Factorising Polynomial for Control Engineering Problem

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SUMMARY

The polynomial s^3 + 16s^2 + 85s + 250 can be factored using its roots, which are -10, -3+4j, and -3-4j. The correct factorization over the real numbers is (x + 10)(x^2 - 6x + 25), while the factorization over the complex numbers is (x + 10)(x - (3 + 4j))(x - (3 - 4j)). The roots must be conjugates when dealing with real coefficients, confirming that -3+4j's conjugate is -3-4j, not -3-3j as initially stated.

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Homework Statement



As part of a control engineering problem, I need to factorise this polynomial.

s^3 + 16s^2 + 85s + 250

(EDIT: the images failed so had to write it in this form)

Homework Equations


I know that the roots are -10, -3+4j and -3-3j.



The Attempt at a Solution


I'm not quite sure where to begin with this. Should i guess a factor and attempt long division to see if it is one? or what should i do?
 
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You KNOW the roots. Surely you know that if a, b, and c are roots of a cubic (with leading coefficient 1) then it factors as (x- a)(x- b)(x- c).

However, the zeros of your polynomial are NOT -3+4j, -3-3j because complex zeros of a polynomial equation with real coefficents must be conjugates. -3+4j is a solution so the other must be 3- 4j, not -3-3j. That tells you immediately that the factors are (x-(-10))(x-(3+4j))((x-(3-4j)). If you want only real factors then multiply oujt the last two: (x-(3+4j))(x-(3-4j))= ((x-3)- 4j)((x-3)+4j)= (x-3)2+ 16= x2- 6x+9+ 16= x2- 6x+ 25.

x3+ 16x2+ 85x+ 250 factors as (x+ 10)(x2- 6x+ 25) over the real numbers and as (x+ 10)(x-3-4j)(x-3+4j) over the complex numbers.
 
Yea sorry that was a typo for the first roots. shortly after writing this i realized the roots would be part of the factors and worked it out. thanks very much!
 

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