How to show that a 5-th degree polynomial has a root?

• Eclair_de_XII
In summary, the problem asks to show that the given polynomial has a positive root, and using Descartes' rule of signs and the intermediate value theorem, it can be proven that the function has exactly one positive root. Quoting and using the theorem should suffice as a proof.
Eclair_de_XII

Homework Statement

"Show that for some ##x\in ℝ##, that ##x^5+2x^4+3x^3+2x^2+x=1##."

The Attempt at a Solution

Okay, so I know from Descartes' rule of sign that the function ##f(x)=x^5+2x^4+3x^3+2x^2+x-1## has exactly one positive root, since the sign of the coefficients change exactly once in the entire function. But I am asked to show that it has a positive root, and I am not fully sure that this explanation would cut it.

I also have that ##x^5+2x^4+3x^3+2x^2+x=\frac{1}{2}x^5+2x^4+3x^3+2x^2+\frac{1}{2}x+\frac{1}{2}x(1+x^4)=\frac{1}{2}x\sum_{k=0}^4 \binom 4 k x^k+\frac{1}{2}x(1+x^4)=\frac{1}{2}x[(1+x)^4+(1+x^4)]##. But I have to show that this expression is identically one, and I haven't an idea of how to do this.

intermediate value theorem. i.e. show it is somewhere less than 1 and somewhere greater than 1, and then use the theorem that the set of values is an interval.

Notice they do not ask you to actually find the value x where f(x) = 1, merely to prove there is one.

(I missed this problem freshman year in college. i didn't realize you are supposed to actually read and internalize the theorems that are presented in the course. I thought you were just supposed to brainstorm every problem on your own. the moral is that the theory presented in the course has uses.)

Similarly every odd degree polynomial has a root.

Gotcha. Thanks for the tip.

Eclair_de_XII said:

Homework Statement

"Show that for some ##x\in ℝ##, that ##x^5+2x^4+3x^3+2x^2+x=1##."

The Attempt at a Solution

Okay, so I know from Descartes' rule of sign that the function ##f(x)=x^5+2x^4+3x^3+2x^2+x-1## has exactly one positive root, since the sign of the coefficients change exactly once in the entire function. But I am asked to show that it has a positive root, and I am not fully sure that this explanation would cut it.
Yes, it would "cut it", presuming you have had Descartes' rule of signs.

Show that for some ##x\in ℝ##, that ##x^5+2x^4+3x^3+2x^2+x=1##."

Odd degree functions looks kind of like a “S” laying on its side,so there must be an x-intercept

Eclair_de_XII said:

Homework Statement

"Show that for some ##x\in ℝ##, that ##x^5+2x^4+3x^3+2x^2+x=1##."

The Attempt at a Solution

Okay, so I know from Descartes' rule of sign that the function ##f(x)=x^5+2x^4+3x^3+2x^2+x-1## has exactly one positive root, since the sign of the coefficients change exactly once in the entire function. But I am asked to show that it has a positive root, and I am not fully sure that this explanation would cut it.

I also have that ##x^5+2x^4+3x^3+2x^2+x=\frac{1}{2}x^5+2x^4+3x^3+2x^2+\frac{1}{2}x+\frac{1}{2}x(1+x^4)=\frac{1}{2}x\sum_{k=0}^4 \binom 4 k x^k+\frac{1}{2}x(1+x^4)=\frac{1}{2}x[(1+x)^4+(1+x^4)]##. But I have to show that this expression is identically one, and I haven't an idea of how to do this.

If quoting and using a theorem does not "cut it", what would cut it for you?

1. How do you determine the degree of a polynomial?

To determine the degree of a polynomial, you need to look at the term with the highest exponent. For example, in the polynomial 5x^3 + 2x^2 - 7x + 3, the highest exponent is 3, so the degree of the polynomial is 3.

2. What does it mean for a polynomial to have a root?

A root of a polynomial is a value that, when plugged into the polynomial, will make the polynomial equal to 0. In other words, it is a value that satisfies the equation and makes it true.

3. How can I show that a 5-th degree polynomial has a root?

One way to show that a 5-th degree polynomial has a root is by using the Intermediate Value Theorem. This theorem states that if a continuous function takes on two values, a and b, at two points, then it must also take on every value between a and b at some point. So, if you can find two points where the polynomial takes on different values, one positive and one negative, then there must be a root between those two points.

4. Can a 5-th degree polynomial have more than one root?

Yes, a 5-th degree polynomial can have up to 5 roots. This is because the degree of a polynomial tells us the maximum number of roots it can have.

5. How do you find the roots of a 5-th degree polynomial?

Finding the roots of a 5-th degree polynomial can be done using different methods such as factoring, graphing, or using the Rational Root Theorem. However, for polynomials with higher degrees, it may not always be possible to find the exact roots, and approximations or numerical methods may be used instead.

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