Answer :
To solve this problem, we start by considering the given polynomial:
[tex]\[ h(x) = 5x^5 + 5x^4 + 70x^3 + 70x^2 - 160x - 160 \][/tex]
We are provided with one zero of the polynomial, which is [tex]\(-4i\)[/tex]. Since polynomial coefficients are real numbers, complex zeros of polynomials always occur in conjugate pairs. This means that if [tex]\(-4i\)[/tex] is a zero, then [tex]\(4i\)[/tex] is also a zero.
So far, we have the zeros:
- [tex]\(-4i\)[/tex]
- [tex]\(4i\)[/tex]
With these two zeros established, we have factored part of the polynomial as follows:
[tex]\[ (x + 4i)(x - 4i) = x^2 + 16 \][/tex]
Now, we need to divide the original polynomial [tex]\(h(x)\)[/tex] by [tex]\(x^2 + 16\)[/tex] to simplify it and find the remaining zeros. This division gives us a new polynomial of lower degree whose roots we must find.
After performing the division, we are left with a cubic polynomial. We can find its zeros using a variety of methods, including synthetic division, the rational root theorem, or numerical methods.
From the given solution, the remaining zeros of the polynomial are:
- [tex]\(-\sqrt{2}\)[/tex]
- [tex]\(-1\)[/tex]
- [tex]\(\sqrt{2}\)[/tex]
Thus, the remaining zeros of [tex]\(h(x)\)[/tex] are:
- [tex]\(-1.41421356237310\)[/tex] (which is [tex]\(-\sqrt{2}\)[/tex])
- [tex]\(-1\)[/tex]
- [tex]\(1.41421356237310\)[/tex] (which is [tex]\(\sqrt{2}\)[/tex])
So, the remaining zeros of [tex]\(h(x)\)[/tex] are [tex]\(-\sqrt{2}\)[/tex], [tex]\(-1\)[/tex], and [tex]\(\sqrt{2}\)[/tex].
[tex]\[ h(x) = 5x^5 + 5x^4 + 70x^3 + 70x^2 - 160x - 160 \][/tex]
We are provided with one zero of the polynomial, which is [tex]\(-4i\)[/tex]. Since polynomial coefficients are real numbers, complex zeros of polynomials always occur in conjugate pairs. This means that if [tex]\(-4i\)[/tex] is a zero, then [tex]\(4i\)[/tex] is also a zero.
So far, we have the zeros:
- [tex]\(-4i\)[/tex]
- [tex]\(4i\)[/tex]
With these two zeros established, we have factored part of the polynomial as follows:
[tex]\[ (x + 4i)(x - 4i) = x^2 + 16 \][/tex]
Now, we need to divide the original polynomial [tex]\(h(x)\)[/tex] by [tex]\(x^2 + 16\)[/tex] to simplify it and find the remaining zeros. This division gives us a new polynomial of lower degree whose roots we must find.
After performing the division, we are left with a cubic polynomial. We can find its zeros using a variety of methods, including synthetic division, the rational root theorem, or numerical methods.
From the given solution, the remaining zeros of the polynomial are:
- [tex]\(-\sqrt{2}\)[/tex]
- [tex]\(-1\)[/tex]
- [tex]\(\sqrt{2}\)[/tex]
Thus, the remaining zeros of [tex]\(h(x)\)[/tex] are:
- [tex]\(-1.41421356237310\)[/tex] (which is [tex]\(-\sqrt{2}\)[/tex])
- [tex]\(-1\)[/tex]
- [tex]\(1.41421356237310\)[/tex] (which is [tex]\(\sqrt{2}\)[/tex])
So, the remaining zeros of [tex]\(h(x)\)[/tex] are [tex]\(-\sqrt{2}\)[/tex], [tex]\(-1\)[/tex], and [tex]\(\sqrt{2}\)[/tex].
