Question

Find The Rational Roots Of x^4+5x^3+7x^2-3x-10=0

Please Help :) Will Give Metal :) Thank You Loves!!

Answer 1

Do you know about the rational root theorem? It will allow you to find the list of possible roots. Then you can use synthetic substitution (or plain old substitution) to test each possible root to find out which ones are valid and which are not.

Answer 2

I know very little about this, need full explanation haha

Answer 3

As the leading coefficient of the polynomial is 1 and the trailing coefficient is -10, all of the potential roots will be in the list of positive and negative factors of -10.

Unfortunately I have to go out for a bit right now, but I'll check back when I return.

Answer 4

okay thank you..

Answer 5

@phi , do you get the problem?

Answer 6

@zepdrix , im kinda desperate here.. hahah :)

Answer 7

okay, I'm back. You've had plenty of time to find the positive and negative factors of 10, what are they? :-)

Answer 8

Umm.. I told you, I know nothing of this hahah ughh :(

Answer 9

you don't know how to find the factors of 10?

which pairs of numbers can be multiplied together to give you 10?

Answer 10

10 and 1, 2 and 5 ?

Answer 11

right. that wasn't so hard, was it? :-)

Answer 12

Haha teach me like you would a 5 year old :) haha

Answer 13

Now, watch my hands carefully :-)

(x+a)(x+b) = x^2 + bx + ax + ab = x^2 + (a+b)x + ab

notice how the last term depends only on the values of a and b?

Let's continue:
(x+c)(x^2 + ax + bx + ab) = x^3 + ax^2 + bx^2 + abx + cx^2 + acx + bcx + abc
= x^3 + (a+b+c)x^2 + (ab + ac + bc)x + abc

again, last term depends only on the values of a, b and c

Answer 14

the lesson here is that so long as we are multiplying things like (x+a) where the coefficient of x = 1, the last term is going to be just the product of all of those +a, +b, +c

Answer 15

now, something else that we know about polynomials is that they can be written in factored form as
(x-r1)(x-r2)...(x-rn) where r1, r2, ... rn are the roots. the roots are the places where the polynomial has value 0, so at x = r1, we'll have a (x-r1)=0 in the product and the result will be 0, no matter what the rest of the polynomial's factors are. similarly, at x = r2, we have (x-r2) = 0 in the product, and so on.

Answer 16

After you said 'Lets continue' on the first one, where did you get (x+c) ?

Answer 17

so, all of those (x+a)(x+b)(x+c)'s can be guessed at pretty easily by factoring that last term.

Answer 18

oh, I was just continuing the pattern. if we multiplied 4 things together, I'd write
(x+a)(x+b)(x+c)(x+d) = <lots of stuff> + abcd
and so on.

Answer 19

I'm just trying to illustrate why it is that we can guess the root candidates by factoring that last term.

Answer 20

now, it's more complicated if the first term doesn't have 1 as a coefficient. we'll leave that for some other time, and maybe some other helper :-)

Answer 21

Omg this is confusing haha try to teach me using the problem, so I can see fixed numbers instead of a bunch of letters haha

Answer 22

so let's try some simple examples before we tackle your big hairy problem.

let's say we have a polynomial with roots at x = -1 and x =2. we could write that in factored form as (x-(-1))(x-2) = (x+1)(x-2) = x^2 -2x + 1x -2 = x^2 -x -2

let's evaluate it at x = -1 and x = 2, just to check if we did that right. if we did, the polynomial will equal 0.

x^2 - x - 2 at x = -1
(-1)^2 - (-1) -2 = 1 + 1 - 2 = 0

x^2 - x - 2 at x = 2
(2)^2 - (2) - 2 = 4 - 2 - 2 = 0

okay, we did it right.

now forget that we know the roots are x = -1 and x = 2. say we just got this polynomial dropped in our lap, with instructions to find the roots.

Answer 23

we say "hmm, ends in -2, and the first term has coefficient 1" and now figure out the possible positive and negative factors of -2. what are they?

Answer 24

(figure out the factors of 2, and write each number with both a + and a - sign)

Answer 25

hmm, +-2, or +-1?

Answer 26

right! now, we get more possible factors than we will have roots. there are a couple of reasons for this.

first, if the last term is positive, we don't know if it was gotten by multiplying two positive numbers or two negative numbers. both (x-2)(x-2) and (x+2)(x+2) end with a +4 when you multiply them out.

(x-2)(x-2) = x^2-2x-2x+4 = x^2 - 4x + 4

(x+2)(x+2) = x^2 + 2x + 2x + 4 = x^2 + 4x + 4

second, if the last term has more than two factors, there are different ways we could have gotten there:

(x+1)(x+4) = x^2 + 1x + 4x + 4 = x^2 + 5x + 4
(x-1)(x-4) = x^2 -1x -4x + 4 = x^2 - 5x + 4
(x-2)(x-2) = x^2-2x-2x+4 = x^2 - 4x + 4
(x+2)(x+2) = x^2 + 2x + 2x + 4 = x^2 + 4x + 4

see how we had a bunch of different ways of ending up with "+4" at the end? we don't know which one is right until we plug some candidates into the polynomial and see if they make it equal 0.

Answer 27

Does that make sense? We're essentially doing a little puzzle here to find out the answers, and the rational root theorem gives us some clues as to where we should look.

Answer 28

Yes That makes sense :) So we list all the possibilities and plug them in, to get 0?

Answer 29

sorry, had a power failure :-(

Answer 30

yes, we make our list of candidates and try them, one by one. There's a wrinkle: when we find that makes the polynomial 0, we can divide the polynomial with long or synthetic division by (x-root) to get a simpler polynomial which will have all the same remaining roots. At that point, we try the same number again, just in case we have a repeated root (maybe the factored polynomial was something like (x+a)^2(a+b), for example).

When we get to a polynomial that we can't factor, we probably have a quadratic equation for which we can use the quadratic formula to find the complex roots.

Answer 31

:O