Polynomial with five roots: determine the roots of the equation x^5+ax^4+bx^3+cx^2+dx+e=0 as functions of a,d and e

• MHB
• lfdahl
In summary, the general formula for finding the roots of a polynomial with five roots is known as the quintic formula, which involves all the coefficients of the polynomial. It is not commonly used due to its complexity, and there are more efficient methods available. The roots cannot be expressed as functions of only the coefficients a, d, and e, as the coefficients b and c must also be considered. A polynomial with five roots can have a maximum of five real roots, but this number can vary depending on the coefficients. There are faster methods for finding the roots, such as the rational root theorem, synthetic division, and the quadratic formula. The roots can be verified by plugging them back into the original equation or by graphing the polynomial.
lfdahl
Gold Member
MHB
I am so sorry for having posted this challenge/puzzle with a serious typo:

The roots of the equation should be functions of $a, d$ and $e$. In my old version I wrote $a, b$ and $e$.

I will see to, that future challenges are properly debugged before posting.For $e \ne 0$, determine the roots of the equation $x^5+ax^4+bx^3+cx^2+dx+e = 0$
as functions of $a, d$ and $e$, given that the equation has two roots whose product is $1$ and
two other roots whose product is $−1$.

Last edited:
Here´s the suggested solution:
Let the roots be $p, \frac{1}{p}, q, -\frac{1}{q}$ and $r$.

(1). $p \cdot \frac{1}{p} \cdot q \cdot( -\frac{1}{q})\cdot r = -e$ so $r = e$.

(2). $(p + \frac{1}{p}) + (q - \frac{1}{q})+r = -a$ so $(p + \frac{1}{p}) + (q - \frac{1}{q}) = -(a+e)$.

(3). $-\frac{e}{p}-ep-\frac{e}{q}+eq-1 = d$ so $(p + \frac{1}{p}) - (q - \frac{1}{q}) = -\frac{d+1}{e}$.

From this, we get:

$$p+\frac{1}{p} = \frac{1}{2}(-(a+e)-\frac{d+1}{e}) = -\frac{1}{2e}(e^2+ae+d+1) = A,$$

$$q - \frac{1}{q} = \frac{1}{2}(\frac{d+1}{e}-(a+e)) = -\frac{1}{2e}(e^2+ae-d-1) = B.$$

Then $p^2-Ap+1 =0$ or $p = \frac{1}{2}(A\pm \sqrt{A^2-4})$ and $\frac{1}{p}=A-p = \frac{1}{2}(A\mp \sqrt{A^2-4})$.
We may use the plus sign for $p$ and the minus sign for $\frac{1}{p}$.
Similarly, we have $q = \frac{1}{2}(B + \sqrt{B^2+4})$ and $\frac{1}{q} = \frac{1}{2}(B - \sqrt{B^2+4})$.

1. What is the general formula for determining the roots of a polynomial with five roots?

The general formula for finding the roots of a polynomial with five roots is known as the quintic formula, which is a complex formula involving the coefficients of the polynomial. It is not commonly used in practice due to its complexity and the availability of more efficient methods for finding roots.

2. Can the roots of a polynomial with five roots be expressed as functions of only the coefficients a, d, and e?

No, the roots of a polynomial with five roots cannot be expressed as functions of only the coefficients a, d, and e. The coefficients b and c must also be taken into account in order to determine the roots.

3. How many real roots can a polynomial with five roots have?

A polynomial with five roots can have a maximum of five real roots, but it is also possible for it to have fewer or even no real roots. The number of real roots depends on the values of the coefficients in the polynomial.

4. Is there a quicker method for determining the roots of a polynomial with five roots?

Yes, there are several more efficient methods for finding the roots of a polynomial with five roots. These include using the rational root theorem, synthetic division, and the quadratic formula. These methods are faster and more practical than using the quintic formula.

5. How can the roots of a polynomial with five roots be verified?

The roots of a polynomial with five roots can be verified by plugging them back into the original equation. If the resulting value is equal to zero, then the root is valid. Additionally, graphing the polynomial can also help verify the roots, as they will be the x-intercepts on the graph.

• General Math
Replies
1
Views
783
• General Math
Replies
3
Views
1K
• General Math
Replies
7
Views
1K
• General Math
Replies
1
Views
1K
• General Math
Replies
2
Views
2K
• General Math
Replies
9
Views
2K
• General Math
Replies
16
Views
3K
• General Math
Replies
1
Views
833
• General Math
Replies
1
Views
1K
• General Math
Replies
1
Views
1K