Given that root 2 is a zero of cubic polynomial 6x Cube + root 2 x squ...
Finding the other two zeros of a cubic polynomial
To find the other two zeros of the given cubic polynomial, we need to use the fact that the sum of the zeros of a polynomial is equal to the negative of the coefficient of the second term divided by the coefficient of the highest term, and the product of the zeros is equal to the constant term divided by the coefficient of the highest term.
Finding the sum of the zeros
The coefficient of the second term is √2, and the coefficient of the highest term is 6. Therefore, the sum of the zeros of the polynomial is:
- (√2) / 6 = - (1 / (3√2))
Finding the product of the zeros
The constant term is -4√2, and the coefficient of the highest term is 6. Therefore, the product of the zeros of the polynomial is:
(-4√2) / 6 = - (2√2) / 3
Using the sum and product of the zeros to find the other two zeros
Let the other two zeros be a and b. We know that:
a + b = - (1 / (3√2))
ab = - (2√2) / 3
We can solve these two equations for a and b using substitution or elimination. For example, we can use the fact that:
(a + b)² = a² + 2ab + b²
Substituting the values of a + b and ab, we get:
(a + b)² = a² - (2√2) / 3 + b²
Simplifying, we get:
a² + b² = (1 / (9√2)) + (2√2) / 3
We also know that:
(a - b)² = a² - 2ab + b²
Substituting the value of ab, we get:
(a - b)² = a² + (4√2) / 3 + b²
Simplifying, we get:
a² + b² = (4 / 3) - (4√2) / 3
Setting the two expressions for a² + b² equal to each other, we get:
(1 / (9√2)) + (2√2) / 3 = (4 / 3) - (4√2) / 3
Simplifying, we get:
2√2a + 2√2b = 1
Substituting the value of a + b, we get:
2√2a + 2√2(- (1 / (3√2)) - a) =