1. ## 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!

2. 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 $a^2+b^2+c^2+d^2$.

3. still confused i cant seem to put it in context with my example. thanks

4. Originally Posted by skystar
still confused 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.

5. small typo

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