A cubic polynomial with rational coefficients has the roots 6 +square root sixand 2 over 3. Find one additional root.
(1 point)
Responses
–6 –square root six
–6 – Image with alt text: square root six
6 +square root six
6 + Image with alt text: square root six
–6 +square root six
–6 + Image with alt text: square root six
6 –square root six
In order for a cubic polynomial with rational coefficients to have irrational roots, it must also have a rational root. We are given that the roots are 6 + √6 and 2/3.
To find the additional root, we will use the fact that the sum of the roots of a cubic polynomial is equal to the negation of the coefficient of the quadratic term divided by the coefficient of the cubic term.
Let the cubic polynomial be represented as:
(x - r)(x - s)(x - t) = 0
Where r = 6 + √6, s = 2/3, and t is the additional root.
The sum of the roots is r + s + t, which should be equal to the negation of the coefficient of the quadratic term divided by the coefficient of the cubic term.
The coefficient of the quadratic term is equal to the product of the pairwise sums of the roots, which is (r + s)(r + t) + (s + t)(r + t) + (r + s)(s + t).
Expanding these products, we have:
(r + s)(r + t) = r^2 + (r + t)s + st
(s + t)(r + t) = rs + rt + t^2
(r + s)(s + t) = rs + (r + t)s + st
Substituting the given roots, we have:
(r + s)(r + t) = (6 + √6 + 2/3)((6 + √6) + t)
(s + t)(r + t) = ((2/3) + t)(6 + √6 + t)
(r + s)(s + t) = (6 + √6 + 2/3)(2/3 + t)
The coefficient of the quadratic term is the sum of these three products:
(r + s)(r + t) + (s + t)(r + t) + (r + s)(s + t) =
[(6 + √6 + 2/3)((6 + √6) + t)] + [((2/3) + t)(6 + √6 + t)] + [(6 + √6 + 2/3)(2/3 + t)]
We must set this sum equal to the negation of the coefficient of the cubic term, which is 1.
[(6 + √6 + 2/3)((6 + √6) + t)] + [((2/3) + t)(6 + √6 + t)] + [(6 + √6 + 2/3)(2/3 + t)] = -1
Simplifying this equation will give us the value of t, which is the additional root.