Factoring Cubic Equation With no Calculator?

Questions
1

Hello, can someone help with the following question without using rational zero theorem?

The equation x^3 - 12x^2 + 26x -48 can be factored in the form (x-a)(x-b)(x-c). Where a,b and c are real numbers.
What is the exact value of of a^2 + b^2 + c^2?

Apparently the answer is 92. Could someone provide an explanation?

2

Since x^3 - 12x^2 + 26x -48 cannot be factored in the form (x-a)(x-b)(x-c) where a,b and c are real numbers

FactorTheCubic

the explanation seems it must be that there is an error. Finding that error is the place to start.

3

In[37]:= NSolve[x^3 - 12 x^2 + 26 x - 48 == 0, x]

Out[37]= {{x -> 1.07192 - 1.929 I}, {x -> 1.07192 + 1.929 I}, {x -> 
   9.85617}}

Clearly the roots are not all real numbers. But it is still true that

In[39]:= (1.071915073061755` - 
    1.9290007822934752` I)^2 + (1.071915073061755` + 
    1.9290007822934752` I)^2 + 9.856169853876489`^2

Out[39]= 92. + 0. I
4

Cubics can be solved exactly and this has been known for a while… http://en.wikipedia.org/wiki/Cubic_function):

Tr[(x /. Solve[x^3 - 12 x^2 + 26 x - 48 == 0, x])^2] // FullSimplify

92

Or more generally:

In[1]:= Tr[(x /. 
    Solve[a x^3 + b x^2 + c x + d == 0, x])^2] // FullSimplify

Out[1]= (b^2 - 2 a c)/a^2

In[2]:= (b^2 - 2 a c)/a^2 /. {a -> 1, b -> -12, c -> 26, d -> -48}

Out[2]= 92
5

The roots: x1 (real {if the the degree of a polynom is odd, than there exists at least one real root}), x2=a+bi & x3=a-bi (complex conjugate of x2).

Then x^3 - 12x^2 + 26x -48= (x-x1) * (x-x2) * (x-x3) in C(omplex).

Comparing the coefficients:
x^3+ax^2+bx+c=0 and (x-x1) * (x-x2) * (x-x3)=0

a=-12=-(x_1+x_2+x_3), therefore x_1+x_2+x_3=12, b=26=x1 * x2 + x1 * x3 + x2 * x3, c =- x1 x2 x3

Since x1^2+x2^2+x3^2=(x1+x2+x3)^2 - 2*(x1 x2+x1 x3+x2 x3)=144-52=92 (and it’s real because x1 is real and x2 & x3 are conjugates, so x2+x3 & x2 * x3 is real, and not because each of the three roots would be real)

Hope, it is clear :).