Suppose the polynomial f(x) has the following roots: 1+6sqrt2, 2−sqrt6, and 6+sqrt2. If f(x) has only rational coefficients, the Irrational Root Theorem indicates that f(x) has at least three more roots.
What are the three additional roots that f(x) must have?
If you have x + y sqrt 2 you better have x- y sqrt 2
because you need sqrt 2 * sqrt 2 to get those original rational coefficients
thank you so much Damon
Since f(x) has a root of 1+6√2, then it must also have a root of 1-6√2. This is because complex roots always appear in conjugate pairs when the polynomial has rational coefficients.
Similarly, since f(x) has a root of 2-√6, it must also have a root of 2+√6.
Lastly, since f(x) has a root of 6+√2, it must also have a root of 6-√2.
Therefore, the three additional roots that f(x) must have are 1-6√2, 2+√6, and 6-√2.
To find the additional roots that f(x) must have, we can make use of the Conjugate Root Theorem. According to this theorem, if a polynomial with rational coefficients has an irrational root of the form a + b√c, where a, b, and c are rational numbers, then the conjugate root, a - b√c, must also be a root of the polynomial.
In this case, the given roots of f(x) are as follows:
1 + 6√2
2 − √6
6 + √2
To find the corresponding conjugate roots, we need to change the sign of the irrational term. Thus, the conjugate roots of the given roots are:
1 - 6√2
2 + √6
6 - √2
Therefore, the three additional roots that f(x) must have are:
1 - 6√2
2 + √6
6 - √2