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

2 Answers
Jun 2, 2016

Answer:

#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

Jun 2, 2016

Answer:

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#