# quartic polynomial roots

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

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

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

..my thoughts... polynomial can be written as $(x-\alpha)(x-\beta)(x-\gamma)(x-\delta)$ 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 $\alpha , \beta, \gamma, \delta$ be the roots of the polynomial equation.

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

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

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

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

Now solve for $a^2+b^2+c^2+d^2$.
• 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 $(x-a)(x-b)(x-c)(x-d)$.
Now expand the polynomial,
$x^4-(a+b+c+d)x^3$
$+(ab+ac+ad+bc+bd+cd)x^2$
$-(abc+abd+acd+bcd)x +abcd$

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

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

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