Form a polynomial f(x) with real coefficients having the given degree of zeros. Can you advise on how to figure this out?

Degree 5; zeros: #-7; -i; -6+i#

1 Answer
George C.
Feb 10, 2018

#f(x) = x^5+19x^4+122x^3+278x^2+121x+259#

Explanation:

If a polynomial has real coefficients then any non-real complex zeros occur in complex conjugate pairs.

So in our example, the polynomial will have zeros:

#-7#, #-i#, #i#, #-6+i#, #-6-i#

Each zero #a# corresponds to a factor #(x-a)#, so we can write #f(x)# as:

#f(x) = (x+7)(x+i)(x-i)(x+6-i)(x+6+i)#

#color(white)(f(x)) = (x+7)(x^2-i^2)((x+6)^2-i^2)#

#color(white)(f(x)) = (x+7)(x^2+1)(x^2+12x+37)#

#color(white)(f(x)) = (x+7)(x^4+12x^3+38x^2+12x+37)#

#color(white)(f(x)) = x^5+19x^4+122x^3+278x^2+121x+259#

Any polynomial with these zeros will be a multiple (scalar or polynomial) of this #f(x)#.

Related questions
Impact of this question
1002 views around the world
You can reuse this answer
Creative Commons License