How do you find a third degree polynomial given roots #3# and #2-i#?

1 Answer
Hammer
Jun 18, 2018

We are given roots

#x_1=3#
#x_2=2-i#

The complex conjugate root theorem states that, if #P# is a polynomial in one variable and #z=a+bi# is a root of the polynomial, then #bar z=a-bi#, the conjugate of #z#, is also a root of #P#.

As such, the roots are

#x_1=3#
#x_2=2-i#
#x_3=2-(-i)=2+i#

From Vieta's formulas, we know that the polynomial #P# can be written as:

#P_a(x)=a(x-x_1)(x-x_2)(x-x_3)#

Where #a# is a constant.

#:. P_a(x) = a(x-3)(x-2+i)(x-2-i)#

#:. P_a(x) =a(x^3-7x^2+17x-15)#

One such polynomial would be

#P_1(x) = x^3-7x^2+17x-15#

Related questions
See all questions in Complex Conjugate Zeros
Impact of this question
7077 views around the world
You can reuse this answer
Creative Commons License