How do you list all possible roots and find all factors of #x^4+2x^3-8x^2+16x-23#?
1 Answer
Here is a solution for
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Explanation:
Assuming typographical errors in the problem, let us solve:
#x^4-2x^3+8x^2+16x-23=0#
By the rational roots theorem, any rational zeros of this quartic are expressible in the form
That means that the only possible rational zeros are:
#+-1, +-23#
Note that
Then:
#x^4-2x^3+8x^2+16x-23= (x-1)(x^3-x^2+7x+23)#
To find the zeros for the remaining cubic, proceed as follows:
#f(x) = x^3-x^2+7x+23#
Discriminant
The discriminant
#Delta = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd#
In our example,
#Delta = 49-1372+92-14283-2898 = -18412#
Since
Tschirnhaus transformation
To make the task of solving the cubic simpler, we make the cubic simpler using a linear substitution known as a Tschirnhaus transformation.
#0=27f(x)=27x^3-27x^2+189x+621#
#=(3x-1)^3+60(3x-1)+682#
#=t^3+60t+682#
where
Cardano's method
We want to solve:
#t^3+60t+682=0#
Let
Then:
#u^3+v^3+3(uv+20)(u+v)+682=0#
Add the constraint
#u^3-8000/u^3+682=0#
Multiply through by
#(u^3)^2+682(u^3)-8000=0#
Use the quadratic formula to find:
#u^3=(-682+-sqrt((682)^2-4(1)(-8000)))/(2*1)#
#=(682+-sqrt(465124+32000))/2#
#=(682+-sqrt(497124))/2#
#=341+-3sqrt(13809)#
Since this is Real and the derivation is symmetric in
#t_1=root(3)(341+3sqrt(13809))+root(3)(341-3sqrt(13809))#
and related Complex roots:
#t_2=omega root(3)(341+3sqrt(13809))+omega^2 root(3)(341-3sqrt(13809))#
#t_3=omega^2 root(3)(341+3sqrt(13809))+omega root(3)(341-3sqrt(13809))#
where
Now
#x_1 = 1/3(1+root(3)(341+3sqrt(13809))+root(3)(341-3sqrt(13809)))#
#x_2 = 1/3(1+omega root(3)(341+3sqrt(13809))+omega^2 root(3)(341-3sqrt(13809)))#
#x_3 = 1/3(1+omega^2 root(3)(341+3sqrt(13809))+omega root(3)(341-3sqrt(13809)))#
Then the complete factorisation of our original quartic takes the form:
#x^4-2x^3+8x^2+16x-23 = (x-1)(x-x_1)(x-x_2)(x-x_3)#