find p(x) of degree 3 having zeros of x=1/2, x-1i, w/integer coefficient

Question

 I need help finding the polynomial having the zeros displayed in the question

Answer 1

We have roots at x = 1/2 and x = i

For any polynomial with a nonreal, complex, root at a+bi

the complex conjugate a-bi is also a root.

If x=i is a root, then so is x=-i

 

Your polynomial is thus:

(x-1/2)(x-i)(x+i)

= (x-1/2)(x² +1)

= x³ - (1/2)x² + x - 1/2

 

This is, of course, the same answer as Josue H gave you...

But with no need to find that 2nd root in the manner in

which Josue did it.

 

If p(x) has a root at 3+4i   then it must have a root at 3-4i

and so forth..

 

i = 0 + 1i

Conjugate:  0 - 1i = -i

 

So the roots were immediately known to be x = 1/2, i , -i

Answer 2

The zeros of your polynomial are x=1/2 and x=i

 

Starting with x=1/2, we get the following:

 

x=1/2

x-1/2=0 (subtracting 1/2 from both sides)

 

This tells us that one of the factors is (x-1/2)

 

Now, let's work with x=i:

 

x=i

x²=i²

x²=-1 (i² is equal to -1)

x²+1=0

 

This tells us that the other factor is (x²+1)

 

To find the polynomial, all we have to do is muliply both factors together:

 

p(x)=(x-1/2)(x²+1)

p(x)=x³-(1/2)x²+x-1/2

 

I hope this helped!