[Math] Why does degree determine the amount of zeros

linear algebrapolynomialsroots

We just learned about complex numbers in my math class and I have a question. Why does the degree of a polynomial equal the amount of zeros it has?

The degree of $f(x) = x^3 – x^2 + x – 1$ is $3$, but there is only $1$ real zero, $x=1$.

There are 3 complex zeros, $x=1$,$i$,$-i$, which equals the degree number. I just don't understand why there isn't a case where a fifth-degree polynomial has the zeros $x=1$,$i$,$-i$ but none other. Why should it have to have five zeros?

I asked my teacher and she said, "polynomials are closed under the complex numbers," but I don't know what that means. ._.

Best Answer

Hint: $\alpha$ is a root of $f(x)$ if and only if $(x-\alpha)$ divides $f(x)$.

The complex numbers $\mathbb{C}$ are algebraically closed (which is what your teacher probably meant).

This mean that a polynomial $p(x)$ of degree $n$with coefficients in $\mathbb{C}$ always "factors completely" as \begin{equation} p(x)=a(x-\alpha_{1})\dots(x-\alpha_{n}), \end{equation} where $a,\:\alpha_{1},\dots,\:\alpha_{n}\in\mathbb{C}$ and the roots of course can be repeated.

Related Solutions

[Math] How to find the roots of a quadratic function

The quadratic functions are generally in the form $f(x)=Ax^2+Bx+C$. The zeroes of this function is then the solution to the equation $Ax^2+Bx+C=0$.$($where $A\ne 0)$
Hence we try to solve this equation.

If you try to solve the equation (just like you'd solve a linear equation), you'd run into trouble when you realize that the $x$s can't be as easily isolated like in the case with linear equations. So we use a different approach, what is called the "completion of squares".

If you are familiar with the method of "completing the squares" (and it isn't very difficult,) you'd know that our goal is to somehow "arrange" the equation in the form $(\text{linear binomial})^2+\text{constant term}$. When you arrange the equation in this way, isolating the $x$s is very easy. I'll not show you the entire process of completing the square for our given function but I presume that you are somewhat familiar with it (if not, I suggest you look it up on the internet).

The end result of our method will yield a very familiar expression, what we are used to calling as the "quadratic formula". You should end up with something like this: $$x=\frac{-B}{2\cdot A} \pm \frac{\sqrt{B^2-4\cdot A\cdot C}}{2\cdot A}$$ where $A$, $B$ and $C$ are the coeffients of our function $f(x)$.

The relation above should give you the zeroes of any quadratic function with coefficients $A$, $B$ and $C$ ( Also note that $A$, $B$, $C \in \mathbb R$). Note that this particular relation has many take-aways. First it hints you of the existence of $2$ zeroes (notice the $\pm$ sign in our relation). Second it kind of explains why $ A \ne 0$ (if $A=0$ then the whole relation goes meaningless, for when you divide any thing by $0 $; in our case $2 \times 0 $, the result is undefined.)

The third, and probably the most important one, is that it tells you when $x$ is a Real number or when it is not (i.e when it is a Complex number). Remember that while $A$, $B$ and $C$ must always be Real numbers, it isn't necessary for $x$ to be Real. It is a well known fact square roots of negative numbers are not Real number, instead, we call them the imaginary numbers $($ and the complex of Real and Imaginary numbers are called the Complex Numbers, usually in the form $a+b\mathbb i $ and $\mathbb i =\sqrt{-1})$.
So,

The expression under the square root sign, if positive, we have two zeroes that are real (more commonly known as two distinct and Real roots).
If the expression under the square root sign is negative, we will get two zeroes that are Complex Numbers (more commonly known as two distinct but Complex roots).
You might ask what happens if the expression under the square root sign equals $0$? In that case, our relation would reduce down to $$x=\frac{-B}{2\cdot A} \pm 0$$ OR, $$x=\frac{-B}{2\cdot A}$$ So if the expression under the square root sign equals zero then you'd still get Real values for the zeroes but you'd get just a single "zero" for the equation (more commonly known as repeated and Real roots).

The geometric interpretation of these three cases is also pretty interesting and gives a deeper insight at what is going on (Although your question has already been answered, i suggest you keep reading, I'm almost done :)).
You should know that the zeroes of a function are those values of $x$ which results in $f(x)$ being equal to zero. Or, if you graph the function, It is that point on the graph(or that value for $x$ ), where the function crosses the x-axis. So, what do these three cases imply on the graph then?

If the roots are distinct and real, or, if the discriminant (btw, that is the name for the expression under the square root) is positive, then, the function should cross the x axis twice.Or
Similarly, if the roots are Real and repeated, (i.e the discriminant is zero), the function should have only one point crossing (or in this case touching) the x-axis. Or
Now if the roots are not real, then what? Because there are two distinct roots (but not real), it might be tempting to think that it also crosses the x- axis twice but, good luck finding the point where it actually crosses the x-axis! (where will you find a number $2+3\mathbb i $ on the x axis?). So the next logical deduction could be that it doesn't actually cross the x-axis at all, which is very plausible. And this is exactly what happens. OR,
And with that I have come to the end of my uber-long answer. (Phew!)

As far as your question is concerned, If you'd like to know if the zeroes are Real or not, check if $B^2 -4AC$ is positive, negative or zero, the nature of the roots (zeroes) should be apparent.

I bothered to write down such long an answer because I really wanted you to know what's going on behind the scenes. You might already know this stuff but still, I Hope it helps

[Math] A polynomial of degree 3 that has three real zeros, only one of which is rational.

Yes, your answer is correct.

Another way to come up with some is to use the form $x^3-nx \;\;\forall\; n\in\Bbb N \land \sqrt n \notin \Bbb N$ (i.e. where $n$ isn't a perfect square but natural).

For the quartic, consider the equation $x^4-(a+b)x^2+(ab)\;\;\forall \;a,b \in \Bbb N \land \sqrt a, \sqrt b \notin \Bbb N$ (i.e. where $a+b$ is the sum of two non-perfect square natural numbers and $ab$ is their product).

Best Answer

Related Solutions

[Math] How to find the roots of a quadratic function

[Math] A polynomial of degree 3 that has three real zeros, only one of which is rational.

Related Question