# Polynomial degree and root relationship

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Why is it that for a 7th degree polynomial, the number of real roots is either 1, 3, 5, or 7?

## Answers and Replies

jedishrfu
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Here's the Khan Academy discussion on it:

https://www.khanacademy.org/math/al...ntal-theorem-of-algebra/v/possible-real-roots

It has to do with non-real roots coming in conjugate pairs that cancel out the imaginary part of the when multiplied together so there won't be any coefficients in the polynomial that have imaginary components ie the polynomial will have only real coefficients and won't have any complex or imaginary coefficients.

Graphically, how many times can it cross the x-axis (ie at y=0)?

Each of those crossings is a real root of the equation.

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Mark44
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Why is it that for a 7th degree polynomial, the number of real roots is either 1, 3, 5, or 7?
There are a couple of reasons.

One: The most fundamental 7th degree polynomila function is ##y = x^7##, which is similar to the graphs of ##y = x##, ##y = x^3##, and ##y = x^5##. The basic behavior of all of these odd-degree polynomial functions is that if x < 0, then y < 0, and if x > 0, then y > 0. The graph must rise more or less steadily through the 3rd quadrant, continuing on into the first quadrant.

Two: Complex roots always occur in conjugate pairs, so if x = a + bi is a root, then x = a - bi must also be a root. If there are 0 complex roots, there are 7 real root. If there are two complex roots, there must be 5 real roots, and so on, eliminating the possibility of an even number of real roots.

The fundamental theorem of algebra states that any polynomial has at least one root. But one can extend this by using the polynomial remainder theorem and say that any polynomial of degree n has exactly n number of roots, counting the multiplicity. There's another theorem, the complex conjugate root theorem with states that if a+bi is a root, then its own conjugate a-bi is also a root (this only works for polynomials with real coefficients). This means that if there are non-real roots, they come in pairs. For a seventh degree polynomial (with real coefficients), the number of complex roots could only be 0, 2, 4, or 6, therefore the number of real roots could be 1, 3, 5, or 7, as stated.

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fresh_42
Mentor
Why is it that for a 7th degree polynomial, the number of real roots is either 1, 3, 5, or 7?
Let ##p(x)\in \mathbb{R}[x]## be our polynomial. If we had a root, say ##r_1##, then with the Euclidean algorithm, a long division, we get ##p(x)=q(x) \cdot (x-r_1)## since ##p(r_1)=0##. There is at least one real root, as the graph of ##p(x)## has to cross the ##x-##axis at least once, because it comes from ##+\infty## and goes to ##-\infty## or vice versa. Now we can go on with ##q(x)## which has degree ##6##. Either has ##q(x)## also a real root, which does not have to the case, or it has not. If it has, say ##r_2##, we continue the division by ##(x-r_2)##. But then we get a polynomial of degree ##5##, which thus again has to have a real root, because the degree is odd and the graph has again to cross the ##x-##axis. At the end, we will get an odd number of roots.

Another way is to directly look at possible factorizations ##p(x)=q_1(x)\cdot \ldots \cdot q_n(x)##. The only polynomials, which cannot be factored further are either of the form ##x+a## which give us roots ##-a##, or quadratic. Now gathering all factors, we get
$$p(x)=(x-r_1)\cdot \ldots \cdot (x-r_m)\, \cdot \, q_1(x) \cdot \ldots \cdot q_k(x)$$
with quadratic polynomials ##q_i(x)##. So the degree equation reads: ##\deg p(x)=2\cdot k + m##. So if ##\deg p(x)=7## or any other odd number, ##m## has to be odd, too.