Question

WILL GIVE MEDALS!! find the rational roots of x^4+5x^3+7x^2-3x-10=0

Answer 1

@mathmale can you help on this one

Answer 2

First, DB, are you familiar with synthetic division? If not, then are you fam. with long div.?

Answer 3

no neither

Answer 4

Then I'll need to ask you what methods of factoring you've used in the past, when factoring polynomials.
There are other ways to factor, but synth. div. and log div. are the mainstays of that when you're just starting algebra.

Answer 5

can you teach me the eayiest

Answer 6

Wow, DB, that really is a big order if you haven't experienced either one. Do you have a textbook for your math course? Are you good at Internet searches?

Answer 7

I just hope you do have a reference available. Synth. div. is fun, but it does take a while to learn.

Answer 8

well use the rational root theorem

the ration roots will come from

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

Answer 9

next you can use the factor theorem to check is if a root

which means

f(c) = 0 then c is a root and (x - c) is a factor

Answer 10

so start by checking

f(1)

Answer 11

Campbell, good suggestion, but remember that DB will have to be able to check each possible root to see whether it is indeed a root; I prefer synth. div. for this.
OK, Campbell, explain what f(1) means, so that it becomes clear to DB what he has to do.

Answer 12

and then you could use polynomial division or keep checking

Answer 13

DB, please comment. Where are you?

Answer 14

i get what your saying

Answer 15

f(1) means you substitute 1 for x

so f(1) = (1)^4 + 5x(1)^3 + 7(1)^2 - 3(1) - 10

if it is equal to zero then

x = 1 is a zero and (x - 1) is a factor...

Answer 16

Well, with Campbell's suggestion, you now have 3 ways to determine whether 1 (or any other possible solution) actually is a solution: Evaluate the function f(x) at x = 1 (Campbell is absolutely right above).
We've already discussed synth division and long div, which you say you're not familiar with.

Answer 17

but like any system you need to start with a value... be it synthetic divison polynomial division or any other method

the rational root theorem and factor theorems are the starting point

Answer 18

Campbell wrote out f(1) for you, DB. Please finish the arithmetic and determine whether or not f(1) is zero. Can you do that?

Answer 19

you should try to find a 2nd root using the factor theorem.

if you use division with 1 factor then you are still looking at a cubic...

so try to find 2 factors before starting division,

divide by the 1st factor

then divide by the 2nd...

that way the quotient will be a quadratic... and easy to determine if it has rational roots...

hope it all helps

Answer 20

Campbell: DB needs to evaluate f(1) first. You got him started; I'd like to see him finish.

Answer 21

f(1) is a zero i need the next one now

Answer 22

I'm just telling him... what is a good management strategy for the problem

1. use the rational root theorem to identify possible rational roots

2. use the factor theorem to identify 2 roots

3. use division of some sort... the quotient will be a quadratic

4. identify if the quadratic has rational roots...

good luck.

Answer 23

Great! Because f(1) is zero, 1 is a root (or a zero) of the given polynomial. Nice work!

DB: Please look at the last term of the given polynomial. It's -10. You need to identify all possible factors of -10. Note that your 1 is a factor. What are some other possible factors of -10? Unfortunately, you'll have to test each one by evaluating f(x) at each.

Answer 24

you don't need to test them all.... its a bit tedious... just find 2...

Answer 25

To get you started: Here's a possible set of factors of -10: {1,-1,2,-2,5,-5,.... }.

Answer 26

Choose one of those and evaluate f(x) at it. If the result is zero, you've got another root of the polynomial; otherwise you don't, and must move on to evaluate f(x) at some other value from the set I gave you.

Answer 27

wouldn't the other zero be -2

Answer 28

It's really important that YOU check that out yourself, for the practice.

Answer 29

yes well done

Answer 30

Great. But, DB, how did you obtain -2?

Answer 31

its listed as a possible zero twice... above...

Answer 32

OK. So, DB just chose -2 at random from that list and evaluated f(x) at x=-2?

OK.

Answer 33

i factored it like he said to

Answer 34

For now, DB, take my word for this: Once you've identified 1 and -2 as zeros, you need to find the zeros of 1x^2+4x+5. Mind trying that?
Hint: you could go back to the set of possible roots I gave you and try out some new possibilities from that set.

Answer 35

DB: factoring is certainly another valid method for identifying roots.

Can you now find the zeros of 1x^2+4x+5? You could try factoring this first if you like.

Answer 36

ok... so using @mathmale quotient

\[Q(x) = x^2 + 4x + 5\]

just use the discriminant to determine if the quaratic has real rational roots..

Answer 37

Excuse me, DB, I was wrong in my last statement; you cannot easily factor x^2+4x+5.
You could, however, apply the quadratic equation to find the roots of this function.

Answer 38

Campbell: well said, good suggestion. Would you care to explain that to DB? Thanks.

Answer 39

DB: What are you thinking now? What do you most wish Campbell or I could help you with or tell you right now? We've explored a variety of ways to find the roots/zeros of the given polynomial.

Answer 40

DB: I see you're looking at someone else's post. Hope Campbell and I haven't lost you.

Answer 41

Appreciate your participating and contributing, Campbell. Thanks.