X- coordinates

To find the x-coordinates of the points where  y = 5x -1  intersects  y  =  2x³ + x² + 1,

set the equations equal to each other:      5x - 1  =  2x³ + x² + 1,

and solve:                                                             0  =  2x³ + x² - 5x + 2.

If there are rational solutions to this equation, they come from the set:  { +2, -2, +1, -1, + 1/2, -1/2 }.

If x = 2:  2(2)³ + (2)² - 5(2) + 2  =  12   --->  2 is not a root.

If x = -2:  2(-2)³ + (-2)² - 5(-2) + 2  = 0   --->   -2 is a root  

If x = 1:  2(1)³ + (1)² - 5(1) + 2  = 0   --->   1 is a root  

If x = -1:  2(-1)³ + (-1)² - 5(-1) + 2  = 6   --->   -1 is a not a root  

If x = 1/2:  2(1/2)³ + (1/2)² - 5(1/2) + 2  = 0   --->   1/2 is a root

If x = -1/2:   2(-1/2)³ + (-1/2)² - 5(-1/2) + 2  = 4.5   --->   -1/2 is not a root

The x-coordinates of the points of intersection are:  -2, 1/2, and 1.

[If not all of these worked, then there still could have been irrational roots.]

Find the x-coordinates of the points where the line y=5x-1 meets the curve y= 2x^3 + x^2 +1.

I keep solving but I can't seem to get anything right,

especially when I set these two next to each other.

Intersections:

$\begin{array}{|rcll|} \hline y_{\text{line}} &=& y_{\text{curve}} \\ 5x-1 &=& 2x^3 + x^2 +1 \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline 2x^3 + x^2 +1 &=& 5x-1 \quad & | \quad -5x+1 \\ 2x^3 + x^2 +1 -5x+1 &=& 0 \\ 2x^3 + x^2 -5x +2 &=& 0 \quad & | \quad :2 \\ x^3 + \frac12 x^2 -\frac52x +\color{red}{1} &=& 0 \\ \hline \end{array}$

We get the rational solutions, when we test all dividers from the absolut term 1:

We must test +1 or -1.

For +1 we get the first root $x_1$: $1^3 + \frac12 \cdot 1^2 -\frac52 \cdot 1 + 1 = 1+ \frac12 -\frac52+1 = 0$

$\begin{array}{|rcll|} \hline x_1 = 1 \\ \hline \end{array}$

We know that $(x-x_1)$ is a divider of  $x^3 + \frac12 x^2 -\frac52x +1$

$( x^3 + \frac12 x^2 -\frac52x +1) : (x-1) = (x^2+1.5x-1 )$

Because $( x^3 + \frac12 x^2 -\frac52x +1) = (x-1)\cdot (x^2+1.5x-1 ) = 0$
we find the roots $x_2$ and $x_3$, if we set  $x^2+1.5x-1 = 0$

$\begin{array}{|rcll|} \hline x^2+1.5x-1 &=& 0 \\ x &=& \frac{-1.5\pm \sqrt{1.5^2-4\cdot 1 \cdot (-1) } }{2} \\ x &=& \frac{-1.5\pm \sqrt{2.25+4} }{2} \\ x &=& \frac{-1.5\pm \sqrt{6.25} }{2} \\ x &=& \frac{-1.5\pm 2.5 }{2} \\\\ x_2 &=& \frac{-1.5+ 2.5 }{2} \\ x_2 &=& \frac{1 }{2} \\ \mathbf{x_2} &\mathbf{=}& \mathbf{0.5} \\\\ x_3 &=& \frac{-1.5 - 2.5 }{2} \\ x_3 &=& \frac{-4 }{2} \\ \mathbf{x_3} &\mathbf{=}& \mathbf{-2}\\ \hline \end{array}$

The x-coordinates of the intersection-points are: $x_1 = 1, x_2 = 0.5, ~ \text{and}~ x_ 3 = -2$