Suppose that $\newline$ $R(x)$ is a polynomlal of degree $9$ whose coefficients are real numbers. $\newline$ Also, suppose that $\newline$ $R(x)$ has the following zeros, $\newline$ $-4,\quad5,\quad-1-5i,\quad i$ $\newline$ Answer the following: $\newline$ (a) Find another zero of $\newline$ $R(x)$ . $\newline$ (b) What is the maximum number of real zeros that $\newline$ $R(x)$ can have? $\newline$ (c) What is the maximum number of nonreal zeros that $\newline$ $R(x)$ can have?

View step-by-step help

Home

Math Problems

Precalculus

Complex conjugate theorem

Full solution

Q. Suppose that

\newline

R(x)

is a polynomlal of degree

9

whose coefficients are real numbers.

\newline

Also, suppose that

\newline

R(x)

has the following zeros,

\newline

-4,\quad5,\quad-1-5i,\quad i

\newline

Answer the following:

\newline

(a) Find another zero of

\newline

R(x)

\newline

(b) What is the maximum number of real zeros that

\newline

R(x)

can have?

\newline

\newline

R(x)

can have?

Complex Zeros Conjugate Pairs: The complex zeros of a polynomial with real coefficients come in conjugate pairs. Since $-1 - 5i$ is a zero, its conjugate $-1 + 5i$ must also be a zero.
Additional Complex Zeros: The zero $i$ is also complex, and its conjugate $-i$ must also be a zero of the polynomial.
Total Number of Zeros: Now we have the following zeros: $-4$ , $5$ , $-1 - 5i$ , $-1 + 5i$ , $i$ , and $-i$ . This gives us a total of $6$ zeros.
Maximum Real Zeros: Since $R(x)$ is a polynomial of degree $9$ , it can have at most $9$ zeros, counting multiplicity.
Maximum Nonreal Zeros: We have identified $6$ zeros so far. Because the degree of the polynomial is $9$ , there can be $3$ more zeros, which could be real or nonreal.
Maximum Nonreal Zeros: We have identified $6$ zeros so far. Because the degree of the polynomial is $9$ , there can be $3$ more zeros, which could be real or nonreal.The maximum number of real zeros a polynomial can have is equal to its degree. Since we already have two real zeros ( $-4$ and $5$ ), the maximum number of additional real zeros $R(x)$ can have is $7$ .
Maximum Nonreal Zeros: We have identified $6$ zeros so far. Because the degree of the polynomial is $9$ , there can be $3$ more zeros, which could be real or nonreal. The maximum number of real zeros a polynomial can have is equal to its degree. Since we already have two real zeros ( $-4$ and $5$ ), the maximum number of additional real zeros $R(x)$ can have is $7$ . The maximum number of nonreal zeros a polynomial can have is also determined by its degree. Since nonreal zeros must come in conjugate pairs, and we have two pairs of nonreal zeros ( $-1 - 5i$ with $-1 + 5i$ , and $i$ with $9$ $0$ ), the maximum number of nonreal zeros $R(x)$ can have is $9$ $2$ , as long as the total number of zeros does not exceed the degree of the polynomial.