How do you find all the zeros of f(x)=x^4-x^3-4x^2-3x-2f(x)=x4−x3−4x2−3x−2?
1 Answer
Here's a sketch of how you can solve this algebraically...
Explanation:
By the rational roots theorem, any rational zeros of
So the only possible rational zeros are:
+-1±1 ,+-2±2
None of these work, so
Tschirnhaus transformation
Let
256f(x) = 256x^4-256x^3-1024x^2-768x-512256f(x)=256x4−256x3−1024x2−768x−512
color(white)(256f(x)) = (4x-1)^4-70(4x-1)^2-328(4x-1)-771256f(x)=(4x−1)4−70(4x−1)2−328(4x−1)−771
color(white)(256f(x)) = t^4-70t^2-328t-771256f(x)=t4−70t2−328t−771
color(white)(256f(x)) = (t^2-at+b)(t^2+at+c)256f(x)=(t2−at+b)(t2+at+c)
color(white)(256f(x)) = t^4+(b+c-a^2)t^2+a(b-c)t+bc256f(x)=t4+(b+c−a2)t2+a(b−c)t+bc
Hence:
{ (b+c = a^2-70), (b-c = -328/a), (bc=-771) :}
So:
(a^2-70)^2 = (b+c)^2 = (b-c)^2+4bc = (-328/a)^2-4(771)
Multiplied out:
a^4-140a^2+4900 = 107584/a^2-3084
Multiplied through by
(a^2)^3-140(a^2)^2+7984(a^2)-107584 = 0
Solve this cubic using Cardano's method (see https://socratic.org/s/azh362cn) to find:
a^2 = 1/3(140+8 root(3)(1628+12sqrt(20589))+8 root(3)(1628-12sqrt(20589))) ~~ 18.885
Without loss of generality we can choose the positive square root as our value of
{ (b = 1/2(a^2-70-328/a)), (c = 1/2(a^2-70+328/a)) :}
leaving two quadratics to solve:
t^2-at+b = 0
t^2+at+c = 0
Finally, we can recover the zeros of our original quartic using:
x = 1/4(t+1)
