Information is given about a polynomial f(x) whose coefficients are real numbers. Find the remaining zeros of f.
Degree 5; zeros: 9, 3+5i, -9i
complex roots always come in conjugate pairs
so roots are 9, 3 ± 5i, ± 9i
fo f(x) = (x-9)(x-3-5i)(x-3+5i)(x+9i)x-9i)
= (x-9)(x^2 - 6x + 34)(x^2 + 91)
expand if need be
To find the remaining zeros of the polynomial f(x), we can use the complex conjugate theorem.
The complex conjugate theorem states that if a polynomial has complex coefficients, and a + bi is a zero of the polynomial (where a and b are real numbers), then its conjugate, a - bi, is also a zero of the polynomial.
Given that the zeros of f(x) are: 9, 3+5i, and -9i, we can find the remaining zeros using the complex conjugate theorem.
Since 9 is a real number, it does not have a corresponding complex conjugate.
For the complex number 3+5i, its complex conjugate is 3-5i. So, the remaining zero of f(x) is 3-5i.
For the complex number -9i, its complex conjugate is 9i. So, the remaining zero of f(x) is 9i.
Therefore, the remaining zeros of f(x) are 3-5i, 9i.
To find the remaining zeros of a polynomial, we can use the fact that complex zeros come in conjugate pairs.
Given that the degree of the polynomial is 5 and the zeros provided are 9, 3+5i, and -9i, we can use the conjugate pair property to determine the remaining zeros.
Since 3+5i is a zero, its conjugate, which is 3-5i, must also be a zero of the polynomial. Similarly, since -9i is a zero, its conjugate, which is 9i, must also be a zero of the polynomial.
Thus, the remaining zeros of the polynomial are 3-5i and 9i.
To summarize:
- The given zeros are 9, 3+5i, and -9i.
- The remaining zeros are 3-5i and 9i.
Therefore, the complete set of zeros of the polynomial f(x) is {9, 3+5i, -9i, 3-5i, 9i}.