Andrew M. answered • 02/14/18

Tutor

New to Wyzant
Mathematics - Algebra a Specialty / F.I.T. Grad - B.S. w/Honors

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)(x

^{2}+1)= x

^{3}- (1/2)x^{2}+ x - 1/2This 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

Harry D.

02/13/18