# Thread: Relations between Coefficents of Polynomials

1. ## Relations between Coefficents of Polynomials

Given ax^3 + bx^2 + cx + d = 0

and roots are e, 2e, 3e

Find relation between coefficients.

I approached this firstly with addition of roots

6e = -b/a
e = -b/6a

Then subbed into equation to attain 5b^3 - 36acb + 216a^2d = 0

However, the solution proposes 11ad = bc, by solving via sum of roots, sum of roots 2 at a time and product of roots then combining together.

Which solution is correct or are they the same?

2. Since the roots are $\displaystyle e, 2e, 3e$, if you were to substitute them into the polynomial, you would get $\displaystyle 0$.

So substituting each of them in gives a set of equations...

$\displaystyle ae^3 + be^2 + ce + d = 0$
$\displaystyle a(2e)^3 + b(2e)^2 + c(2e) + d = 0$
$\displaystyle a(3e)^3 + b(3e)^2 + c(3e) + d = 0$

$\displaystyle ae^3 + be^2 + ce = -d$
$\displaystyle 8ae^3 + 4be^2 + 2ce = - d$
$\displaystyle 27ae^3 + 9be^2 + 3ce = -d$.

Apply $\displaystyle R_2 - 8R_1 \to R_2$ and $\displaystyle R_3 - 27R_1 \to R_3$ to get

$\displaystyle ae^3 + be^2 + ce = -d$
$\displaystyle -4be^2 - 6ce = 7d$
$\displaystyle -18be^2 - 24ce = 26d$

Apply $\displaystyle R_3 - \frac{9}{2}R_2 \to R_3$ to give

$\displaystyle ae^3 + be^2 + ce = -d$
$\displaystyle -4be^2 - 6ce = 7d$
$\displaystyle 3ce = -\frac{11}{2}d$.

Solving the third equation for $\displaystyle c$:

$\displaystyle c = -\frac{11d}{6e}$.

Substituting into the second equation and solving for $\displaystyle b$:

$\displaystyle -4be^2 + 11d = 7d$

$\displaystyle -4be^2 = -4d$

$\displaystyle b = \frac{d}{e^2}$.

Substituting into the first equation and solving for $\displaystyle a$:

$\displaystyle ae^3 + d - \frac{11d}{6} = -d$

$\displaystyle ae^3 - \frac{5d}{6} = -d$

$\displaystyle ae^3 = -\frac{d}{6}$

$\displaystyle a = -\frac{d}{6e^3}$.

Therefore the solution is:

$\displaystyle a = -\frac{d}{6e^3}, b = \frac{d}{e^2}, c = -\frac{11d}{6e}$.

3. Sorry it was probably my question being too obscure, but it is the relation between coefficents of polynomials, that is the eradication of e, and linking a,b,c,d. Now ive gotten 2 answers as i stated,
5b^3 - 36acb + 216a^2d = 0 and 11ad = bc, but I am not sure if the 1st identity is valid as they are not the same.

4. The relationship is through $\displaystyle d$ and $\displaystyle e$.

5. Mabey you attacking the question in far greater depth than I am. I merely want the relationship between the coefficents, i.e. a,b,c,d. I think what your syaing is that this relationship between d and e will satisfy the two solutions is stated? Is that correct?

6. Originally Posted by Lukybear
Sorry it was probably my question being too obscure, but it is the relation between coefficents of polynomials, that is the eradication of e, and linking a,b,c,d. Now ive gotten 2 answers as i stated,
5b^3 - 36acb + 216a^2d = 0 and 11ad = bc, but I am not sure if the 1st identity is valid as they are not the same.
Do you know Vieta's formulas? We know that If $\displaystyle x_{1}, x_{2}$, and $\displaystyle x_{3}$ are the roots of the equation $\displaystyle a_{3}x^3+a_{2}x^2+a_{1}x+a_{0}$, then $\displaystyle -\frac{a_{2}}{a_{3}} = x_{1}+x_{2}+x_{3}$, $\displaystyle \frac{a_{1}}{a_{3}} = x_{1}x_{2}+x_{1}x_{3}+x_{3}x_{2}$ , and $\displaystyle -\frac{a_{0}}{a_{3}} = x_{1}x_{2}x_{3}$. So if $\displaystyle e, 2e, 3e$ are the roots of $\displaystyle f(x) = ax^2+bx^2+cx+d$, then the Vieta relations are: $\displaystyle e+2e+3e = 6e = -\frac{b}{a}[/Math], call it (1), [Math] (e)(2e)+(e)(3e)+(2e)(3e) = 11e^2 = \frac{c}{a}$, call it (2), and $\displaystyle (e)(2e)(3e) = 6e^3 = -\frac{d}{a}$, call it (3). Put (1) into (2), and we have $\displaystyle \frac{36a^2c}{11b^2} = a$. Now put (1) into (3) and we get $\displaystyle \frac{36a^3d}{b^3} = a$. But $\displaystyle a = \frac{36a^2c}{11b^2}$, so we have $\displaystyle \frac{36a^2c}{11b^2} = \frac{36a^3d}{b^3}$. That is the relation between the coefficients.

