Question

What are the possible rational zeros of f(x) = x4 - 4x3 + 9x2 + 5x + 14?

± 1, ± 1 over 2, ± 1 over 7, ± 1 over 14
± 1, ± 2, ± 7, ± 14
± 1, ± 1 over 4,± 1 over 5, ± 1 over 9, ± 1 over 14
± 1, ± 4, ± 5, ± 9, ± 14

Answer 1

can someone explain how to do this? :/

Answer 2

@vishweshshrimali5 @tkhunny @ganeshie8 @Compassionate @dumbcow :)

Answer 3

well theres going to 4 answers

Answer 4

yes

Answer 5

so c and d is out

Answer 6

dumcow will tell you the rest

Answer 7

its the factors of last coefficient over factors of first coefficient

Answer 8

Please stop tagging the world.

You will have to read the "Rational Root Theorem". Is this familiar?

Answer 9

Is your function correct ?

Answer 10

1 is not a root

Answer 11

It appears to be incorrect

Answer 12

it doesn't make much sense. I have read it. I understand how that we need 4 answers.

Answer 13

Due to what @RaphaelFilgueiras said, I suspect that you did some mistake in writing the function

Answer 14

that is exacting what my question states :/

Answer 15

This function has no real root

Answer 16

http://www.wolframalpha.com/input/?i=factorise+x%5E4-4x%5E3%2B9x%5E2%2B5x%2B14

Answer 17

http://en.wikipedia.org/wiki/Rational_root_theorem#Example

Answer 18

that's a little odd...

Answer 19

Well it appears that there *is* some mistake.

Answer 20

hmmmm

Answer 21

It asks for possible rational roots. This is understood to be a priori and the existence of actual rational roots is of no consequence. The correct answer is given in the choices. There is no observable error.

Answer 22

But there is no possible rational root @tkhunny

Answer 23

so its between answers A and B

Answer 24

What does possible rational root mean ?

Answer 25

Because using \(\pm\) would mean 2 solutions. Each option would give 8 solutions

Answer 26

ahh this theorem is becoming obsolete now with graphing calculators and wolfram

Answer 27

this was in my lesson, I don't get this:
To find all of the possible rational zeros of the polynomial function, divide each factor of p by each factor of q. In other words, find all possible combinations of p over q.

p over q = ±1 over 1, ±2 over 1, ±3 over 1, ±6 over 1
When each of the fractions is simplified, the possible rational zeros are found:

±1, ±2, ±3, ±6

Keep in mind that ±1, ±2, ±3, and ±6 are only the possible rational zeros. Irrational and complex zeros cannot be identified by the p over q values. Also, not all of these eight possible zeros will actually be zeros. Remember, the Fundamental Theorem of Algebra tells us that the function has exactly five zeros.

To find the zeros of the function, test each of the possible zeros in the synthetic division process to discover which results in a remainder of 0.

Answer 28

Ohh now I get it.. Sorry about my all above comments :)

Answer 29

But what are p and q ?

Answer 30

It's ok... from the lesson this is what they are referring to....f(x) = 1x5 + 2x4 – 4x3 – 4x2 – 5x – 6

Answer 31

1) Look at an equation.
2) Determine the possible numbers of Positive and Negative Real roots.
3) Determine the possible Rational Roots.
4) Combine the information from 2) and 3) to narrow down the list a little.
5) Actually search for rational roots.
6) Actually search for Real Roots.
7) Finish up by finding Complex Roots.

What the problem statement is talking about is Step #3. In Step #3, we do not yet know that there are no Rational Roots. We are just listing possibilities - where we should look if there happen to be some. Later, we may discover that none actually IS a root, but we don't care about that in Step #3 - we don't know enough, yet.

Answer 32

From the lesson, "Also, not all of these eight possible zeros will actually be zeros. "

Answer 33

that is what I am stuck on...

Answer 34

Well we can find out the step 2 answer using Descarte's theorem.

But I think as @tkhunny said, the question is asking about step 3

Answer 35

In the lesson it says "divide each factor of p by each factor of q. In other words, find all possible combinations of p over q."

Are p and q some coefficients or some functions or anything else ?

