# How many possible positive, negative and complex solutions are there in this equation x^3-10x^2+21x-210

Question

anonymous · Answer

are they asking you to find the exact number of solutions that are positive negative and complex...or do they want to know how many solutions there can be?

anonymous · Answer

Equals zero?  Total number of real and complex roots is three, based on the degree of the function.  At least one will be real.

anonymous · Answer

Factor by grouping to find out the particulars of this problem.

.sam. · Answer

3

amistre64 · Answer

its a Dsign question

anonymous · Answer

Yeah factor and find as many solutions...if there are still some solutions left that you could not find through factoring then the leftover solutions will be complex

.sam. · Answer

x^(3)-10x^(2)+21x-210=0

(x^(2)+21)(x-10)=0

x^(2)+21=0
x-10=0

x=i sqrt(21),-i sqrt(21),10

amistre64 · Answer

x^3 -10x^2 +21x -210
   1   |    1       |   1  |   0   ; 3 sign changes in the normal manner

there are possibly 3 or 1 positive zeros

amistre64 · Answer

swap signs on odd powers

-x^3 -10x^2 -21x -210
            0                       |

there are no sign changes in the negative version; there are no negative roots

amistre64 · Answer

there are possibly 2 complex roots which is why we had to step the first one by 2

anonymous · Answer

$$x^3-10x^2+21x-210=0\rightarrow x^2(x-10)+21(x-10)$$$$=(x^2+21)(x-10)=0\rightarrow x \in \left\{ 10,\pm i \sqrt{21} \right\}$$

anonymous · Answer

okay so there are possibly 3 or 1 positive zeros and 2 complex but no negative?

amistre64 · Answer

correct

anonymous · Answer

thank you

amistre64 · Answer

youre welcome