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divide. 7) \\(\frac{24x^3 + 24x^2 + 24x}{6x^3}\\) 8) \\(\frac{6k^5 + k^…

Question

divide.

  1. \\(\frac{24x^3 + 24x^2 + 24x}{6x^3}\\)
  2. \\(\frac{6k^5 + k^4 + 6k^3}{6k^3}\\)
  1. \\(\frac{2n^3 + 8n^2 - 20n}{2n}\\)
  2. \\(\frac{-9h + 27}{h - 3}\\)
  1. \\(\frac{(5p - 20)(p - 3)}{p - 4}\\)
  2. \\(\frac{(3q + 12)(2q - 1)}{(2q - 1)(q + 4)}\\)

Explanation:

7) Step1: Split the fraction

$\frac{24x^3 + 24x^2 + 24x}{6x^3} = \frac{24x^3}{6x^3} + \frac{24x^2}{6x^3} + \frac{24x}{6x^3}$

7) Step2: Simplify each term

$4 + \frac{4}{x} + \frac{4}{x^2}$

8) Step1: Split the fraction

$\frac{6k^5 + k^4 + 6k^3}{6k^3} = \frac{6k^5}{6k^3} + \frac{k^4}{6k^3} + \frac{6k^3}{6k^3}$

8) Step2: Simplify each term

$k^2 + \frac{k}{6} + 1$

9) Step1: Split the fraction

$\frac{2n^3 + 8n^2 - 20n}{2n} = \frac{2n^3}{2n} + \frac{8n^2}{2n} - \frac{20n}{2n}$

9) Step2: Simplify each term

$n^2 + 4n - 10$

10) Step1: Factor numerator

$\frac{-9h + 27}{h - 3} = \frac{-9(h - 3)}{h - 3}$

10) Step2: Cancel common factor

$-9$ (where $h
eq 3$)

11) Step1: Factor numerator term

$\frac{(5p - 20)(p - 3)}{p - 4} = \frac{5(p - 4)(p - 3)}{p - 4}$

11) Step2: Cancel common factor

$5(p - 3) = 5p - 15$ (where $p
eq 4$)

12) Step1: Factor numerator term

$\frac{(3q + 12)(2q - 1)}{(2q - 1)(q + 4)} = \frac{3(q + 4)(2q - 1)}{(2q - 1)(q + 4)}$

12) Step2: Cancel common factors

$3$ (where $q
eq -\frac{1}{2}, q
eq -4$)

Answer:

  1. $\boldsymbol{4 + \frac{4}{x} + \frac{4}{x^2}}$
  2. $\boldsymbol{k^2 + \frac{k}{6} + 1}$
  3. $\boldsymbol{n^2 + 4n - 10}$
  4. $\boldsymbol{-9}$ (with $h

eq 3$)

  1. $\boldsymbol{5p - 15}$ (with $p

eq 4$)

  1. $\boldsymbol{3}$ (with $q

eq -\frac{1}{2}, q
eq -4$)