Answer :
Sure! Let's find the remaining zeros of the polynomial given the following zeros and the condition that the polynomial has rational coefficients.
The given zeros are:
1. [tex]\( 5 \)[/tex]
2. [tex]\( -4 + 3i \)[/tex]
3. [tex]\( 6 - \sqrt{7} \)[/tex]
Since the polynomial has rational coefficients, any non-rational zeros (complex or involving radicals) must occur in conjugate pairs. Thus, we have the following steps:
1. Identifying the Conjugates:
- For the complex zero [tex]\( -4 + 3i \)[/tex], the conjugate is [tex]\( -4 - 3i \)[/tex].
- For the irrational zero [tex]\( 6 - \sqrt{7} \)[/tex], the conjugate is [tex]\( 6 + \sqrt{7} \)[/tex].
2. Listing All Zeros:
Thus, besides the given zeros, we know the polynomial must also have:
- [tex]\( -4 - 3i \)[/tex] (the conjugate of [tex]\( -4 + 3i \)[/tex])
- [tex]\( 6 + \sqrt{7} \)[/tex] (the conjugate of [tex]\( 6 - \sqrt{7} \)[/tex])
So, the other zeros of the polynomial are:
[tex]\[ -4 - 3i, 6 + \sqrt{7} \][/tex]
Therefore, the other zeros are exactly [tex]\( -4 - 3i \)[/tex] and [tex]\( 6 + \sqrt{7} \)[/tex].
The given zeros are:
1. [tex]\( 5 \)[/tex]
2. [tex]\( -4 + 3i \)[/tex]
3. [tex]\( 6 - \sqrt{7} \)[/tex]
Since the polynomial has rational coefficients, any non-rational zeros (complex or involving radicals) must occur in conjugate pairs. Thus, we have the following steps:
1. Identifying the Conjugates:
- For the complex zero [tex]\( -4 + 3i \)[/tex], the conjugate is [tex]\( -4 - 3i \)[/tex].
- For the irrational zero [tex]\( 6 - \sqrt{7} \)[/tex], the conjugate is [tex]\( 6 + \sqrt{7} \)[/tex].
2. Listing All Zeros:
Thus, besides the given zeros, we know the polynomial must also have:
- [tex]\( -4 - 3i \)[/tex] (the conjugate of [tex]\( -4 + 3i \)[/tex])
- [tex]\( 6 + \sqrt{7} \)[/tex] (the conjugate of [tex]\( 6 - \sqrt{7} \)[/tex])
So, the other zeros of the polynomial are:
[tex]\[ -4 - 3i, 6 + \sqrt{7} \][/tex]
Therefore, the other zeros are exactly [tex]\( -4 - 3i \)[/tex] and [tex]\( 6 + \sqrt{7} \)[/tex].
