Polynomial (degree 3) with only 1 root

[TheOAP]

Assume a degree 3 polynomial. Assume the graph (and my program) indicate there is only 1 Root.
According to Dr Richard Feynman there must be 2 solution. So the other 2 must be a pair of complex "roots" ?
Assume equation : x3 +bx2+ c7x + 12
and assume the "root" found was x = 3. To make the ("root" times constant) equal to 12 then the Complex values (surely!) must be (x + 2i)(x - 2i)
I hope this makes sense as I don't see otherwise how the Complex values can be calculated.
Pete
Regards

 

[MathLover1]

x^3 +bx^2+ cx + 12=0

assume the "root" found was x = 3

as a product of factors, polynomial is

f(x)=(x-x1)(x-x2)(x-x3)

so, one factor is (x-3)
assuming, as you state, other two factors are (x + 2i) and (x - 2i)

f(x)=(x-3)(x + 2i)(x - 2i)
f(x)=x^3 - 3x^2 + 4x - 12

=>b=3 and c=4

check the graph:

as you can see, there is only one real root at x=3

 

[nycmathguy]

Assume a degree 3 polynomial. Assume the graph (and my program) indicate there is only 1 Root.
According to Dr Richard Feynman there must be 2 solution. So the other 2 must be a pair of complex "roots" ?
Assume equation : x3 +bx2+ c7x + 12
and assume the "root" found was x = 3. To make the ("root" times constant) equal to 12 then the Complex values (surely!) must be (x + 2i)(x - 2i)
I hope this makes sense as I don't see otherwise how the Complex values can be calculated.
Pete
Regards

Don't forget to say thank you to MathLover1 for answering YOUR thread.

 

[nycmathguy]

x^3 +bx^2+ cx + 12=0

assume the "root" found was x = 3

as a product of factors, polynomial is

f(x)=(x-x1)(x-x2)(x-x3)

so, one factor is (x-3)
assuming, as you state, other two factors are (x + 2i) and (x - 2i)

f(x)=(x-3)(x + 2i)(x - 2i)
f(x)=x^3 - 3x^2 + 4x - 12

=>b=3 and c=4

check the graph:

as you can see, there is only one real root at x=3

I love it. Thank you. This is precalculus, right?