Solve the equation $x^4-2x^3-21x^2+22x+40=0$ whose roots are in A.P.

Question

Solve the equation $x^4-2x^3-21x^2+22x+40=0$ whose roots are in A.P. (arithmetic progression).

I don't understand this solution.

Why are the terms of AP considered as mentioned in the question and not in the form of a, a+d, a+2d... and so on?

I am not able to understand why the terms of the AP are of that specific form and not in the general form of a, a+d, a+2d and a+3d. The aforementioned sequence is also in the form of AP and should also provide the answer but I am not able to get to the answer by the approach I am trying.

Also explain the approach as well.

Also the reasoning and thinking behind the approach used to find the roots if explained will also be helpful or any sources for the same.

Approach

Consider the 4 roots to be {a, a+d, a+2d and a+3d}. By using Vieta's formula:\

$a+(a+d)+(a+2d)+(a+3d)= \frac{-b}{a}$
$4a+6d=2$
$2a+3d=1$
$a= \frac{(1-3d)}{2}$

Similarly, $a*(a+d)*(a+2d)*(a+3d)= \frac{c}{a}$
$(a^2+ad)*(a^2+3ad+2ad+6d^2)=40$

Then substituting the value of $a$ in the above equation:
$221d^4-90d^3-66d^2+6d-639=0$

There may be some errors but it will result in a quartic equation nonetheless and it goes back again being similar to the original equation. Thus my question is then is there no other way of solving this equation other than considering the terms of the Arithmetic Progression to be of a particular sequence.

I am new to math.stack exchange so if there are any error in the phrasing of the question, I am sorry. Also any help is welcomed kindly. Thanks and regards. Special thanks to everyone who pointed out the mistakes in paraphrasing of my initial question and guiding me.

Answer 1

Generally when considering AP's with 3,4,5 terms respectively, it makes it easier if we write the terms as: (which makes calculating terms of the AP much faster(especially sum and product) )

• $3: a-r,a,a+r$
• $4: a-3r,a-r,a+r,a+3r$
• $5: a-2r,a-r,a,a+r,a+2r$

You can then exploit using Vietta's Relations of Sum and Product of roots to get values of a and r respectively, as outline in the solution provided by them.

Using $a,a+r,a+2r,a+3r$ will make the calculations much more tedious as you will get a linear relation between a and r and a 4th power one which will not easily cancel.

Answer 2

Let the equation have a rational root $x=\frac{m}{n}$, then $$\frac{m^4}{n^4}-\frac{2m^3}{n^3}-\frac{21m^2}{n^2}+\frac{22m}{n}+40=0, \; m\in\mathbb{Z}, \; n\in \mathbb{N}$$ $$40n^4+22mn^3-21m^2n^2-2m^3n+m^4=0$$ $$m^4=n\left ( 40n^3+22mn^2-21m^2n-2m^3 \right )$$

A natural number has no common prime divisors, then $n$ is a divisor of $1$, so $n$ equals one $$40n^4=22mn^3-21m^2n^2-2m^3n+m^4$$ $$40n^4=m\left ( 22n^3-21mn^3-2m^2n+m^3 \right )$$

A natural number has no prime divisors, then $m$ is a divisor of $40$, hence, $m$ is equal to one of the numbers $\pm 5$, $\pm 2$, $\pm 1$. Note that $1$ cannot be a root of the equation since the sum of the coefficients of our equation is not zero, but $-1$ is a root of the equation since $1+2-21-22+40=0\Rightarrow x_1=-1$

Our equation then takes the form $$(x+1)\left ( x^3 - 3 x^2 - 18 x + 40 \right )=0$$

Do the same with the cubic equation and find that $x_2=2$, hence $$(x+1)(x-2)\left ( x^2-x-20 \right )=0$$

This is already a simple quadratic equation, hence $x_{3,4}=\left \{ -4.5 \right \}$

Answer 3

The declaration that the zeroes of a polynomial are in arithmetic progression automatically implies that the zeroes (which have multiplicity 1) are arranged symmetrically about their average. As you found, their sum is $ \ 4a + 6d \ = \ 2 \ \ , \ $ making their average $ \ a + \frac32·d \ = \ \frac12 \ \ . \ $ So if we consider these numbers relative to the average, we have $ \ -\frac32·d \ \ , \ \ -\frac12·d \ \ , \ \ +\frac12·d \ \ , \ \ +\frac32·d \ \ $ "centered" on $ \ \frac12 \ \ . \ $ (The provided solution could well have explained this a bit more thoroughly.)

The product of the zeroes is then $$ \ \left(\frac12 - \frac32·d \right)·\left(\frac12 - \frac12·d \right)·\left(\frac12 + \frac12·d \right)·\left(\frac12 + \frac32·d \right) $$ $$ = \ \ \left(\frac14 - \frac14·d^2 \right)·\left(\frac14 - \frac94·d^2 \right) \ \ = \ \ 40 \ \ , $$ leading to the biquadratic equation shown in the solution. (The version shown here has real solutions $ \ d \ = \ \pm 3 \ \ , \ $ which will lead to the same result. The remark in the solution that $ \ d \ = \ -3 \ $ is "inadmissible" is not really correct, since a negative difference simply means we have a "descending" arithmetic progression, with initial term $ \ a' \ = \ \frac{2 \ - \ 6·[-3]}{4} \ \ = \ 5 \ \ $ and the same set of zeroes for our polynomial.) So the use of $ \ a - 3d \ , \ a - d \ , \ a + d \ , \ a + 3d \ \ $ is not simply a "choice of convenience", but arises directly from the inherent symmetry of the arithmetic progression.

