AJ L. answered 03/19/23
Patient and knowledgeable Algebra Tutor committed to student mastery
x=-1 is the root of x+1=0
x=±3i are the roots of x2+9=0
Hence, one such polynomial could be P(x) = (x2+9)(x+1)
Anni M.
asked 03/19/23AJ L. answered 03/19/23
Patient and knowledgeable Algebra Tutor committed to student mastery
x=-1 is the root of x+1=0
x=±3i are the roots of x2+9=0
Hence, one such polynomial could be P(x) = (x2+9)(x+1)
Mark M. answered 03/19/23
Mathematics Teacher - NCLB Highly Qualified
P(x) = (x + 1)(x + 3I)(X - 3I)
Get a free answer to a quick problem.
Most questions answered within 4 hours.
Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.