 x^2 - 144} \div \frac{2 x^4}{3(x - 4)} = \frac{6 x^3}{9 x^2 - 144} \times \frac{3 (x - 4)}{2 x^4}
\][/tex]
2. Simplify the constants separately:
[tex]\[
\frac{6}{2} = 3 \quad \text{and} \quad \frac{3}{1} = 3
\][/tex]
3. Combine the expressions:
[tex]\[
3 \times \frac{x^3}{9 x^2 - 144} \times \frac{x - 4}{x^4}
\][/tex]
4. Factorize the denominator in the first fraction:
[tex]\[
9 x^2 - 144 = 9 (x^2 - 16) = 9 (x - 4)(x + 4)
\][/tex]
Thus, the fraction becomes:
[tex]\[
\frac{x^3}{9 (x - 4)(x + 4)}
\][/tex]
5. Plug the factorized form back into the multiplication:
[tex]\[
3 \times \frac{x^3}{9 (x - 4)(x + 4)} \times \frac{x - 4}{x^4}
\][/tex]
6. Cancel common factors:
Cancel [tex]\(x - 4\)[/tex]:
[tex]\[
3 \times \frac{x^3}{9 (x + 4)} \times \frac{1}{x^4} = 3 \times \frac{x^3}{9 (x + 4) x^4}
\][/tex]
7. Simplify the fraction:
[tex]\[
3 \times \frac{x^3}{9 x^4 (x + 4)} = \frac{3 x^3}{9 x^4 (x + 4)} = \frac{3}{9 x (x + 4)} = \frac{1}{3 x (x + 4)}
\][/tex]
8. Cancel the common factor of 3:
[tex]\[
\frac{1}{3 x (x + 4)}
\][/tex]
Now as you can see, the expression [tex]\(\frac{6 x^3}{9 x^2 - 144} \div \frac{2 x^4}{3 (x - 4)}\)[/tex] simplifies to [tex]\(\frac{1}{x (x + 4)}\)[/tex], where [tex]\(r = 4\)[/tex].
Therefore, we have shown that [tex]\(\frac{6 x^3}{9 x^2 - 144} \div \frac{2 x^4}{3 (x - 4)}\)[/tex] can indeed be written in the form [tex]\(\frac{1}{x (x + r)}\)[/tex] with [tex]\(r = 4\)[/tex].
