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Sagot :
To find the remaining zero(s) of a polynomial of degree 4 with rational coefficients, we need to consider the properties of polynomials with such coefficients. Specifically, if a polynomial has rational coefficients and an irrational number as a zero, its conjugate must also be a zero of the polynomial.
Given zeros:
- -1 (a rational number)
- [tex]\(\sqrt{2}\)[/tex] (an irrational number)
- [tex]\(\frac{5}{3}\)[/tex] (a rational number)
Since [tex]\(\sqrt{2}\)[/tex] is an irrational number, its conjugate [tex]\(-\sqrt{2}\)[/tex] must also be a zero of the polynomial to ensure that the polynomial has rational coefficients.
Therefore, the four zeros of the polynomial are:
- [tex]\(-1\)[/tex]
- [tex]\(\sqrt{2}\)[/tex]
- [tex]\(\frac{5}{3}\)[/tex]
- [tex]\(-\sqrt{2}\)[/tex]
Putting it all together, the complete list of zeros for the polynomial is:
[tex]\[ -1, \sqrt{2}, \frac{5}{3}, -\sqrt{2} \][/tex]
Given zeros:
- -1 (a rational number)
- [tex]\(\sqrt{2}\)[/tex] (an irrational number)
- [tex]\(\frac{5}{3}\)[/tex] (a rational number)
Since [tex]\(\sqrt{2}\)[/tex] is an irrational number, its conjugate [tex]\(-\sqrt{2}\)[/tex] must also be a zero of the polynomial to ensure that the polynomial has rational coefficients.
Therefore, the four zeros of the polynomial are:
- [tex]\(-1\)[/tex]
- [tex]\(\sqrt{2}\)[/tex]
- [tex]\(\frac{5}{3}\)[/tex]
- [tex]\(-\sqrt{2}\)[/tex]
Putting it all together, the complete list of zeros for the polynomial is:
[tex]\[ -1, \sqrt{2}, \frac{5}{3}, -\sqrt{2} \][/tex]
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