# How Do You Solve High Degree Polynomials in Calculus Problems?

• b2386
In summary, the problem requires finding the maximum value of r for this equation:r^{2}e^{\frac{-r}{a}}(1-\frac{r}{a}+\frac{r^2}{4a^{2}})I took the derivative and set it equal to zero but kept getting a high degree polynomial that I am not sure how to solve for r. I tried Newton's method but it didn't work. I am in precalc and need help figuring out where I am mathematically.
b2386
Hi everyone,

I am having trouble finding an answer for this problem.

The problem requires me to find the maximum value of r for this equation:

$$r^{2}e^{\frac{-r}{a}}(1-\frac{r}{a}+\frac{r^2}{4a^{2}})$$

I took the derivative and set it equal to zero but keep getting a high degree polynomial that I am not sure how to solve for r.

Any help would be appreciated.

b2386 said:
Hi everyone,

I am having trouble finding an answer for this problem.

The problem requires me to find the maximum value of r for this equation:

$$r^{2}e^{\frac{-r}{a}}(1-\frac{r}{a}+\frac{r^2}{4a^{2}})$$

I took the derivative and set it equal to zero but keep getting a high degree polynomial that I am not sure how to solve for r.

Any help would be appreciated.

You will get a polynomial of degree 4. You can factor it and solve for each of the factors individually. This is not necessarily enjoyable, but you can do it. You could use Newton's method, or you could graph it and solve numerically.

What did you get when you differentiated it?

Once I took out all the extra fluff, I am left with this:

$$2-\frac{r}{a}-3r+\frac{r^2}{a^2}+4r^2-\frac{r^3}{4a^3}=0$$

Could you explain how to factor this polynomial using Newtons method? I am not familiar with it.

Also, I know r is equal or close to 5a

b2386 said:
Once I took out all the extra fluff, I am left with this:

$$2-\frac{r}{a}-3r+\frac{r^2}{a^2}+4r^2-\frac{r^3}{4a^3}=0$$

Could you explain how to factor this polynomial using Newtons method? I am not familiar with it.

Also, I know r is equal or close to 5a

Also, if you think r=5a is one solution, then plug that in into the derivative and you will get zero if it is a solution.

Newton's method is a root finding algorithm. You can look it up and try it if you would like, but it would probably be better to wait on that. Are you in precalc? Cause you are doing calculus right now... so I'm a little unsure about where you are mathematically.

b2386 said:
Once I took out all the extra fluff, I am left with this:

$$2-\frac{r}{a}-3r+\frac{r^2}{a^2}+4r^2-\frac{r^3}{4a^3}=0$$

Could you explain how to factor this polynomial using Newtons method? I am not familiar with it.

Also, I know r is equal or close to 5a

"Newton's method" is not a method for factoring- it's a method for finding numerical approximations to solutions to equations.

In any case, your polynomial is incorrect.
Multiplying that "r2" into the polynomial part, you have
$$y= e^{-\frac{r}{a}}\left(\frac{r^4}{4a^2}-\frac{r^3}{a}+ r^2\right)$$
$$y'= -\frac{1}{a}e^{-\frac{r}{a}}\left(\frac{r^4}{4a^2}-\frac{r^3}{a}+ r^2\right)+ e^{-\frac{r}{a}}\left(\frac{r^3}{a^2}-\frac{3r^2}{a}+ 2r\right)$$
Setting that equal to 0, you can, of course, divide through by the exponential to get rid of it and then you have
$$-\frac{r^4}{a^3}+ \frac{r^3}{a^3}- \frac{r}{a}+ \frac{r^3}{a^3}-\frac{3}{a}r^2+ r= 0$$
(Notice that equal powers of r have equal powers of a in the denominator!)
You immediately factor out an "r" (remembering that one solution is r= 0) and multiply through by [itex]-a^3[/tex] to get
$$r^3- 2ar^2+ 4a^2r- 2a^3= 0$$

You can get rid of the a by letting x= r/a so r= ax and the equation becomes
$$a^3x^3= -2a^3x^2+ 4a^3x-2a^3= 0$$
or
$$x^3- 2x^2+ 4x- 2= 0$$
If that has any rational roots, they can only be 1, -1, 2, or -2, and it's easy to check that none of those work. This clearly has an irrational root between 0 and 1 and, graphing, it appears that that is the only real root.

Last edited by a moderator:

## 1. What is the definition of a derivative of a polynomial?

The derivative of a polynomial is the mathematical concept that represents the rate of change of the polynomial at a particular point. It is the slope of the tangent line to the curve of the polynomial at that point.

## 2. How do you find the derivative of a polynomial?

To find the derivative of a polynomial, you need to use the power rule. This rule states that the derivative of a term with a variable raised to a power is equal to the coefficient of the term multiplied by the power and then the variable raised to one less power. You can apply this rule to each term in the polynomial and then combine the derivatives to get the overall derivative of the polynomial.

## 3. Why is the derivative of a polynomial important?

The derivative of a polynomial is important because it helps us understand the behavior of the polynomial. It tells us how quickly the polynomial is changing at a particular point, which can be useful in many applications such as calculating velocity, acceleration, and optimization problems.

## 4. Can the derivative of a polynomial be negative?

Yes, the derivative of a polynomial can be negative. This means that the polynomial is decreasing at that particular point. It is also possible for the derivative to be zero, which indicates a horizontal tangent line and a point of inflection for the polynomial.

## 5. Are there any special cases for finding the derivative of a polynomial?

The only special case for finding the derivative of a polynomial is when the polynomial has a constant term, such as f(x) = c. In this case, the derivative is simply 0, because the rate of change of a constant is always 0.

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