Final answer:
The error made by the student involves irrational roots of a polynomial equation with rational coefficients; these types of roots must appear in conjugate pairs. The student likely miscalculated √3 as a root without including its conjugate -√3. Therefore correct option is D
Step-by-step explanation:
A student claims that a fifth-degree polynomial equation with rational coefficients has roots -5, -3, 1, 2, and √3. The possible error the student may have made is captured in one of the provided options. To determine which is the correct error, let's evaluate each one:
- Option A suggests that if a polynomial has rational coefficients, at least one of the roots must be a fraction. This is incorrect because a polynomial with rational coefficients can have integer roots.
- Option B implies that an odd-degree polynomial must have at least one pair of complex roots. This is inaccurate since an odd-degree polynomial can have all real roots.
- Option C states that a fifth-degree polynomial should have at most four distinct roots. This is incorrect because a polynomial of degree n should have n roots, counting multiplicity.
- The correct answer is Option D, which states that the irrational roots of a polynomial equation with rational coefficients always come in conjugate pairs. Since √3 is an irrational root, its conjugate, -√3, must also be a root if the polynomial has rational coefficients. Therefore, the student either incorrectly calculated √3 as a root or omitted the root -√3.
It is important to remember that in polynomial equations with rational coefficients, roots that are irrational should come in pairs, as complex conjugates if the roots are complex, or as pairs of surds, like √3 and -√3, if the roots are real irrational numbers. This is a consequence of the Complex Conjugate Root Theorem, which applies to polynomials with real coefficients as well.