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

##### 1 Answer

#### 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# ,#+-2#

None of these work, so

**Tschirnhaus transformation**

Let

#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)^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)#