-2, 1 - 5i; degree 3 polynomial

Question

I don't understand how to incorporate the 1-5i to get a third degree polynomial.. I know that -2 is (x+2) but I am stumbling over 1-5i. Please Help

Answer 1

If the coefficients are all real, then the remaining root is 1+5i.

 

We can derive a quadratic whose roots are 1±5i

 

x = 1 ± 5i

 

x - 1 = ±5i

 

(x - 1)² = (±5i)²

 

x² - 2x + 1 = -25

 

x² - 2x + 26 = 0

 

So the cubic you are looking for is

 

(x + 2)(x² - 2x + 26)

 

= x³ - 2x² + 26x + 2x² - 4x + 52

 

= x³ + 22x + 52

 

Note that any multiple of this polynomial by a constant also has the same set of roots.

 

 

Answer 2

The key to this is to remember that, if the coefficients of this polynomial are real numbers, then, if 1-5i is a root, then its "conjugate", 1+5i, must also be a root. Se we have the three roots:  -2, 1-5i, and 1+5i

So the 3rd degree polynomial would be  (x+2)(x-(1-5i))(x-(1+5i)).  (Note the careful use of parentheses.)