Question

show that zeroes of cubic polynomial x^3 - 3 x^2 - 2 x + 6 are - root of 2 , root of 2 and 3 if sum of its two zeroes is zero .

Answer 1

Answer:

three Zeroes are  √2 , -√2  & 3

Step-by-step explanation:

show that zeroes of cubic polynomial x^3 - 3 x^2 - 2 x + 6 are - root of 2 , root of 2 and 3 if sum of its two zeroes is zero .

Let say two roots whose sum = 0  are

α & - α  & third root = β

(x - α)(x - (-α))(x - β) = x³ - 3x² - 2x + 6

=> (x² - α²)(x - β) = x³ - 3x² - 2x + 6

=> x³ - βx² - xα² + α²β = x³ - 3x² - 2x + 6

Equating powers

=> -β = -3

=> β = 3

-α² = - 2

=> α²  = 2

=> α = ±√2

α²β = (2)(3) = 6  ( also fulfilling )

so three Zeroes are  √2 , -√2  & 3