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(𝑥⁴-16)÷𝑥³+2𝑥²+4𝑥+8

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We have, (x^{4} - 16) \div x^{3} + 2 x^{2} + 4 x + 8 = \frac{x^{4} - 16}{x^{3} + 2 x^{2} + 4 x + 8} Consider, x^{4} - 16 = {(x^{2})}^{2} - 4^{2} Now consider, x^{3} + 2 x^{2} + 4 x + 8 = x^{2} (x + 2) + 4 x + 8 = x^{2} (x + 2) + 4 (x + 2) = (x^{2} + 4) (x + 2) So, \frac{x^{4} - 16}{x^{3} + 2 x^{2} + 4 x + 8} = \frac{{(x^{2})}^{2} - 4^{2}}{(x^{2} + 4) (x + 2)} = \frac{(x^{2} - 4) (x^{2} + 4)}{(x^{2} + 4) (x + 2)} [using a^{2} - b^{2} = (a + b) (a - b)] = \frac{x^{2} - 4}{x + 2} = \frac{x^{2} - 2^{2}}{x + 2} = \frac{(x + 2) (x - 2)}{x + 2} [using a^{2} - b^{2} = (a + b) (a - b)] = x - 2

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