6x^4 + x^3 + 10x^2 + x + 6 = 0
(solve for x)
2006-11-02 15:43:09 · 5 answers · asked by blue skies 2 in Science & Mathematics ➔ Mathematics
If you could give the process it would be greatly appreciated. Thank you.
2006-11-02 15:44:47 · update #1
sorry I don't think it factors out into that. Because when I expanded it the result was 3x^4 + 2 x^3 + 6x^2 - 3x + 6.
2006-11-02 16:21:12 · update #2
my mistake: it expands into 3x^4 - x^3 + 7x^2 + x + 6 when I multiplied the factor together.
2006-11-02 16:26:32 · update #3
I have factored it please refer to.
From this solution can be found
2006-11-03 01:48:47 · answer #1 · answered by Mein Hoon Na 7 · 0⤊ 0⤋
6x^4 + x^3 + 10x^2 + x + 6 = (x^2 - x + 2) (3x^2 +2x + 3) = 0
and the two quadratic formulas just have complex solutions,
so if you are looking for real roots, the answer is, this polynomial does not have real roots
s
2006-11-02 16:15:31 · answer #2 · answered by Anonymous · 0⤊ 1⤋
6 x^4 + x^3 + 10x^2 + x + 6 factorizes into (x^2 - x + 2)(3x^2 +2x + 3), of which both quadratics have only complex roots, being: 1/4(1- i sqrt(15)), 1/4(1 + i sqrt(15)), 1/3(-1 - 2i sqrt(2)), and 1/3(-1 + 2i sqrt(2)).
2006-11-02 16:10:45 · answer #3 · answered by Scythian1950 7 · 0⤊ 0⤋
if it somewhat is your calculus homestead paintings????? in case you're in calculus and don't comprehend the thank you to do this then i propose you to drop calculus precise now. a million. remedy for x. 2. remedy for x. 3. subtract 3 from the two sides. use factoring and remedy for x.
2016-12-28 11:34:12 · answer #4 · answered by dustman 3 · 0⤊ 0⤋
A general quartic eqn, which this appears to be, is more than you'll get for free from me; try this:
http://en.wikipedia.org/wiki/Quartic_equation
2006-11-02 15:51:13 · answer #5 · answered by Steve 7 · 0⤊ 1⤋