7. Yes, which is 11ad = bc simplified, and if e + 2e + 3e = 6e.

But since 6e = -d/c
e=-d/6c

f(-d/6c) would also render a relationship between coefficents will it not? since e satisfies f(e) = 0, by definition. This renders 5b^3 - 36acb + 216a^2d = 0 .

Just wondering whether the two relations are equivalent.

8. Hello, Lukybear!

Given: .$\displaystyle ax^3 + bx^2 + cx + d \:= \:0$ .with roots: $\displaystyle e, 2e, 3e$

Find relation between coefficients.

Do you know the rest of that theorem?

The equation is: .$\displaystyle x^3 + \dfrac{b}{a}x^2 + \dfrac{c}{a}x + \dfrac{d}{a} \:=\:0$ .

If the roots are $\displaystyle \,p,q,r$, then :

. . $\displaystyle \begin{array}{ccccc}p + q + r &=&\text{-}\dfrac{b}{a} & \text{(Roots "taken one at a time")}\\ \\[-3mm] pq + qr + pr &=& \dfrac{c}{a} & \text{(Roots "taken two at a time")}\\ \\[-3mm] pqr &=& \text{-}\dfrac{d}{a} & \text{(Roots "taken three at a time")} \end{array}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

$\displaystyle \begin{array}{cccccccc}\text{1-at-a-time:} & e + 2e + 3e &=& 6e &=& \text{-}\dfrac{b}{a} & [1] \\ \\[-3mm] \text{2-at-a-time:} & (e)(2e) + (2e)(3e) + (e)(3e) &=& 11e^2 &=& \dfrac{c}{a} &[2] \\ \\[-3mm] \text{3-at-a-time:} & (e)(2e)(3e) &=& 6e^3 &=& \text{-}\dfrac{d}{a} &[3] \end{array}$

. . $\displaystyle \begin{array}{ccccccccc}\text{From [1], we have:} & 6e &=& \text{-}\dfrac{b}{a} & \Rightarrow & a &=& \text{-}\dfrac{b}{6e} & [4] \\ \\[-3mm] \text{From [2], we have:} & 11e^2 &=& \dfrac{c}{a} & \Rightarrow& a & =& \dfrac{c}{11e^2} & [5] \\ \\[-3mm] \text{From [3], we have:} & 6e^3 &=& \text{-}\dfrac{d}{a} & \Rightarrow & a &=& \text{-}\dfrac{d}{6e^3}& [6] \end{array}$

Equate [4], [5], [6]: .$\displaystyle a \:=\:\text{-}\dfrac{b}{6e} \:=\:\dfrac{c}{11e^2} \:=\:\text{-}\dfrac{d}{6e^3}$

Multiply by $\displaystyle 66e^3\!:\;\;66ae^3 \:=\:\text{-}11be^2 \:=\:6ce \:=\:\text{-}11d$

We have: .$\displaystyle \begin{Bmatrix}66ae^3 &=& \text{-}11be^2 & \Rightarrow & e &=& \text{-}\dfrac{b}{6a} & [7] \\ \\[-3mm] 6ce &=& \text{-}11d & \Rightarrow & e &=& \text{-}\dfrac{11d}{6c} & [8] \end{Bmatrix}$

Equate [7] and [8]: .$\displaystyle \text{-}\dfrac{b}{6a} \:=\:\text{-}\dfrac{11d}{6c} \quad\Rightarrow\quad \boxed{bc \:=\:11ad}$

Edit: TheCoffeeMachine beat me to it . . . *sigh*

9. $\displaystyle a(x-e)(x-2e)(x-3e)$
$\displaystyle =ax^3-6eax^2+11ae^2x-6ae^3$

$\displaystyle a=a, b=-6ea, c=11ae^2, d=-6ae^2$
$\displaystyle be^2-d=0 \Rightarrow e^2=\frac{d}{b}$
$\displaystyle c=11a(\frac{d}{b})$

10. Originally Posted by Lukybear
Just wondering whether the two relations are equivalent.
I don't think they are equivalent. But that's odd -- they have to be. Let's ask Soroban.
Originally Posted by Soroban
TheCoffeeMachine beat me to it . . . *sigh*
Hey, Soroban, don't sigh. I might have been quicker but I can never match your painstaking clarity.

What do you think: is the relationship given by $\displaystyle f(-\frac{b}{6a})$ equavalent to the one that you have found?

11. Well obviously not. But it is still valid, as it is a relationship between coefficents.