How do you factor and solve 3x^3 + 4x^2 - 15x = 0?

Jun 29, 2016

$x = 0 , x = \frac{5}{3} , x = - 3$

Explanation:

$f \left(x\right) = x y = x \left(3 {x}^{2} + 4 x - 15\right) = 0$.
Factor y by using the new AC Method (Socratic Search)
Converted trinomial: $y ' = {x}^{2} + 4 x - 45$.
Find p' and q' that have opposite signs (ac < 0), and that has their
sum (b = 4) and their product (c = -45).
They are; p' = -5 and q' = 9.Back to trinomial y, $p = \frac{p '}{a} = - \frac{5}{3} = - \frac{5}{3}$, and $q = \frac{q '}{a} = \frac{9}{3} = 3$
Factored form of y:
$y = 3 \left(x - \frac{5}{3}\right) \left(x + 3\right) = \left(3 x - 5\right) \left(x + 3\right) .$
Finally, the real roots of f(x) = x(3x - 5)(x + 3) are:
$x = 0 , x = \frac{5}{3} , \mathmr{and} x = - 3$