Stuck on how to factor $2x^3+3x^2-8x+3$ without trial/error

Question

I know this is pretty basic, but I am unable to factor $2x^3+3x^2-8x+3$ without the use of trial/error. Even the rational roots theorem requires trial/error, as you need to implement the rational roots theorem to find at least one of the roots. The only other way (besides the rational root theorem) I found to factor this was to add two terms and subtract two terms. However, this is also pretty useless as it too requires number sense and trial/check. I know a solution to my problem would be graphing, but that would take too much time (even by finding the derivative and critical values) and I have no graphing calculator.

My question is, what method would I implement to factor without trial/error? Additionally, if this expression would be set equal to zero and by chance this expression has all complex/irrational roots ($x^3-2=0$), would this method work?

Answer 1

$$ 2+3+3 = 8 $$ .........................

Answer 2

$$\begin{align*}2x^3+\color{red}{3x^2}-8x+3 & =2x^3\color{red}{-2x^2+5x^2}-8x+3\\ & =2x^2(x-1)+(5x-3)(x-1)\\ & =(x-1)(2x^2+5x-3)\\ & =(x-1)(2x-1)(x+3)\end{align*}$$

Answer 3

Your rejection of the rational roots theorem seems irrational to me since the number of possibilities is quite small.

Here's how I did it without any trial or error:

I gave it to Wolfy and was told that the roots are -3, 1/2, and 1.

The alternative is to use the explicit formula (after eliminating the quadratic term).