We can alternatively apply the "shift" to the quartic polynomial $ \ f(x) \ = \ x^4 - 2x^3 - 21x^2 + 22x + 40 \ $ by taking $ \ x \ = \ u + \frac12 \ \ $ thus $$ f \left(u + \frac12 \right) \ \ = \ \ \left(u + \frac12 \right)^4 \ - \ 2\left(u + \frac12 \right)^3 \ - \ 21\left(u + \frac12 \right)^2 \ + \ 22\left(u + \frac12 \right) \ + \ 40 \ $$ to produce the "depressed" polynomial $$ u^4 \ - \ \frac{45}{2}·u^2 \ + \ \frac{729}{16} \ \ = \ \ \left(u^2 - \frac{45}{4} - 9 \right)·\left(u^2 - \frac{45}{4} + 9 \right) \ \ = \ \ \left(u^2 - \frac{81}{4} \right)·\left(u^2 - \frac{9}{4} \right) \ \ . $$ The zeroes are then $$ u \ \ = \ \ x \ - \ \frac12 \ \ = \ \ -\frac92 \ \ , \ \ -\frac32 \ \ , \ \ +\frac32 \ \ , \ \ +\frac92 \ \ \Rightarrow \ \ x \ \ = \ \ -4 \ \ , \ \ -1 \ \ , \ \ +2 \ \ , \ \ +5 \ \ . $$

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ADDENDUM (5/30) -- We can generalize this discussion to say that if the zeroes of a polynomial are in arithmetic progression and

• the polynomial is of even degree $ \ 2n \ \ ( \ c_{2n}x^{2n} + c_{2n \ - \ 1}x^{2n \ - \ 1} + \ \ldots \ + c_1x + c_0 \ ) \ \ , \ $ then the average of its zeroes is $ \ a \ = \ \large{-\frac{c_{2n \ - \ 1}}{2n \ · \ c_{2n} } } \ \ $ and the difference $ \ d \ $ between zeroes in the sequence is found from $$ \left(a^2 - \frac14·d^2 \right)·\left(a^2 - \frac94·d^2 \right)· \ \ldots \ · \left(a^2 - \frac{(2n \ - \ 1)^2}{4}·d^2 \right) \ \ = \ \ \frac{c_0}{c_{2n}} \ \ . \ $$ Alternatively, the polynomial may be transformed as $ \ f(x) \ \rightarrow \ f(u + a) \ \ , \ $ which then has even symmetry with zeroes being $ \ u \ = \ \pm \frac12·d \ \ , \ \ \pm \frac32·d \ \ , \ \ldots \ , \ \ \pm \frac{(2n \ - \ 1) }{2}·d \ \ ; \ \ $ or

• the polynomial is of odd degree $ \ 2n + 1 \ \ ( \ c_{2n \ + \ 1}x^{2n \ + \ 1} + c_{2n }x^{2n } + \ \ldots \ + c_1x + c_0 \ ) \ \ , \ $ then the average of its zeroes is $ \ a \ = \ \large{-\frac{c_{2n }}{(2n \ + \ 1) \ · \ c_{2n \ + \ 1} } } \ \ $ is one of the zeroes, and the difference $ \ d \ $ between zeroes in the sequence is found from $$ \left(a^2 - d^2 \right)·\left(a^2 - 4d^2 \right)· \ \ldots \ · \left(a^2 - n^2d^2 \right) \ \ = \ \ \frac{-c_0}{a \ · \ c_{2n \ + \ 1}} \ \ . \ $$ We may also transform the polynomial as $ \ f(x) \ \rightarrow \ f(u + a) \ \ , \ $ which has odd symmetry with the zeroes being $ \ u \ = \ 0 \ \ , \ \ \pm d \ \ , \ \ \pm 2·d \ \ , \ \ldots \ , \ \ \pm n·d \ \ . $

As an illustration of the odd-degree case, we are given that the zeroes for $ \ 27x^5 + 45x^4 - 210x^3 - 230x^2 + 263x + 105 \ $ are in arithmetic progression. One of the zeroes is the average of the zeroes, $ \ a \ = \ -\frac{45}{5 \ · \ 27} \ = \ -\frac13 \ \ . \ $ We may find the difference between the zeroes in the progression from $$ \left( \ \left[-\frac13 \right]^2 - d^2 \ \right)·\left( \ \left[-\frac13 \right]^2 - 4d^2 \ \right) \ \ = \ \ \frac{-105}{(-1/3)·27} $$ $$ \rightarrow \ \ 4·d^4 \ - \ \frac59·d^2 \ - \ \frac{944}{81} \ \ = \ \ 4·\left( \ d^2 \ - \ \frac{5 \ + \ 123}{72} \ \right)·\left( \ d^2 \ - \ \frac{5 \ - \ 123}{72} \ \right) $$ $$ = \ \ 4·\left( \ d^2 \ - \ \frac{16}{9} \ \right)·\left( \ d^2 \ + \ \frac{59}{36} \ \right) \ \ = \ \ 0 \ \ , $$ for which the real roots are $ \ d \ = \ \pm \frac43 \ \ . \ $ If we instead work with the transformed polynomial, we obtain $$ f \left(u + \frac13 \right) \ \ = \ \ 27u^5 \ - \ 240u^3 \ + \ \frac{1024}{3}·u \ \ = \ \ \frac{u}{3}·(9u^2 \ - \ 64)·(9u^2 \ - \ 16) \ \ . $$ Either method yields the arithmetic sequence of zeroes $$ -\frac13 - 2·\frac43 \ = \ -3 \ \ , \ \ -\frac13 - \frac43 \ = \ -\frac53 \ \ , \ \ -\frac13 \ \ , $$ $$ -\frac13 + \frac43 \ = \ +1 \ \ , \ \ -\frac13 + 2·\frac43 \ = \ +\frac73 \ \ . $$