What are all the zeroes of f(x) = 2x^3 + 3x^2 + 4x - 10?

2 Answers
Jan 11, 2016

1.060419 and -1.280210+-i*1.753940

Explanation:

Rewriting the function:
f(x)=2*(x^3+1.5x^2+2x-5)

The easy way (trying as root +-1, +-5 and +-1/5) doesn't help us to find the zeros.

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

a x^3 + b x^2 + c x + d = 0

x_k = -(1/(3a)) (b+u_k C+Delta_0/(u_k*C)), k=1,2, 3

where
u_1 = 1, u_2 = (-1 + i*sqrt(3))/2, u_3 =(-1 - i*sqrt(3))/2

C = root(3)((Delta_1 + sqrt(Delta_1^2 - 4*Delta_0^3))/2) with
Delta_0 = b^2-3 a c
Delta_1= 2 b^3-9 a b c+27 a^2 d

In case:
Delta_0=1.5^2-3*1*2=2.25-6=-3.75
Delta_1=2*3.375-9*1*1.5*2+27*1*(-5)=6.75-27-135=-155.25

C=root(3) ((-155.25+sqrt((-155.25)^2-4*(-3.75)^3))/2)
C=root(3) ((-155.25+sqrt(24313.5))/2)=root(3)(.338934)=.697223

For k=1 and u_1=1
x_1=(-1/3)(1.5+.697223+(-3.75)/.697223)=(-1/3)(2.197223-5.378480) => x_1=1.060419

We could continue with this formula, but dividing (x^3+1.5x^2+2x-5) by (x-1.060419) we get (x^2+2.560419x+4.715117), whose roots we can obtain in this way:
(x^2+2.560419x+4.715117)=0
Delta=6.555745-18.860468=-12.304723
x=(-2.560.419+-i*3.507809)/2 => x_2=-1.280210+i*1.753904, x_3=-1.280210-i*1.753904

Jan 16, 2016

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 2 non-Real Complex zeros, I like to use Cardano's method...

Given f(x) = 2x^3+3x^2+4x-10

Let t_1 = 2x

Let t_2 = t_1+1

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 t_2 = u+v

0 = (u+v)^3+5(u+v)-46

= u^3+v^3+(3uv+5)(u+v)-46

To eliminate the term in (u+v) add the constraint v = -5/(3u)

= u^3 - (5/(3u))^3-46

= u^3 - 125/(27u^3)-46

Multiply through by 27u^3 and rearrange to obtain this quadratic in u^3:

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 is given by:

u = root(3)((621+-6sqrt(10806))/27) =1/3 root(3)(621+-6sqrt(10806))

Then since the derivation was symmetric in u and v we can deduce that one of these values is u and the other v.

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 u with omega u or omega^2 u, where omega = -1/2+sqrt(3)/2i is the primitive cube root of 1 ...

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)))