# How do you find a polynomial function f with real coefficients of the indicated degree that possesses the given zeros: degree 2; 4+3i?

##### 2 Answers

#### Answer:

#### Explanation:

We know that the function is a degree-2 polynomial, so the function has two solutions, one of which is given as

Know that **if one imaginary species is a solution to a polynomial equation, its conjugate is also a solution**.

Therefore, the other solution to this equation is the conjugate of

Now that we know the solutions to the polynomial equation, let's derive a function, by setting them equal to zero like so:

which is the same as

Factoring out the polynomial gives us

Combining like terms..

The equation of the function with zeros

#### Answer:

# x^2-8x+25= 0 #

#### Explanation:

Complex roots of real valued polynomials always appear in conjugate pairs, so if one root of

# alpha = 4+3i #

Then another root is:

# beta = 4-3i #

Se we are looking for a polynomial of degree

So then:

Sum

#=alpha+beta#

#" " = (4+3i) + (4-3i)#

#" " = 8# Product

#=alphabeta#

#" " = (4+3i)(4-3i)#

#" " = 16-9i^2#

#" " = 25#

And so the required equation is:

# x^2-("sum of roots")x + ("product of roots") = 0 #

# :. x^2-8x+25= 0 #