$$x^{3}+2x^{2}-4x; 8\div2x^{3}+7x^{2}; \underline{+4x-4}$$
$4x\left(x-1\right)\left(2x^{5}-3x^{4}-22x^{3}+20x^{2}-16x+32\right)$
$$x\left(x^{2}+2x-4\right)=x\left(x-\left(\sqrt{5}-1\right)\right)\left(x-\left(-\sqrt{5}-1\right)\right)$$ $$4x-4=4\left(x-1\right)$$
$$4x\left(x-1\right)\left(x-\left(-\sqrt{5}-1\right)\right)\left(x-\left(\sqrt{5}-1\right)\right)\left(2x^{3}-7x^{2}-8\right)$$
$$8x^{7}-20x^{6}-76x^{5}+168x^{4}-144x^{3}+192x^{2}-128x$$
Show Solution
Hide Solution
$x\left(x^{2}+2x-4\right),\ 8+7x^{2}-2x^{3},\ 4\left(x-1\right)$
