Linear Factors Theorem and Conjugate Zeros Theorem

Question

Suppose that R(x) is a polynomial of degree 15 whose coefficients are real numbers.

Also, suppose that R(x) has the following zeros.

 

1-i, -3i, 4(with multiplicity 4)

Answer the following:

a) Find another zero of R(x)

b) What is the maximum number of real zeros (not necessarily distinct) that R(x) can have?

c) What is the maximum number of nonreal zeros that R(x) can have?

 

Answer 1

With a polynomial of degree 15 there are

15 roots.  These roots not necessarily distinct,

as with the indicated root of 4 with multiplicity 4 

meaning that (x-4)⁴ is part of the factored p(x).

 

If a nonreal, complex, number a+bi is a root, then

so it's conjugate a-bi

 

The roots we have accounted for are thus:

1-i, 1+i, -3i, 3i, 4 with multiplicity 4...

This is 8 of 15 roots leaving 7 unaccounted for.

 

Maximum number real zeroes = 11 since we

know there are at least 4 nonreal zeroes.

 

Maximum number of nonreal zeroes:

We have 4 real zeroes identified leaving

11 possible zeroes. .  As discussed,

nonreal zeroes come in pairs and the total

zeroes is 15... Maximum number of nonreal

zeroes is 10 in 5 sets of complex conjugate pairs.

Note that 4 of those 10 are already identified.