Understanding Polynomials: Explaining Degree and Real Roots

[nobahar]

Hello again my favourite helpers (or should I say, saviours!)
Long time no speak, but I am in more need of an explanation than an answer.
In this mathematics textbook I have, it gives an explanation under the heading Polynomials in general.
It goes as follows:

"If f(x)= axⁿ+bx^n-1+cx^n-2+...+k, a does not equal zero, then f(x) is said to be a polynomial of degree (or order) n in the variable x.
If we consider the equation f(x) = 0 this will have up to n real roots.
In particular, if f(x) is a cubic function ax³+bx²+cx+d then the equation f(x) = 0 can have up to three real roots. The number of real roots will depend upon the values of a, b, c and d."

(Sadler, A. J. and Thorning, D. W. S. (2007) Understanding pure mathematics, Glasgow : Oxford University Press.)

Then it gives graphs of the roots if a is greater than 0.
Basically, I was hoping someone could elaborate for me!
Especially the part about f(x) = 0 and n real roots. What is meant by real roots?
Thanks, as always!

P.S. If I have to include the three section titles, let me know as oppose to just deleting the thread if possible, it's just that they had no relevance here... :)

 

[CompuChip]

Well, first of all, I assume that you understand the condition a not equal to zero in the definition (after all, we do not want x² to be called a degree 3 polynomial, do we ).

Do you know anything about complex numbers?
If not, then you can leave out the "real" and just read it as: "f(x) can have up to three roots". As a simpler example consider quadratic functions a f(x) = x² + b x + c. They can have no roots at all (e.g. x² + 1), just one root (e.g. x²) or two roots (e.g. x² - 1) but never more.
If you really want to know what is meant by "real" numbers, find a resource on complex numbers.

 

[nobahar]

Thanks Compuchip.
So something of degree 4 will have 4 real roots (since the n^th term has n roots)? Just checking, and since the use of real is explicit, that implies there are going to be complex roots?
Following Compuchips advice, I found a source that says an n^th term has n roots of the form a+bi, which is a complex number. How does this solve polynomials? Since the complex number is formed from an imaginary part. Are the real roots 'actual' solutions, since the polynomial could have 'actual' - or applied- uses?
Again, thanks in advance.

 

[HallsofIvy]

Thanks Compuchip.
So something of degree 4 will have 4 real roots (since the n^th term has n roots)? Just checking, and since the use of real is explicit, that implies there are going to be complex roots?

No. A polynomial of degree 4 can have up to 4 real roots.

Following Compuchips advice, I found a source that says an n^th term has n roots of the form a+bi, which is a complex number. How does this solve polynomials?

It doesn't. However, the "fundamental theorem of algebra" says that any polynomial equation has a solution in the complex numbers- which, in turn, means that it has a linear factor of the form (x- (a+ bi)). Since the other factor is now a polynomial of degree one less, we can no apply the theorem to that to get another root (possibly the same as the first) and so another factor. Any n^th degree polynomial can be factored into n linear factors, not necessarily different, using complex numbers. Since any of those complex number could happen to be real, an n^th degree polynomial can have anywhere from 0 to n real roots.

Since the complex number is formed from an imaginary part. Are the real roots 'actual' solutions, since the polynomial could have 'actual' - or applied- uses?
Again, thanks in advance.

That depends on the application. If there exist some "non-mathematical" reason why only real number solutions could work in the application, then, yes, you would reject any non-real solution. But there are plenty of applications in which complex number solutions are perfectly plausible.

 

[nobahar]

The "fundamental theorem of algebra" says that any polynomial equation has a solution in the complex numbers- which, in turn, means that it has a linear factor of the form (x- (a+ bi)). Since the other factor is now a polynomial of degree one less, we can no apply the theorem to that to get another root (possibly the same as the first) and so another factor.

Hi halls, the help is much appreciated, but an example would be good , I can't see where the x- (a+bi) comes from...

 

[CompuChip]

Let's consider the simple example
$$f(x) = x^2 + x - 2$$
One root (solution of f(x) = 0) is x = 1, as you can easily check. That means that we can write f(x) as (x - 1) * g(x), with g(x) some other function, which guarantees that if you plug in x = 1, you get zero. You can also easily see, by doing the multiplication that, g(x) = x - 2. This has as a solution x = 2, so f(x) is (x - 1)(x - 2).

If the roots are not real, but complex, say "2 + 3i" is a root, the same idea holds, can you can write your original function F(x) as (x - (2 + 3i)) * G(x), with G(x) some other function.

It is quite obvious, hopefully, that if f(x) is some polynomial with root x0, and you write it as f(x) = (x - x0) g(x), that then g(x) is again a polynomial and its degree is exactly one lower than that of f(x). So if you repeat the trick by writing g(x) = (x - x1) h(x), and h(x) as h(x) = (x - x2) k(x), eventually you will end up with
f(x) = (x - x0) (x - x1) (x - x2) ... z(x)
where z(x) is a polynomial of degree 0, i.e.: a constant.

The roots x0 may be real numbers such as 1 or $1 + \sqrt{5}$ or $3 + \frac{\sqrt{12}}{7}$, or complex numbers such as $1, i, 3 + 2i, \frac{12}{3 + 5i}$, etc.

 

[nobahar]

I know this is going to be wrong but...
Doesn't f(x)=x²+x-2=(x-1)(x+2)? As oppose to (x-1)(x-2)?

 

[Dick]

I know this is going to be wrong but...
Doesn't f(x)=x²+x-2=(x-1)(x+2)? As oppose to (x-1)(x-2)?

Sure. The other solution is x=(-2). Typo. The explanation is fine though.

 

[nobahar]

Of course, I wasn't disputing that!
It's excellent and I'm extremely grateful, I just wanted to make sure I had interpreted it correctly. Thanks Dick for the response; and although it wasn't mentioned in the last message: thanks Compuchip, much appreciated.