Question

divide
\\(\frac{-48x^6 + 32x^4 - 40x^2}{-8x^4}\\)

Explanation:

Step1: Split the fraction

Split the numerator into three separate fractions: $\frac{-48x^6}{-8x^4} + \frac{32x^4}{-8x^4} - \frac{40x^2}{-8x^4}$

Step2: Simplify each fraction

• For $\frac{-48x^6}{-8x^4}$: Divide the coefficients $-48\div(-8) = 6$, and subtract the exponents of $x$: $x^{6 - 4}=x^2$, so it becomes $6x^2$.
• For $\frac{32x^4}{-8x^4}$: Divide the coefficients $32\div(-8) = -4$, and $x^4\div x^4 = 1$, so it becomes $-4$.
• For $\frac{-40x^2}{-8x^4}$: Divide the coefficients $-40\div(-8) = 5$, and subtract the exponents of $x$: $x^{2 - 4}=x^{-2}=\frac{5}{x^2}$, so it becomes $\frac{5}{x^2}$.

Step3: Combine the results

Combine the simplified fractions: $6x^2 - 4 + \frac{5}{x^2}$

Answer:

$6x^2 - 4 + \frac{5}{x^2}$