1. Solving for 0

how do i systematically solve $\displaystyle 2x^5-5x^2+1=0$

If i use the rational root theorum, then I would factor out either (x+1) or (x+1/2) but when I use those factors to divide into $\displaystyle 2x^5-5x^2+1=0$, i get an long confusing answer with remainders.

2. Originally Posted by dorkymichelle
how do i systematically solve $\displaystyle 2x^5-5x^2+1=0$

If i use the rational root theorum, then I would factor out either (x+1) or (x+1/2) but when I use those factors to divide into $\displaystyle 2x^5-5x^2+1=0$, i get an long confusing answer with remainders.
There are no rational solutions. Where has the question come from? Were the instructions to solve it exactly?

3. how can you tell if theres no rational solutions?
it's a graphing problem for calculus, i was trying to find the x intercepts.

4. Originally Posted by dorkymichelle
how can you tell if theres no rational solutions?
it's a graphing problem for calculus, i was trying to find the x intercepts.
Use technology.

solve 2x&#x5e;5 - 5x&#x5e;2 &#x2b; 1 &#x3d; 0 - Wolfram|Alpha

5. so there are roots,
x ~~ -0.439794
x ~~ 0.45594
x ~~ 1.301545
But I have a test on Tuesday, and I'm going to fail very bad if I can't figure out how to solve for x intercepts in these type of equations the long way.

6. Originally Posted by dorkymichelle
so there are roots,
x ~~ -0.439794
x ~~ 0.45594
x ~~ 1.301545
But I have a test on Tuesday, and I'm going to fail very bad if I can't figure out how to solve for x intercepts in these type of equations the long way.
It cannot be done 'the long way'. Discuss this with your teacher.

7. this is because the roots are only estimates and are decimals that keep going on and on right?
but how do you tell when you saw the equation that there were no roots?

8. Probably the rational root theorem, which in this case states that the only possible rational roots are $\displaystyle \pm1,\,\pm\frac12$, none of which are actually roots. Therefore there are no rational roots.

9. why can't +- 1/2 or +- 1 be roots?
what rule or theorum says that?

10. Originally Posted by dorkymichelle
why can't +- 1/2 or +- 1 be roots?
what rule or theorum says that?
Because plugging them in the function does not yield zero.