What are all the zeroes of f(x) = 2x^3 + 3x^2 + 4x - 10?
2 Answers
Explanation:
Rewriting the function:
The easy way (trying as root
Then we can go the hard way as described in
https://en.wikipedia.org/wiki/Cubic_function#General_formula_for_roots
So for the general cubic equation
where
In case:
For
We could continue with this formula, but dividing
x_1 = 1/6(-3 + root(3)(621+6sqrt(10806)) + root(3)(621-6sqrt(10806)))
x_2=1/6(-3 + omega root(3)(621+6sqrt(10806)) + omega^2 root(3)(621-6sqrt(10806)))
x_3=1/6(-3 + omega^2 root(3)(621+6sqrt(10806)) + omega root(3)(621-6sqrt(10806)))
Explanation:
When a cubic has one Real and
Given
Let
Let
Then:
4f(x) = 8x^3+12x^2+16x-40
=(2x)^3+3(2x)^2+8(2x)-40
=t_1^3+3t_1^2+8t_1-40
=(t_1^3+3t_1^2+3t_1+1) + (5t_1+5)-46
=(t_1+1)^3+5(t_1+1)-46
=t_2^3+5t_2-46
Next let
0 = (u+v)^3+5(u+v)-46
= u^3+v^3+(3uv+5)(u+v)-46
To eliminate the term in
= u^3 - (5/(3u))^3-46
= u^3 - 125/(27u^3)-46
Multiply through by
27(u^3)^2-1242(u^3)-125 = 0
Use the quadratic formula to find:
u^3 = (1242+-sqrt(1242^2+(4*27*125)))/54
=(1242+-sqrt(1556064))/54
=(1242+-12sqrt(10806))/54
=(621+-6sqrt(10806))/27
So the Real value of
u = root(3)((621+-6sqrt(10806))/27) =1/3 root(3)(621+-6sqrt(10806))
Then since the derivation was symmetric in
t_2 = u+v
= 1/3 root(3)(621+6sqrt(10806)) + 1/3 root(3)(621-6sqrt(10806))
Hence the Real root of the original cubic is:
x_1 = (t_2-1)/2
=1/6(-3 + root(3)(621+6sqrt(10806)) + root(3)(621-6sqrt(10806)))
The Complex roots come from replacing the Real value of
x_2=1/6(-3 + omega root(3)(621+6sqrt(10806)) + omega^2 root(3)(621-6sqrt(10806)))
x_3=1/6(-3 + omega^2 root(3)(621+6sqrt(10806)) + omega root(3)(621-6sqrt(10806)))