Question

Is it possible to factor `y=x^3+2x^2-18x+4`? If so, what are the factors?

Answer 1

`y =(x-0.228698)(x+5.44241)(x-3.21371)`

Explanation:

According to Gauss, (Gauss, Carl Friedrich - 1799)
"...every non-zero, single-variable, degree `n` polynomial with complex coefficients has, counted with multiplicity, exactly `n` roots". (wikipedia).

So the polynomial

`p_3(x)=x^3 + 2 x^2 - 18 x + 4` has exactly `3` roots `x_1,x_2,x_3`

so the factors are

`p_3(x) = (x-x_1)(x-x_2)(x-x_3)`

Use Cardano's rule to find the roots giving

`{x_1 =0.228698, x_2= -5.44241, x_3= 3.21371}`

https://socratic.org/questions/how-do-you-find-all-the-real-and-complex-roots-of-x-3-x-2-x-2#270714

Answer 2

Use a trigonometric method to find:

`x^3+2x^2-18x+4 = (x-x_0)(x-x_1)(x-x_2)`

where

`x_0 ~~ 3.21370759428`

`x_1 ~~ -5.44240577684`

`x_2 ~~ 0.22869818256`

Explanation:

Given:

`f(x) = x^3+2x^2-18x+4`

The discriminant `Delta` of a cubic `ax^3+bx^2+cx+d` is given by the formula:

`Delta = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd`

In our example `a=1`, `b=2`, `c=-18`, `d=4` and we find:

`Delta = 1296+23328-128-432-2592=21472 > 0`

Since `Delta > 0` this cubic has `3` Real zeros.

In general Cardano's method will not be very helpful to find the zeros in such a case, since the resulting algebraic solution will include cube roots of Complex numbers.

Use a trigonometric method instead:

First simplify to remove the square term:

`27f(x) = 27x^3+54x^2-486x+108`

`=(3x+2)^3-174(3x+2)+448`

Let `t = 3x+2`

We want to solve:

`t^3-174t+448= 0`

Let `t = 2sqrt(58) cos theta`

The multiplier `2sqrt(58)` is chosen to get `4 cos^3 theta - 3 cos theta` below...

Then:

`0 = t^3-174t+448`

`=(2sqrt(58) cos theta)^3 - 174(2sqrt(58) cos theta) + 448`

`=464 sqrt(58) cos^3 theta - 348 sqrt(58) cos theta + 448`

`=116sqrt(58)(4 cos^3 theta - 3 cos theta) + 448`

`=116sqrt(58)cos 3 theta + 448`

So `cos 3 theta = -448/(116 sqrt(58)) = -112/(29 sqrt(58))`

So `3 theta = +-arccos(-112/(29 sqrt(58)))+2kpi`

So `theta = +-1/3 (arccos(-112/(29 sqrt(58)))+2kpi)`

So `cos theta = cos(1/3 (arccos(-112/(29 sqrt(58)))+2kpi))`

So `t = sqrt(232) cos(1/3 (arccos(-112/(29 sqrt(58)))+2kpi))`

So `x = 1/3(sqrt(232) cos(1/3 (arccos(-112/(29 sqrt(58)))+2kpi)) - 2)`

which takes distinct values for `k = 0, 1, 2.`

These are our `3` Real zeros `x_k`

Then: `x^3+2x^2-18x+4 = (x-x_0)(x-x_1)(x-x_2)`

`x_0 ~~ 3.21370759428`

`x_1 ~~ -5.44240577684`

`x_2 ~~ 0.22869818256`