Find an nth degree polynomial function.

Question

Find an nth degree polynomial function with real coefficients satisfying the given conditions. n=3; -6 and i are zeros; f(-3)=60

Answer 1

The zeros are -6, i and -i (complex roots always come in conjugate pairs)

f(x) = (x + 6)(x - i)(x + i)

f(x) = (x + 6)(x² + 1)

f(-3) = (-3 + 6)((-3)² + 1)

f(-3) = 3(9 + 1)

(f(-3) = 30

 

f(x) = 2(x + 6)(x² + 1)

Answer 2

If i is a root, then its complex conjugate -i is also a root.  Then we have the 3 required roots for n=3.  They are -6, i, and -i.

 

Therefore,

 

f(x) = A(x + 6)(x - i)(x + i)

 

Note that

 

(x - i)(x + i) = x² - i² = x² - (-1) = x² + 1

 

Then so far we have

 

f(x) = A(x + 6)(x² + 1)

 

f(-3) = 60 = A(-3 + 6)[(-3)² + 1] = 3A(10) = 30A

 

60 = 30A ⇒ A = 60/30 = 2

 

Finally,

 

f(x) = 2(x + 6)(x² + 1)