Question

divide.
$\frac{4x^{4}-4x^{2}-8x+16}{2x-2}$
$2x^{3}+2x^{2}-4+\frac{?}{2x-2}$

Explanation:

Step1: Rearrange division formula

Let the unknown remainder be $R$. We use the relationship:
$$\text{Dividend} = \text{Divisor} \times \text{Quotient} + R$$
So,
$$R = (4x^4 - 4x^2 - 8x + 16) - (2x - 2)(2x^3 + 2x^2 - 4)$$

Step2: Expand the product term

Calculate $(2x - 2)(2x^3 + 2x^2 - 4)$:

$$\begin{align*} &2x(2x^3 + 2x^2 - 4) - 2(2x^3 + 2x^2 - 4)\\ =&4x^4 + 4x^3 - 8x - 4x^3 - 4x^2 + 8\\ =&4x^4 - 4x^2 - 8x + 8 \end{align*}$$

Step3: Compute the remainder $R$

Subtract the expanded product from the dividend:

$$\begin{align*} R&=(4x^4 - 4x^2 - 8x + 16) - (4x^4 - 4x^2 - 8x + 8)\\ &=4x^4 - 4x^2 - 8x + 16 - 4x^4 + 4x^2 + 8x - 8\\ &=8 \end{align*}$$

Answer:

$8$