Answer 36

Were you given anything related to p and q in the lesson ?

Answer 37

i belive that you need to do are look at the factors of the constant 14...q

±1,±2,±7,±14


and the coefficient of the leading term which is

±1


and the possible rational roots can be
qp

Answer 38

It said the factors of P=+-1, +-2, +-3, +-6
and the factors of q: +-1

Answer 39

alright.. I guess I'll try to figure this one out on my own. thank you for your suggestions guys. :)

Answer 40

i believe its B though

Answer 41

;)

Answer 42

factors of 14: 1,2,7,14
factors of 1: 1

--> +- 1/1 ,2/1 , 7/1, 14/1

its not hard once you grasp the concept

Answer 43

What are you still figuring out?

f(x) = 1x5 + 2x4 – 4x3 – 4x2 – 5x – 6

In standard form,

The leading coefficient is 1. All the factors of 1 are 1. That's easy. These are your q's.

The constant term is -6. All the factors of 6 are 1, 2, 3, 6. That's easy. These are your p's.

Now, we just make a collection of all combinations. Since q presents only 1, we will get only integers.

+/-1, +/-2, +/-3, +/-6 -- Done

Answer 44

Slightly more complicated:

(x) = 2x5 + 2x4 – 4x3 – 4x2 – 5x – 8

In standard form,

The leading coefficient is 2. All the factors of 2 are 1, 2. That's easy. These are your q's.
The constant term is -8. All the factors of 8 are 1, 2, 4, 8. That's easy. These are your p's.

Now, we just make a collection of all combinations.

+/-1, +/-2, +/-4, +/-8 -- Using q = 1
+/-1/2, +/-2/2, +/-4/2, +/-8/2 -- Using q = 2

There is some redundancy, since 2/2 = 1 and we have that already. 4/2 = 2 and we have that already. 8/2 = 4 and we have that already. The only new suspect we found with q = 2 is +/- 1/2.

Answer 45

how about for this equation f(x) = x4 - 4x3 + 9x2 + 5x + 14?

Answer 46

You tell me.
Your numerators are factors of 14. Name them.
Your denominators are factors of 1. Name them.

Answer 47

@superhelp101 , are you not seeing my posts??

Answer 48

lol

Answer 49

yes I see until this part....but i am still not understanding the concepts of dividing p by q

Answer 50

relating to the problem i posted

Answer 51

Start simple.

What is the root of this equation?

3x + 5 = 0?

Answer 52

I don't really get how and why you divided these
+- 1/1 ,2/1 , 7/1, 14/1

Answer 53

forget p and q

think last/first if that makes it easier

Answer 54

Solve my little linear problem. You will see it.

Answer 55

rational root theorem

Answer 56

x=-5/3

Answer 57

How did that 5 get into the numerator?
How did that 3 get into the denominator?

That's ALL we're doing. If a root is going to be rational, it has to be the result of some linear factor that looks just like that. The only factors that can look like that are (Some Factor of the leading coefficient * x + Some factor of the Constant Term)

2x^2 +5x - 3 = (2x-1)(x+3) = 0 ==> x = 1/2 or x = -3/1

Notice how the numerators, 1 and 3, are factors of the 3 we had in the first place.
Notice how the denominators, 2 and 1, are factors of the 2 we had in the first place.

This is all we are doing. Playing with linear factors that you naturally do, anyway.

Answer 58

ok

Answer 59

You better show us another one. Is this problem set done? I am not yet filled with warm, fuzzy feelings.

Answer 60

it makes sense, if i have any questions i'll let you know ;)

Answer 61

Past my bedtime. I'll have to try to sleep while I'm worried about you. :-)

Answer 62

plz don't let me bother you, it's 1:30 am here.. past mine too

Answer 63

what??!!!

Answer 64

its 2:35 PM here

Answer 65

wow! I am a hard worker ;)

Answer 66

hmmmm not really. : D

Answer 67

jk

Answer 68

lol, are you near tokyo?

Answer 69

pretty much but not in Japan nor China

Answer 70

where then?

Answer 71

Korea

Answer 72

lol my best friend is from korea ;)