# How do you solve 2x^3-6x^2+x-1<=2x-3?

May 9, 2017

Firstly set it $=$ to $0$ by taking away $2 x - 3$ from both sides.
This leaves us with:
$2 {x}^{3} - 6 {x}^{2} - x + 2 \le 0$
Then we can use the factor theorem to find one of our roots.
Start with $f \left(x\right) = 1$
$2 {\left(1\right)}^{3} - 6 {\left(1\right)}^{2} - \left(1\right) + 2 = - 3$ Not a factor
Try $2$
$2 {\left(2\right)}^{3} - 6 {\left(2\right)}^{2} - \left(2\right) + 2 = - 8$ Not a factor
Try $- 1$
$2 {\left(- 1\right)}^{3} - 6 {\left(- 1\right)}^{2} - \left(- 1\right) + 2 = - 7$ Not a factor
Try $- 2$
$2 {\left(- 2\right)}^{3} - 6 {\left(- 2\right)}^{2} - \left(- 2\right) + 2 = - 36$ Not a factor
Try $3$
$2 {\left(3\right)}^{3} - 6 {\left(3\right)}^{2} - \left(3\right) + 2 = - 1$ Not a factor
Try $4$
$2 {\left(4\right)}^{3} - 6 {\left(4\right)}^{2} - \left(4\right) + 2 = 30$ Not a factor.

Very sorry I think I've done something wrong but I thought I'd send you my working a)because I find sometimes reading what someone else has done, even if wrong, helps to solve things and b)because I cant bring myself to get rid of that working.

It could be possible that you just need a factor higher than I'm using so feel free to try some, and if so you'll need to do some algebraic long division. Happy to show you this if you find a factor!