Solving a cubic?

Question

x^3 - 30(x^2) + 108x - 104 = 0

The question is, does it have three imaginary roots, two imaginary roots, three distinct real roots, or a repeated root? And why?

Answer 1

It obviously has one real root (all odd degree polynomials do). If it has two real roots, then it has to have three. So, if we can prove that it has two real roots, then we are done.

Likewise, if we can prove it doesn't have two real roots, then it must have two imaginary roots.

Consider the derivative of the function (I'm going to call it f).
f'=3x^2-60x+108 =3(x^2-20+36)= 3(x-2)(x-18).
So f is increasing when x<2, decreasing when 218.

Thus x=2 is a local maximum.

f(2)= 2^3-30(2^2)+108•2-104= 8-120+216-104= 0.

This means that 2 is a double root (since it a root and a local maximum). Therefore f has at least 2 real roots (even though not distinct) and must have three. So f has no imaginary roots.


Furthermore, we can easily divide f(x) by (x-2)^2=x^2-4x+4 and get

f(x)=x^3-30x^2+108x-104 =x(x-2)^2+g(x)
thus g(x)= x^3-30x^2+108x- 104-x(x^2-4x+4)= -26x^2+104x-104= -26(x^2-4x+4)=-26g(x)

thus f(x)= xg(x)-26g(x)= (x-26)g(x)= (x-26)(x-2)^2

Therefore the roots are 2 and 26, where 2 is a double root.

Answer 2

If you divide the cubic by (x-2) you will get the answer x^2 - 28x + 52 with no remainder. This means that x = 2 is a root.

You are then left with a simple quadratic x^2 - 28x + 52 = 0. The discriminant (b^2 - 4ac) is positive, therefore the quadratic has two real roots. If the discriminant was negative then the quadratic would have two complex roots. Note the quadratic can have 0 or 2 complex roots, it cannot have 1 complex root.

Factorising the quadratic you get (x-2)(x-26)=0

So the cubic now reads (x-2)(x-2)(x-26)=0

So the cubic has 3 real roots and they are 2, 2 and 26

Answer 3

It can't have 3 imaginary roots because all the coefficients are real.

By the way: 2 2 26

Answer 4

x^3 - 30x^2 + 108x - 104 = (x - 2)(x - 2)(x - 26)

3 real roots

Answer 5

substitute x=2 and the LHS=0
so x-2 is a factor of LHS.
x^3-30x^2+108x-104=0
(x-2)(x^2-28x+52)=0
(x-2)=0 or (x^2-28+52)=0
x=2 is one root.
x^2-28x+52=0
x^2-26x-2x+52=0
x(x-26)-2(x-26)=0
(x-26)(x-2)=0
x=2 or 26.
So, the three roots are 2, 2 and 26.

Answer 6

After giving this all a great deal of thought I have decided that I haven't a clue what you are all on about or which planet I am on!!
Keep up the good work???