Question

If the zeroes of the polynomial x^3-3x^2-6x+8 are of the form a-b, a, a+b find all zeroes

Answer 1

Hii friend,

P(X) = X³-3X²-6X+8

Where,

A = 1 , B = -3 , C = -6 And D = 8

Let Alpha = a-b , Beta = a , Gama = a+b

Sum of zeros = (Alpha + Beta+Gama) = -B/A

(a-b+a+a+b) = 3/1

=> 3a = 3 => a = 3/3 => 1

And,

Alpha × Beta + Beta × Gama + Gama × Alpha = C/A

=> a(a-b)+a(a+b)+(a+b)(a-b) = 6/1

=> 3a²-b² = -6 => 3 × (1)² - b² = 6

=> -b² + 3 = -6

=> b² = 6+3 => b² = 9

=> b = ✓9 = +-3

Therefore,

a = 1 and b = +-3

HOPE IT WILL HELP YOU..... :-)

Answer 2

It's helpful for you☺ Make me brainlest☺☺