Find all real roots of this polynomial

quasar987

How do we do this? I know how to find all RATIONNAL roots but what about the irrationnal ones?

[tex]2x^5-5x^4-11x^3+23x^2+9x-18=0[/itex]

matt grime

If it doesn't factor into smaller degree polys (over Q) then analytic answers may be obtained using elliptic functions, numerical answers using your preferred numerical method.

You don't state if it has any rational solutions. does it? if so you can factor them out.

arildno

It might not be enough to find all roots, but note that both x=1 and x=-1 are roots.

arildno

A third root is x=-2.
Hence, you should be able to find the last two.

HallsofIvy

Actually, all five roots are rational.

quasar987

Could you explain how you can see just by looking at the thing that

1) It has no more no less than 5 roots

2) They are all rationnal


Thanks!

Manchot

He probably didn't just glance at and see that there are five rational roots. The easiest way to find them is just graph the polynomial on a graphing calculator and see that the zeros are all rational. The more tiresome way is to do synthetic/long division. You know that all of the rational roots are of the form [itex]\pm\frac{b}{a}[/itex], where b is the set of factors of 18 and a is the set of factors of 2. From there you could just use trial and error.

quasar987

This is the way I was refering to when I said "I know how to find all RATIONNAL roots". This is also the only way I am allowed to use since this excerice was taken from a real analysis textbook.

So without the use of a calculator, how can we know that a polynomial as irrationnal roots, and how can we find them?

Manchot

Ok, let's assume that you already know the rational roots of a polynomial, which in this case will be a function of x. Take your original polynomial, and divide it by x-a, where a is one of the rational roots that you already know. You can use either long division or synthetic division (since they're really the same thing, anyway). Take the result and divide by x-b, where b is another one of the rational roots. Keep doing this until you've "used up" all of your known roots.

Now, you'll have one of two possibilities. Either you've reduced the polynomial completely, in which case you have no irrational roots, or you haven't, which means that you do. The number of them that you have is equal to the degree of the "leftover" polynomial. Therefore, if it's quadratic, you'll have two, cubic will have three, etc.

To find them, think about it this way. You know that the leftover polynomial must equal zero when x is equal to one of the roots. Therefore, all you have to do is set the polynomial equal to zero, and solve it. If you have a quadratic one, it's relatively easy: just use the quadratic formula. If you have cubic or quartic ones, you have to use the cubic and quartic equations, respectively. However, I doubt that you'll ever be given a problem where this will happen. If you're left with a fifth degree polynomial or higher, good luck. It's been proven that there is no equation which solves these in terms of "simple" functions. You'll have to use inverse functions.

NateTG

A polynomial of degree [tex]n[/tex] has [tex]n[/tex] roots. Sometimes, there are multiple roots with the same value -- for example [tex]x^2-2x+1[/tex] has the roots [tex]{1,1}[/tex], and sometimes roots are imaginary i.e. [tex]x^2+1[/tex] has the roots [tex]{i,-i}[/tex]

By the time that Halls posted, three roots had already been mentioned, so it would be easy to divide out by [tex](x-1)[/tex],[tex](x+1)[/tex] and [tex](x+2)[/tex] leaving a quadratic polynomial which can be attacked using the quardratic formula.

In general, it's usually good to after the easy roots first.

shmoe

Another "easy" root of this polynomial is x=3. Combined with 1,-1 and -2 you have 4 rational roots. You know that the sum of the roots of your polynomial is 5/2. Since you are guaranteed 5 roots (though an even number of them may be complex), you know that the fifth must also be rational (it must also be real in this case). You can also use this to see the remaining root is 3/2.

To show a polynomial has irrational roots, first find all the rational ones and factor them out. If what's left is 5th degree or higher, graph it by hand (otherwise you can use the formulas). Actually, before you do this see if there's any clever simplifications you can do, like turning x^6-2x^3-6 into y^2-2y-6. You should be able to crudely approximate where the remaining (if any) real roots are (they are necessarily irrational), this depends on how crude your graph is. You can prove without a doubt that these roots exist using the intermediate value theorem. You can approximate them using Newtons method as accurately as you like if need be.

quasar987

What is the logic behind this dividing by (x-b) buisness? How come a polynomial of degree 5 divided by (x-b) will give a polynomial of degree 4 with the same roots are that of the degree 5?

Also, from shmoe's last post: How can you tell that the sum of the roots of the polynomial is 5/2 ?

Gokul43201

Let the roots be a, b, c, d, e. That means the polynomial is [itex](x-a)(x-b)(x-c)(x-d)(x-e) = 0[/itex]

Expand this to see that it becomes [itex]x^5 - (a+b+c+d+e)x^4 + . . . - abcde = 0[/itex]

So, the sum and product of roots are easy to identify from the coefficients of the polynomial.

shmoe

If f is a polynomial of degree 5 and f(b)=0, then f(x)=(x-b)g(x), where g(x) is a 4th degree polynomial. Division will pull out this g(x) for you. Since f(x)=(x-b)g(x), any root of g(x) will be a root of f(x), so we keep b in our pocket and can just find the roots of g(x) to get the remaining roots of f(x).

It's important to note that the roots of f and g are not necessarily the same, b may not be a root of g (it could be though). But you are guaranteed that any root of g is a root of f and that any root of f different from b will be a root of g, hence this will give all roots of f.


What Gokul said...note that his polynomial was monic (leading term was 1), yours wasn't, so you have to divide by 2. You can multiply a polynomial by a non-zero constant without affecting it's roots.

quasar987

I SEE! Nice.

O.K. One less problem, 7945 to go !