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Solve for a real number of $$x$$ which satisfies:
$$x^3$$ $$-$$ $$5x^2$$ $$+$$ $$8x$$ $$-$$ $$4$$ = $$0$$

Note by Mục Xiên
2 years, 5 months ago

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$x^3 - 5x^2 + 6x + 2x - 4 = 0$

$x(x^2 - 5x + 6) + 2(x - 2) = 0$

$x( x-2)(x-3) + 2(x - 2) = 0$

$(x-2)(x^2 - 3x + 2) = 0$ · 2 years, 5 months ago