form a polynomial f(x) with real coeffiecients having the given degree and zeros

Question

degree 4; zeros: -2-5i; 3 multiplicity of 2

Answer 1

Hi, Marissa!

 

We need to write out the factors that produce those zeros. To get 3 as a zero, the factor must have been (x-3). Having a multiplicity of two means that particular factor appeared twice in the original polynomial. So we have (x-3)².

 

With imaginary number roots, the only way they could come from real coefficients is if the conjugate is also a root. So when it says -2-5i is a root, then -2+5i must be as well. That gives us these roots:

 

x - (-2-5i) = x+2+5i

x - (-2+5i) = x+2-5i

 

With those roots, we have all the factors we need:

 

f(x) = (x-3)²(x+2+5i)(x+2-5i)

 

They want all real coefficients, so we need to multiply that last guy out...

 

(x+2+5i)(x+2-5i) = x² + 2x - 5xi + 2x + 4 -10i + 5ix +10i - 25i²

 

Note that every term with a single i cancels out, and i² = -1. This leads us to....

 

(x+2+5i)(x+2-5i) 

= x² + 2x - 5xi + 2x + 4 -10i + 5ix +10i - 25i²

= x² + 4x + 4 + 25

= x² + 4x + 29

 

So our original polynomial is...

 

f(x) = (x-3)²(x² + 4x + 29)

 

I assume they want this multiplied out, so...

 

f(x) = (x²-6x+9)(x²+4x+29)

 

Can you take it from there?