Consider $P(x) = 5x^6 - ax^4 - bx^3 - cx^2 - dx - 9$, where $a$; $b$; $c$; $d$ are real. If the roots of $P(x)$ are in AP, find the value of $a$.

Question

Consider $P(x) = 5x^6 - ax^4 - bx^3 - cx^2 - dx - 9$, where $a$; $b$; $c$; $d$ are real. If the roots of $P(x)$ are in arithmetic progression, find the value of $a$.

Although I am sure that this problem requires Vieta's formula I don't know them for 6-degree equation. (I think we factor the 6-degree equation to two 3-degree equations.)

Answer 1

Guide:

• Let $x_i$ be the root of a degree $6$ polynomials that is monic.

$$\prod_{i=1}^6 (x-x_i)=0$$

• Upon expanding we can see that the coefficient of $x^5$ is the negative of the sum of the roots.

• Also, the constant term for the monic polynomial is equal to the product of the roots. For even degree monic polynomial, this is the case, for odd degree, there is a negative sign. Try to expand for various degree to observe that.

• Let the root be

$$m\pm\frac{5d}{2}, m\pm\frac{3d}{2}, m\pm\frac{d}2$$

Using the properties of the sum and the product, you should be able to solve for $m$ and $d$ and recover the whole polynomials if you want.