Question

24. Simplify the complex fraction.

(5x^3/x+2)/30x^2/x+3)

A) x(x+3)/6(x+2)
B) 6x/ (x+2)(x+3)
C) 150x^5/(x+2)(x+3)
D) x+3/2

25. Divide the following rational expression,

16z^5 - 12z^4 + 8z^3 - 6z/2z^2

A) 8z^3 - 6z^2 + 8z - 3z
B) 8z^3 - 6z^2 + 4z - 3/z
C) 8z^3 + 6z^2 + 8z + 3z
D) 10z^5 - 3

Answer 1

@Directrix I'm about to get some sleep, would you mind taking over?

Answer 2

\[\large \frac{\dfrac{5x^3}{x+2}}{\dfrac{30x^2}{x+3}}\] Remember, when you're dividing by 2 fractions, follow this rule: \[\large \frac{\frac{a}{b}}{\frac{c}{d}} \implies \frac{a}{b} \cdot \frac{d}{c}\]

Answer 3

therefore, \( \frac{\dfrac{5x^3}{x+2}}{\dfrac{30x^2}{x+3}}\) becomes...\[=\frac{5x^3}{x+2} \cdot \frac{x+3}{30x^2}\]

Answer 4

Now we just cross cancel like terms, \[\frac{\color{red}{5}\color{blue}{x^3}}{x+2} \cdot \frac{x+3}{\color{red}{30}\color{blue}{x^2}}\]

Answer 5

For your second problem: \(\dfrac{16z^5 - 12z^4 + 8z^3 - 6z}{2z^2}\)

Divide each term in the numerator by the term in the denominator. \[\frac{\color{red}{16}\color{blue}{z^5}}{\color{red}{2}\color{blue}{z^2}} -\frac{\color{red}{12}\color{blue}{z^4}}{\color{reD}{2}\color{blue}{z^2}}+\frac{\color{red}{8}\color{blue}{z^3}}{\color{red}{2}\color{blue}{z^2}}-\frac{\color{red}{6}\color{blue}{z}}{\color{red}{2}\color{blue}{z^2}}\]

Answer 6

@Jhannybean Thanks, I've completed them. :)

Answer 7

Np :)