a cubic polynomial with rational coefficents has the roots 6+square root of 6 and 2/3. find one additional root
Let's call the cubic polynomial:
P(x) = ax^3 + bx^2 + cx + d
We know that the roots of the polynomial are 6+√6, 6-√6, and 2/3.
Using Vieta's formulas, we know:
1) The sum of the roots = - b / a
2) The product of the roots = -d / a
From equation 1, we have:
(6 + √6) + (6 - √6) + (2/3) = - b / a
(12 + 2/3) = - b / a
Simplifying the left side:
12 + 2/3 = 36/3 + 2/3
= 38/3
So, - b / a = 38/3
Now, we can find an additional root by using equation 2. Since we only need to find one additional root, we can calculate the third root by:
Root_3 = -d / a / (6 + √6) / (6 - √6) / (2/3)
Simplifying the right side:
Root_3 = -d / a / (3(6 + √6)(6 - √6) / 2
The terms (6 + √6)(6 - √6) represent the difference of squares, so we can simplify it to:
Root_3 = -d / a / (3(36 - 6) / 2)
Root_3 = -d / a / (3(30) / 2)
Root_3 = -d / a / 45
Since Root_3 is an additional root, it must be different from the given roots of 6+√6 and 2/3. Thus, we know that Root_3 cannot be equal to 6+√6 or 2/3, so we can conclude:
Root_3 ≠ 6 + √6
Root_3 ≠ 2/3
We also know that -b/a ≠ 6 + √6 and -b/a ≠ 2/3, based on the equations we used to find Root_3.
Therefore, we have the additional root:
Root_3 ≠ 6 + √6, 6 - √6, 2/3
It's not possible to determine the exact value of the third root without more information.