Find the roots of the polynomial $x^5-5x^4-35x^3+ax^2+bx+c$, given that the roots form an arithmetic progression.

Question

Find the roots of the polynomial $x^5-5x^4-35x^3+ax^2+bx+c$, given that the roots form an arithmetic progression.

What I've tried :

Letting $\alpha$, $\alpha+d$, $\alpha+2d$, $\alpha+3d$, $\alpha+4d$ be the roots of $p(x)$ .

I got sum of roots = $5\alpha + 10d = 5$ or $\alpha + 2d = 1$

So, $1$ is one of the roots of $p(x)$ .

Equating $p(x) = 0$ , I got $a+b+c = 39$ .

And from product of roots , I got $(1 - 4d^2)(1 - d^2) = c$ .

I don't know how to further deduce the remaining roots of $p(x)$ .

Answer 1

Here's a simpler calculation, knowing the algebraic identity of $(\sum a)^2 - 2 \sum ab = \sum a^2$.

By Vieta's formula, you know that the sum of roots is 5, so the middle term is 1. Let the difference be $d$.
Using the algebraic identity, $$ 5^2 + 2 \times 35 = (1 - 2d)^2 + ( 1- d)^2 + 1^2 + (1+d)^2 + (1+2d)^2 = 5 + 10d^2.$$

Hence, $ d = \pm 3$, and the roots are $ -5, -2, 1, 4, 7$.

Answer 2

You know that $\alpha+2d=1$. And you also know that \begin{multline} \alpha (d+\alpha )+(2 d+\alpha ) (d+\alpha )+(3 d+\alpha ) (d+\alpha )+(4 d+\alpha ) (d+\alpha )+\\+\alpha (2 d+\alpha )+\alpha (3 d+\alpha )+(2 d+\alpha ) (3 d+\alpha )+\alpha (4 d+\alpha )+\\+(2 d+\alpha ) (4 d+\alpha )+(3 d+\alpha ) (4 d+\alpha )=35, \end{multline} which means that $35 d^2+40 \alpha d+10 \alpha ^2=35$. And this, together the fact that $\alpha+2d=1$ will give you that the set of roots is $\{-5,-2,1,4,7\}$.

Answer 3

You found that your roots are $1,1\pm d,1\pm2d.$

• Now (ignoring $a,b,c$), $p(x)=(x-1)(x^4-4x^3-39x^2+\dots),$ so $$\begin{align}-39&=\Bigl((1-2d)+(1+2d)\Bigr)\Bigl((1-d)+(1+d)\Bigr)+(1-2d)(1+2d)+(1-d)(1+d)\\&=4+(1-4d^2)+(1-d^2)\\&=6-5d^2\end{align}$$ hence $d=\pm3,$ so your five roots are $$-5,-2,1,4,7.$$
• Alternatively, if you don't want to factor out $x-1$, you can calculate directly the 2nd elementary symmetric polynomial of the five roots: $$\begin{align}-35&=\Bigl((1-2d)+(1+2d)\Bigr)\Bigl((1-d)+(1+d)\Bigr)+(1-2d)(1+2d)+(1-d)(1+d)+1\Bigl((1-2d)+(1+2d)+(1-d)+(1+d)\Bigr)\\&=4+(1-4d^2)+(1-d^2)+4\\&=10-5d^2\end{align}$$ and the same conclusion follows.
• A third way is to do the change of variable $x=1+y:$ $$\begin{align}p(x)&=(1+y)^5-5(1+y^4)-35(1+y)^3+\dots\\&=(y^5+5y^4+10y^3)-5(y^4+4y^3)-35y^3+\dots\\&=y^5-45y^3+\dots\end{align}$$ hence $$\begin{align}-45&=(2d-2d)(d-d)+(2d)(-2d)+d(-d)=-5d^2. \end{align}$$

Answer 4

This is related to the third method that Anne Bauval suggested:

Having determined that one zero of $ \ x^5 - 5x^4 - 35x^3 + ax^2 + bx + c \ $ must be $ \ 1 \ $ if the five real zeroes are to be in arithmetic progression, we can perform synthetic or polynomial division to find the factorization $ \ (x - 1)·p(x) \ = \ (x - 1)·( \ x^4 - 4x^3 - 39x^2 + [a - 39]·x + [a + b - 39] \ ) \ \ $ (as you wrote in one of your comments).

If we "shift to the left" by $ \ 1 \ $ this reduced polynomial $ \ p(x) \ \ , \ $ its zeroes will be $ \ -2d \ , \ -d \ , \ +d \ , \ + 2d \ \ , \ $ so it is now an even function. This new polynomial is $$ q(x) \ \ = \ \ p(x + 1) \ \ = \ \ x^4 \ - \ 45x^2 \ + \ (a - 125)x \ + \ (2a + b - 120) \ \ . $$
This tells us immediately that $ \ a \ = \ 125 \ \ $ and that this even-symmetry polynomial is $ \ q(x) $ $ = \ x^4 - 45x^2 + (130 + b ) \ \ . \ $ We require that $ \ \pm d \ \ , \ \ \pm 2d \ \ $ be zeroes of $ \ q(x) \ \ . \ $ We could solve for the prospective zeroes of this biquadratic polynomial and find the value of $ \ b \ $ that produces the proper spacing between those zeroes. But it is rather easier to observe that $$ d^4 \ - \ 45d^2 \ + \ (130 + b) \ \ = \ \ (2d)^4 \ - \ 45·(2d)^2 \ + \ (130 + b) $$ $$ \Rightarrow \ \ 15d^4 \ - \ 135d^2 \ \ = \ \ 15d^2·(d^2 - 9) \ \ = \ \ 0 \ \ . $$ As $ \ d \ = \ 0 \ $ is "spurious" since all of the zeroes would be identically zero, the "useful" solution is $ \ d \ = \ \pm 3 \ \ . \ $ The zeroes of $ \ q(x) \ $ are then $ \ -6 \ \ , \ \ -3 \ \ , \ \ +3 \ \ , \ \ +6 \ \ , \ $ giving us the zeroes of the original quintic polynomial as $ \ \boxed{-5 \ \ , \ \ -2 \ \ , \ \ 1 \ \ , \ \ 4 \ \ , \ \ 7} \ \ . \ $

Note that we have never needed to know the unknown coefficients of the quintic other than $ \ a \ \ , \ \ $ although we can now readily determine that polynomial to be $ \ x^5 - 5x^4 - 35x^3 + 125x^2 + 194x - 280 \ \ . $