## anonymous 5 years ago Can you help with an explanation can someone tell me if this problem is correct, list the possible rational roots? f(x)=x^2+4x-5 ans: (x-1)(x+5)=x=1 x=-5 +-1,+-4,+-5=/1=+-1,+-4,+-5 ans:{1,-5

1. amistre64

if you use the enter key; itll make that alot more readable kind of like doing this so that we can see it clearly rathrthansomeconglomoratedmessthatwehavetodecipher

2. Yuki

I asked you if you needed help with the explanation with the rational root theorem

3. anonymous

yes i do should i reenter the problem

4. amistre64

5. Yuki

nope, the problem is not that relevant

6. amistre64

the manners are tho :)

7. Yuki

I will start explaining so pay close attention okay ? i'll go through the idea of the proof so that you can remember it

8. anonymous

ok

9. amistre64

yuki; been down this road; but good luck ;)

10. Yuki

If you let any polynomials of form P(x) = 0, you can solve for the roots. where polynomials are generally of the form $P(x) = {a_n}x^n + a_{n-1}x^{n-1}+ ... + a_1x +a_0$

11. Yuki

the only thing that really matters here is the leading term and the constant term, so I will abbreviate the middle part as Q now, one of the ways to solve for the root systematically, you take the following steps $a_n x^n + Q + a_0 = 0$ $a_n x^n + Q = -a_0$ $x^n +Q/a_n = {-a_0 \over a_n}$ the last term is very important

12. Yuki

I won't go into any more details, but if you keep doing this you will eventually come to a conclusion saying, x^n must be some power of a_0 / a_n thus a possible number of x, must be of form p/q where p = all factors of a_0 q = all factors of a_n

13. Yuki

when all coefficients of P(x) are integers

14. Yuki

Therefore, the main idea is this if a polynomial P(x) have coefficients only made of integers, then the possible rational roots are + or - p/q. where p is the factors of the constant, and q is the factors of the leading coefficient.

15. Yuki

in your problem, the constant term is -5 and the factors are + or - 1 and 5. the leading coefficient is 1, so the factors are + or - 1. since you have + or - for both top and bottom, we will ignore that for now and put it later once we find the positive version of the possible rational roots. they have to have the form p/q where $p = \pm 1, 5 and q = \pm 1$

16. Yuki

thus ${p \over q} = \pm 1 , \pm 5$

17. Yuki

let's say that your leading coefficient was actually, 10, because I want you to see the pattern better. now p is still the same, but q will now be $\pm 1, 2, 5,10$

18. Yuki

so technically, p/q will now be $\pm {1 \over 1},{5 \over 1}$ $\pm {1 \over 2}, {5 \over 2}$ $\pm {1 \over 5}, {5 \over 5}$ $\pm {1 \over 10}, {5 \over 10}$ are all of my combinations

19. Yuki

but as you can see, $\pm {1 \over 1} = \pm {5 \over 5}$ and also $\pm {1 \over 2} = \pm {5 \over 10}$

20. Yuki

so we are double counting. once you eliminate those and simplify for a polynomial that looks like $P(x) = 10x^4 + 3x^2 + x -5$ the possible rational roots are ${p \over q } = \pm 1,5,1/2,5/2,1/5,1/10$

21. Yuki

one last thing. the rational root theorem only tells you the "possible" roots so it doesn't mean that one of those are guaranteed to be an actual solution. for example, $f(x) = x^2 +1$

22. Yuki

it is obvious that this quadratic has no real roots, but the possible rational roots according to the thm are + or - 1. which both will fail. so always keep in mind that it is just a "fairly good guess"

23. Yuki

does that help ?

24. anonymous

yes very much

25. anonymous

thank u

26. Yuki

good luck in whatever you need to do :)