Answer:
(a) Ā Other zeros: Ā 3i Ā and Ā -4-i
(b) Ā 9
(c) Ā 10
Step-by-step explanation:
Given information:
For any complex number Ā [tex]z=a+bi[/tex], the complex conjugate of the number is defined as Ā [tex]z^*=a-bi[/tex] . Ā
If f(z) is a polynomial with real coefficients, and zā is a root of f(z)=0, then its complex conjugate zā* is also a root of f(z)=0.
Therefore, if R(x) is a polynomial with real coefficients, and 3i is a root of R(x)=0, then its complex conjugate -3i is also a root of R(x)=0.
Similarly, if -4+i is a root of R(x)=0, then its complex conjugate -4-i is also a root of R(x)=0.
The degree of the polynomial is the greatest of the exponents of its various terms. Ā
The degree of R(x) with just the given roots (including the complex conjugates) is 6, which includes 2 real zeros.
As the actual degree of R(x) is 13, the maximum number of real zeros would be 7 more than the given real zeros.
Therefore, the maximum number of real zeros that R(x) can have is 9.
For each complex root there is also a complex conjugate root.
We have been given 2 complex zeros and 2 real roots, so when we include the complex conjugates, the degree of R(x) is 6.
The actual degree of R(x) is 13, so the maximum number of non-real zeros is 6 more (as they come in pairs) than the already given 4 non-real zeros. Ā Therefore, the maximum number of non-real zeros that R(x) can have is 10.
