Question

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

Answer 1

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

Answer 2

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

Answer 3

yes i do should i reenter the problem

Answer 4

yes please

Answer 5

nope, the problem is not that relevant

Answer 6

the manners are tho :)

Answer 7

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

Answer 8

ok

Answer 9

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

Answer 10

If you let any polynomials of form

P(x) = 0, you can solve for the roots.
where polynomials are generally of the form

\[P(x) = {a_n}x^n + a_{n-1}x^{n-1}+ ... + a_1x +a_0\]

Answer 11

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

Answer 12

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

Answer 13

when all coefficients of P(x) are integers

Answer 14

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.

Answer 15

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

\[p = \pm 1, 5 and q = \pm 1\]

Answer 16

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

Answer 17

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

\[\pm 1, 2, 5,10\]

Answer 18

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

Answer 19

but as you can see,

\[\pm {1 \over 1} = \pm {5 \over 5}\]

and also

\[\pm {1 \over 2} = \pm {5 \over 10}\]

Answer 20

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

\[{p \over q } = \pm 1,5,1/2,5/2,1/5,1/10\]

Answer 21

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,

\[f(x) = x^2 +1\]

Answer 22

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"

Answer 23

does that help ?

Answer 24

yes very much

Answer 25

thank u

Answer 26

good luck in whatever you need to do :)