How do you solve $x^{3} + 4 x^{2} - 20 x = 80$ $x^3 + 4x^2 - 20x = 80$ ?

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How do you solve $x^{3} + 4 x^{2} - 20 x = 80$ $x^3 + 4x^2 - 20x = 80$ ?

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Answer 1 · 2015-11-28T22:20:35

Subtract $80$ $80$ from both sides, then factor by grouping to find:

$x = \pm 2 \sqrt{5}$ $x = +-2sqrt(5)$ or $x = - 4$ $x = -4$

Explanation:

First subtract $80$ $80$ from both sides to get:

$x^{3} + 4 x^{2} - 20 x - 80 = 0$ $x^3+4x^2-20x-80 = 0$

Factor the left hand side by grouping:

$x^{3} + 4 x^{2} - 20 x - 80$ $x^3+4x^2-20x-80$

$= (x^{3} + 4 x^{2}) - (20 x + 80)$ $=(x^3+4x^2)-(20x+80)$

$= x^{2} (x + 4) - 20 (x + 4)$ $=x^2(x+4)-20(x+4)$

$= (x^{2} - 20) (x + 4)$ $=(x^2-20)(x+4)$

$= (x - \sqrt{20}) (x + \sqrt{20}) (x + 4)$ $=(x-sqrt(20))(x+sqrt(20))(x+4)$

$= (x - \sqrt{2^{2} \cdot 5}) (x + \sqrt{2^{2} \cdot 5}) (x + 4)$ $=(x-sqrt(2^2*5))(x+sqrt(2^2*5))(x+4)$

$= (x - 2 \sqrt{5}) (x + 2 \sqrt{5}) (x + 4)$ $=(x-2sqrt(5))(x+2sqrt(5))(x+4)$

So the roots of $x^{3} + 4 x^{2} - 20 x = 80$ $x^3+4x^2-20x=80$ are $x = \pm 2 \sqrt{5}$ $x = +-2sqrt(5)$ and $x = - 4$ $x=-4$

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How do you solve $x^{3} + 4 x^{2} - 20 x = 80$ $x^3 + 4x^2 - 20x = 80$ ?

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