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

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?

• $1$ is a root. Your polynomial divided by $(x-1)$ is a quadratic. – Will Jagy Jan 29 '18 at 0:02
• How did you discover 1 as a root? Did you do it with a method without trial/error? – Yash Jain Jan 29 '18 at 0:04
• @YashJain If you mean "without trial/error" in the strictest of senses (e.g. not even allowing to "see" that $1$ is a root because $2+3-8+3=0$), then pretty much the only eligible answer is the cubic formula. – dxiv Jan 29 '18 at 0:10
• I realized it is 1, but what would happen if the solutions composed only of irrational/complex roots? – Yash Jain Jan 29 '18 at 0:12
• Without trial/error means with an algorithm, i.e. cubic formula – G Cab Jan 29 '18 at 0:15

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

• What do I do with those numbers? – Yash Jain Jan 29 '18 at 0:06
• I realized it is 1, but what would happen if the solutions composed only of irrational/complex roots? – Yash Jain Jan 29 '18 at 0:12

\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*}

• "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." – Yash Jain Jan 29 '18 at 0:06
• I guess this is adding/subtracting one term, but it still requires trial/check. – Yash Jain Jan 29 '18 at 0:06
• @YashJain Not much guess and check actually... It didn't take me very long as long as you look at the constant term and the linear term, and try to make the quadratic factorable. – Crescendo Jan 29 '18 at 0:13
• @YashJain Alike, if you consider rational root theorem guess and check, then your only bet is the cubic formula. Which is like killing a fly with a sledgehammer: overkill – Crescendo Jan 29 '18 at 0:14

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).