Answer
The solutions are $x = 0, \frac{-1 -\sqrt {37}}{2}, \quad \text{and}\quad \frac{-1 +\sqrt {37}}{2}$.
Work Step by Step
We want to factor this polynomial first so we can find the solutions.
Rewrite the equation so that all terms are on the left side and it is equal to zero on the right side:
$3x^3 + 3x^2 - 27x = 0$
Factor out a $3x$ from each of the terms:
$3x(x^2 + x - 9) = 0$
According to the Zero-Product Property, if the product of two factors $a$ and $b$ equals zero, then either $a$ is zero, $b$ is zero, or both equal zero. Therefore, we can set each factor equal to zero:
First factor:
$3x = 0$
$x = 0$
Second factor:
$x^2 + x - 9 = 0$
To factor a quadratic polynomial in the form $x^2 + bx + c$, we look at factors of the product of $c$ such that, when added together, equal $b$.
For the equation $x^2 + x - 9 = 0$, $c=-9$ so look for factors of $-9$ whose sum is equal $b$ or $1$. These are the possibilities:
$-9=(9)(-1)$
$b = 8$
$-9=(-9)(1)$
$b = -8$
$-9=(3)(-3)$
$b = 0$
No pair of factors give a sum of $1$ so the trinomial cannot be factored.
The only option is to solve the equation using the quadratic formula:
$x = \dfrac{-b \pm \sqrt {b^2 - 4ac}}{2a}$
where $a$ is the coefficient of the $x^2$ term, $b$ is the coefficient of the 1st degree term, and $c$ is the constant.
The equation above has $a=1, b-1, $ and $c=-9$. Substitute these values into the quadratic formula above to obtain:
$x = \dfrac{-1 \pm \sqrt {1^2 - 4(1)(-9)}}{2(1)}$
$x = \dfrac{-1 \pm \sqrt {1 + 36}}{2}$
$x = \dfrac{-1 \pm \sqrt {37}}{2}$
The solutions are $x = 0, -\frac{1 ± \sqrt {37}}{2}$.
