# Factoring Cubic Equation With no Calculator?

Posted 8 years ago
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 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?
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Posted 8 years ago
 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)=0a=-12=-(x1+x2+x3), therefore x1+x2+x3=12, b=26=x1 * x2 + x1 * x3 + x2 * x3, c =- x1 x2 x3Since 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 :).
Posted 8 years ago
 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 
Posted 8 years ago
 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 `
Posted 8 years ago
 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 numbersFactorTheCubicthe explanation seems it must be that there is an error. Finding that error is the place to start.