Answer
The solutions are $x = 0, \frac{5 - 5\sqrt {2}}{2}, \text{ and } \frac{5 + 5\sqrt {2}}{2}$.
Work Step by Step
Rewrite this equation so all the terms are on the left side of the equation and that the equation equals $0$.
$12x^3 - 60x^2 - 75x = 0$
Factor out $3x$:
$3x(4x^2 - 20x - 25) = 0$
Use the Zero-Product Property by equating each factor to $0$, then solve each equation:
First factor:
$3x = 0$
$x = 0$
Second factor:
$4x^2 - 20x - 25 = 0$
We cannot factor this polynomial, so we resort to using the quadratic formula, which is:
$x = \dfrac{-b \pm \sqrt {b^2 - 4ac}}{2a}$
where $a=4, b=-20,$ and $c=-25$.
Let us plug in the numbers into the formula:
$x = \dfrac{-(-20) \pm \sqrt {(-20)^2 - 4(4)(-25)}}{2(4)}$
$x = \dfrac{20 \pm \sqrt {400 + 400}}{8}$
$x = \dfrac{20 \pm \sqrt {800}}{8}$
The number $800$ can be expanded into the factors $400$ and $2$:
$x = \dfrac{20 \pm \sqrt {400(2)}}{8}$
$x = \dfrac{20 \pm 20\sqrt {2}}{8}$
$x = \dfrac{5 \pm 5\sqrt {2}}{2}$
The solutions are $x = 0, \dfrac{5 ± 5\sqrt {2}}{2}$.
