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: \(x_1 = 1, x_2 = 0.5, ~ \text{and}~ x_ 3 = -2\)
+53 
