# Thread: Forming a Cubic Equation from Roots

1. ## Forming a Cubic Equation from Roots

Stuck on a further pure question, asked one of my maths teachers today but she hadn't done most of the paper in 20 years.
Here's the question:
Given that the roots of the cubic equation
$x^3+4x^2+3x+2=0$
are $\alpha, \beta, \gamma$ determine the cubic equation with roots $\beta \gamma , \gamma \alpha , \alpha \beta$

All I have so far is
$\alpha + \beta + \gamma = -4 \: ; \: \beta \gamma + \gamma \alpha + \alpha \beta = 3 \: ; \: \alpha \beta \gamma = -2$

I assume the new equations will looks like:
$x^3-(\beta \gamma , \gamma \alpha , \alpha \beta)x^2+(\alpha^2 \beta \gamma + \alpha \beta^2 \gamma + \alpha \beta \gamma ^2)x - \alpha^2 \beta^2 \gamma^2$

I'm pretty sure the coefficient of the $x^2$ term will be $\beta \gamma + \gamma \alpha + \alpha \beta$ which equals 3.

And for the constants:
$\alpha \beta \gamma = -2$
$(\alpha \beta \gamma)^2 = 4$
$\alpha^2 \beta^2 \gamma^2 - 2(\beta \gamma + \gamma \alpha + \alpha \beta)=4$
$\alpha^2 \beta^2 \gamma^2 -2 \times 3 = 4$
$\alpha^2 \beta^2 \gamma^2 = 10$

I think I'm correct so far (may be some typos though), but I don't know how to get the x term.

Thanks.

edit

I got it, just factoring the coefficients of the x term:
$\alpha \beta \gamma (\alpha + \beta + \gamma)$
brain ache now.

2. I get

$x^3+4x^2+3x+2 = (x-\alpha)(x-\beta)(x-\gamma)$

$x^3+4x^2+3x+2 = x^3-(\beta+\alpha+\gamma)x^2+(\alpha\beta-\gamma\beta-\alpha\gamma)x-\gamma\alpha\beta$

so

$4 = -(\beta+\alpha+\gamma), 3=\alpha\beta-\gamma\beta-\alpha\gamma, 2 =-\gamma\alpha\beta$

3. Originally Posted by pickslides
I get

$x^3+4x^2+3x+2 = (x-\alpha)(x-\beta)(x-\gamma)$

$x^3+4x^2+3x+2 = x^3-(\beta+\alpha+\gamma)x^2+(\alpha\beta-\gamma\beta-\alpha\gamma)x-\gamma\alpha\beta$

so

$4 = -(\beta+\alpha+\gamma), 3=\alpha\beta-\gamma\beta-\alpha\gamma, 2 =-\gamma\alpha\beta$
Yeah that's what I got, the new equation I formed was:
$
x^3-3x^2 +8x + 1$

4. Hello, alexgeek!

You're wandering a bit, but I'd say you got it (sort of).

$\text{Given that the roots of the equation }x^3+4x^2+3x+2\:=\:0\,\text{ are }\,a, b, c$
$\text{determine the cubic equation with roots: }\:ab, bc, ac$

$\text{All I have so far is: }\;\begin{array}{ccc} a + b + c &=& \text{-}4 \\
ab + bc + ac &=& 3 \\
abc &=&\text{-}2 \end{array}$

$\text{I assume the new equation will looks like:}$
. . $x^3-(ab + bc + ac)x^2 + (a^2bc+ ab^2c + abc^2)x - a^2b^2c^2 \;=\;0$

$\text{The coefficient of the }x^2\text{ term will be }(ab + bc + ac) \:=\:3$

$\text{And the constant: }\:abc \:=\:-2 \quad\Rightarrow\quad a^2b^2c^2 \:=\:4$

The coefficient of the $\,x$-term is:

. . $a^2bc + ab^2c + abc^2 \:=\:abc(a+b+c) \;=\; (-2)(-4) \:=\:8$

Therefore, the cubic equation is: . $x^3 - 3x^2 + 8x - 4 \:=\:0$

5. What were your solutions for $\gamma ,\alpha ,\beta$ ?

6. The individual roots? The question doesn't ask for them, they're some decimal complex numbers and one decimal real solution, http://www.wolframalpha.com/input/?i=x^3%2B4x^2%2B3x%2B2%3D0

7. Actually I think the equation is $
x^{3}-3x^{2}+(-2)(-4)x-(2)^{2}\equiv x^{3}-3x^{2}+8x-4
$

Messed up the constants I think.

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