## mskyeg 2 years ago Find any rational roots of P(x). P(x)=3x^4+2x^3-9x^2+4

1. satellite73

try 1, -1 first then 2, -2 then 4,-4 if these don't work we have to move on to fractions

2. mskyeg

there is more than 1 so even if they work there wtill might be a fraction. Oh and with this i have to show my work and do it the textbook way not by trial and error

3. Freyes

So what exactly is the textbook way for your textbook

4. satellite73

there is no other way to do it you have to check. if that is "trial and error' that is what you need to do

5. Freyes

Well, in practicality my teacher told me to graph the function.. and then you can use synthetic division for the ones that aren't "exact"

6. Freyes

But unless you have some sort of sixth sense of math you really don't have a set method.. unless your textbook does, in which case i'd want to know.

7. mskyeg

ya thats the way i am just having trouble relating it to this specifc problem

8. satellite73

1 works by your eyeballs because if $P(x)=3x^4+2x^3-9x^2+4$ then $P(1)=3+2-9+4=0$

9. Freyes

well for this one i would guess -2

10. Freyes

but its just a guess

11. satellite73

add up the coefficients and see that you get 0 then you can factor as $(x-1)\times (something)$ and then you can either factor the "something" or else use the quadratic formula

12. satellite73

you only need to find one zero, then factor and end up with a quadratic, and you can always find the zeros of any quadratic

13. satellite73

scratch that, this is a polynomial of degree 4, so you factor and end up with a cubic when you do that, you will see that 1 is a zero of the cubic polynomial as well, so you will factor again as $$(x-1)(x-1)(something)$$

14. Freyes

x+2

15. Freyes

but there really isn't an exact method to it is all.