Write a polynomial of least degree with real coefficients and with the root -9–11𝑖

Question

Write your answer using the variable x and in standard form with a leading coefficient of 1

Answer 1

Hi Madi,

 

If one of the roots is complex, then its conjugate is also a root.  So the two roots are x=-9-11i   and   x=-9+11i

 

To turn these into FACTORS, put everything on the left side so that the right side = 0.  Meaning:

 

x+9+11i=0     and     

 

x+9-11i = 0

 

 

The polynomial will be found by multiplying these two factors:

 

(x+9+11i)(x+9-11i)

 

 

There are several ways to do this.  Can "brute force" multiply term by term and gather all the like terms and get the quadratic polynomial that way.   Can see this as the factorization of the difference of squares, and I am sure there are many other ways, these are the two that come to mind.

 

Brute force"

x²+9x-11xi+9x+81-99i+11xi+99i-121i²

 

Some things to know/notice:

 

i² = -1

the -99i and +99i cancel each other

the -11xi and +11xi cancel each other

 

So this simplifies to

 

x² + 18x + 81 - 121(-1)

 

Ans:  x² + 18x +202

 

 

The other way is to "see" it grouped like this:

 

[ (x+9) + 11i ] [ (x+9) - 11i ]

 

When these are multiplied, you get   

 

(x+9)² - (11i)²

 

= x² +18x +81 -121i²

 

=x² + 18x +81+121

 

=x² +18x + 202    (just like the other method)