# How do you find the zeros of a function f(x)= 2x^3 - 3x^2- 12x +20?

May 12, 2015

First, factor the function.
Usually, try simple small numbers as real roots: 1, or -1, or 2.

f(2) = 16 - 12 - 24 + 20 = 36 - 36 = 0, then the function has (x - 2) as a

factor. Next, you may algebraically divide or guess.

$f \left(x\right) = \left(x - 2\right) \left(2 {x}^{2} + x - 10\right) = \left(x - 2\right) . \left(x - p\right) \left(x - q\right)$

Use the new AC Method to factor the trinomial.
Converted trinomial: $f ' \left(x\right) = {x}^{2} + x - 20 = \left(x - p '\right) \left(x - q '\right)$. Compose factor pairs of a.c = -20. Proceed: (-2, 10)(-4, 5). This last sum is (5 - 4 = 1 = b), then p' = -4 and q' = 5. Consequently: p = p'/a = -4/2 = -2 and q = q'/a = 5/2
.
Finally: $f \left(x\right) = \left(x - 2\right) \left(x - 2\right) \left(x + \frac{5}{2}\right) = \left(x - 2\right) \left(x - 2\right) \left(2 x + 5\right)$

The Zeros are: x = 2 (double root);$x = - \frac{5}{2}$