Question

divide.
\\(\frac{42x^4 + 35x^3 + 21x^2}{7x^3}\\)

Explanation:

Step1: Split the fraction

We can split the numerator into three separate fractions: $\frac{42x^4}{7x^3} + \frac{35x^3}{7x^3} + \frac{21x^2}{7x^3}$

Step2: Simplify each fraction

For the first fraction: $\frac{42x^4}{7x^3} = \frac{42}{7} \cdot \frac{x^4}{x^3} = 6x$ (using the rule $\frac{x^a}{x^b}=x^{a - b}$)
For the second fraction: $\frac{35x^3}{7x^3} = \frac{35}{7} \cdot \frac{x^3}{x^3} = 5$
For the third fraction: $\frac{21x^2}{7x^3} = \frac{21}{7} \cdot \frac{x^2}{x^3} = 3 \cdot x^{-1} = \frac{3}{x}$ (using the rule $x^{-n}=\frac{1}{x^n}$)

Step3: Combine the results

Adding the simplified fractions together: $6x + 5 + \frac{3}{x}$

Answer:

$6x + 5 + \frac{3}{x}$