A community for students.
Here's the question you clicked on:
 0 viewing
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+4x5 ans: (x1)(x+5)=x=1 x=5
+1,+4,+5=/1=+1,+4,+5 ans:{1,5
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+4x5 ans: (x1)(x+5)=x=1 x=5 +1,+4,+5=/1=+1,+4,+5 ans:{1,5

This Question is Closed

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0if you use the enter key; itll make that alot more readable kind of like doing this so that we can see it clearly rathrthansomeconglomoratedmessthatwehavetodecipher

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.1I asked you if you needed help with the explanation with the rational root theorem

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes i do should i reenter the problem

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.1nope, the problem is not that relevant

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the manners are tho :)

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.1I will start explaining so pay close attention okay ? i'll go through the idea of the proof so that you can remember it

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0yuki; been down this road; but good luck ;)

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.1If you let any polynomials of form P(x) = 0, you can solve for the roots. where polynomials are generally of the form \[P(x) = {a_n}x^n + a_{n1}x^{n1}+ ... + a_1x +a_0\]

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.1the 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

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.1I 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

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.1when all coefficients of P(x) are integers

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.1Therefore, 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.

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.1in 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 \[p = \pm 1, 5 and q = \pm 1\]

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.1thus \[{p \over q} = \pm 1 , \pm 5\]

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.1let'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 \[\pm 1, 2, 5,10\]

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.1so 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

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.1but as you can see, \[\pm {1 \over 1} = \pm {5 \over 5}\] and also \[\pm {1 \over 2} = \pm {5 \over 10}\]

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.1so 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 \[{p \over q } = \pm 1,5,1/2,5/2,1/5,1/10\]

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.1one 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, \[f(x) = x^2 +1\]

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.1it 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"

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.1good luck in whatever you need to do :)
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.