# quartic polynomial roots

• May 20th 2008, 06:51 AM
skystar
quartic polynomial roots
let $\displaystyle \alpha , \beta, \gamma, \delta$ be the roots of the polynomial equation.

$\displaystyle x^{4}+5x^{3}+3x^{2}-x+2=0$

calculate $\displaystyle \alpha^{2} +\beta^{2} +\gamma^{2} + \delta^{2}$

..my thoughts... polynomial can be written as $\displaystyle (x-\alpha)(x-\beta)(x-\gamma)(x-\delta)$ expand this and equate? but will this be general for every polynomial irrespective of the signs of the coefficients? thanks appreciate its alot of latex!
• May 20th 2008, 07:00 AM
ThePerfectHacker
Quote:

Originally Posted by skystar
let $\displaystyle \alpha , \beta, \gamma, \delta$ be the roots of the polynomial equation.

$\displaystyle x^{4}+5x^{3}+3x^{2}-x+2=0$

calculate $\displaystyle \alpha^{2} +\beta^{2} +\gamma^{2} + \delta^{2}$

Note,
$\displaystyle (a+b+c+d)^2 = (a^2+b^2+c^2+d^2)+2(ab+ac+ad+bc+bd+cd)$.

Now by Viete's theorem, $\displaystyle a+b+c+d = - 5$ and $\displaystyle ab+ac+ad+bc+bd+cd = 3$.

Now solve for $\displaystyle a^2+b^2+c^2+d^2$.
• May 20th 2008, 07:08 AM
skystar
still confused (Doh) i cant seem to put it in context with my example. thanks
• May 20th 2008, 07:13 AM
ThePerfectHacker
Quote:

Originally Posted by skystar
still confused (Doh) i cant seem to put it in context with my example. thanks

Let us see. Say the polynomial factors as $\displaystyle (x-a)(x-b)(x-c)(x-d)$.
Now expand the polynomial,
$\displaystyle x^4-(a+b+c+d)x^3$
$\displaystyle +(ab+ac+ad+bc+bd+cd)x^2$
$\displaystyle -(abc+abd+acd+bcd)x +abcd$

Now line up the coeffcients,
$\displaystyle a+b+c+d = -5$
$\displaystyle ab+ac+ad+bc+bd+cd = 3$
$\displaystyle abc+abd+acd+bcd = 1$
$\displaystyle abcd=2$

EDIT: Mistake fixed.
• May 20th 2008, 07:18 AM
skystar
small typo (Wink)

$\displaystyle$
$\displaystyle -(abc+abd+acd+bcd)x^3 +abcd$

$\displaystyle -(abc+abd+acd+bcd)x +abcd$