1) Let's begin with that, by using the Rational Zero Theorem
![\begin{gathered} f(x)=x^4-14x^3+34x^2+114x+65 \\ \\ (dividers\:of\:the\:independent\:term)/(dividers\:of\:leading\:coefficient)=(\pm1,5,13,65)/(\pm1) \\ \\ Plug\:x=-1 \\ \left(-1\right)^4-14\left(-1\right)^3+34\left(-1\right)^2+114\left(-1\right)+65=0 \\ 0=0\:True! \\ Plug\:x=1 \\ 1^4-14\cdot \:1^3+34\cdot \:1^2+114\cdot \:1+65=0 \\ 200=0\:False \\ Plug\:x=-5 \\ \left(-5\right)^4-14\left(-5\right)^3+34\left(-5\right)^2+114\left(-5\right)+65=0 \\ 2720=0\:False \\ Plug\:x=5 \\ 5^4-14\cdot \:5^3+34\cdot \:5^2+114\cdot \:5+65=0 \\ 360=0\:False! \\ Plug\:x=-13 \\ \left(-13\right)^4-14\left(-13\right)^3+34\left(-13\right)^2+114\left(-13\right)+65=0 \\ 63648=0\:False \\ Plug\:x=13 \\ 13^4-14\cdot \:13^3+34\cdot \:13^2+114\cdot \:13+65=0 \\ 5096=0\:False! \\ Plug\:x=-65 \\ \left(-65\right)^4-14\left(-65\right)^3+34\left(-65\right)^2+114\left(-65\right)+65=0 \\ 21831680=0\:False \\ Plug\:x=65 \\ 65^4-14\cdot \:65^3+34\cdot \:65^2+114\cdot \:65+65=0 \\ 14157000=0\:False \\ \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/cq06epv9g9v75rar6pxs5h53pag1j6lxxs.png)
So, with the Rational Zero Theorem method, we could only find one root.
x=-1
2)Now, let's apply the Descartes' Rule of Signs
![\begin{gathered} Enlisting\:the\:coefficients\:we^(\prime)ve\:got: \\ \\ 1,-14,34,114,65 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/k6dtldyemkfllwmu9tdaawy45p6qzx1ush.png)
Note that there are two sign changes, so there are two or no
positive roots