Perform the division.
-$\frac{\mathrm{x}^{8}-1}{\mathrm{x}^{4}-1}$
A) $x^{4}+12 x-1$
B) $x^{4}-1$
C) $x^{4}$
D) $x^{4}+1$

Question

Quizplus · Accepted Answer

The division $\frac{x^8 - 1}{x^4 - 1}$ can be simplified by recognizing both the numerator and the denominator as difference of squares. $x^8 - 1 = (x^4)^2 - 1^2$ and $x^4 - 1 = (x^2)^2 - 1^2$, which allows us to use the formula $a^2 - b^2 = (a - b)(a + b)$. Applying this to both the numerator and the denominator, we get $\frac{(x^4 - 1)(x^4 + 1)}{x^4 - 1}$. The $x^4 - 1$ terms cancel out, leaving $x^4 + 1$.

Perform the Division x8−1x4−1\frac{\mathrm{x}^{8}-1}{\mathrm{x}^{4}-1}x4−1x8−1​ A) x4+12x−1x^{4}+12 x-1x4+12x−1 B) x4−1x^{4}-1x4−1 C) x4x^{4}x4 D) x4+1x^{4}+1x4+1

Perform the Division $\frac{\mathrm{x}^{8}-1}{\mathrm{x}^{4}-1}$ A) $x^{4}+12 x-1$ B) $x^{4}-1$ C) $x^{4}$ D) $x^{4}+1$