Question

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

Answer 1

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(x) = (x - 2)(2x^2 + x - 10) = (x - 2).(x - p)(x - q)`

Use the new AC Method to factor the trinomial.
Converted trinomial: `f'(x) = x^2 + x - 20 = (x - p')(x - q')`. 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(x) = (x - 2)(x - 2)(x + 5/2) = (x - 2)(x - 2)(2x + 5)`

The Zeros are: x = 2 (double root);`x = -5/2`