Question

How do you find all the zeros of `f(x)=x^4-x^3-4x^2-3x-2`?

Answer 1

Here's a sketch of how you can solve this algebraically...

Explanation:

`f(x) = x^4-x^3-4x^2-3x-2`

By the rational roots theorem, any rational zeros of `f(x)` are expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `-2` and `q` a divisor of the coefficient `1` of the leading term.

So the only possible rational zeros are:

`+-1`, `+-2`

None of these work, so `f(x)` has no rational zeros.

Tschirnhaus transformation

Let `t=4x-1`

`256f(x) = 256x^4-256x^3-1024x^2-768x-512`

`color(white)(256f(x)) = (4x-1)^4-70(4x-1)^2-328(4x-1)-771`

`color(white)(256f(x)) = t^4-70t^2-328t-771`

`color(white)(256f(x)) = (t^2-at+b)(t^2+at+c)`

`color(white)(256f(x)) = t^4+(b+c-a^2)t^2+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` and rearranged a little:

`(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 `a`, then:

`{ (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)`