$$1+4x+4x^{2}-16x^{4},1+8x^{3}$$
$\left(8x^{3}+1\right)\left(16x^{4}-\left(2x+1\right)^{2}\right)$
$$-16x^{4}+\left(2x+1\right)^{2}=-4\left(x-\left(-\frac{1}{4}\sqrt{5}+\frac{1}{4}\right)\right)\left(x-\left(\frac{1}{4}\sqrt{5}+\frac{1}{4}\right)\right)\left(4x^{2}+2x+1\right)$$ $$8x^{3}+1=\left(2x+1\right)\left(4x^{2}-2x+1\right)$$
$$4\left(2x+1\right)\left(x-\frac{1-\sqrt{5}}{4}\right)\left(x-\frac{\sqrt{5}+1}{4}\right)\left(4x^{2}-2x+1\right)\left(4x^{2}+2x+1\right)$$
$$128x^{7}-32x^{5}-16x^{4}-8x^{3}-4x^{2}-4x-1$$
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$-16x^{4}+\left(2x+1\right)^{2},8x^{3}+1$
